%%perfect hash: 1000 known string keys #MLP-sg %50

See also string bucket-sort

Q: similar to FIX tags, you are given a published dictionary of 5000 short string keys, each up to 10-characters. Every message consists of about 100 key/payload pairs. Design a parser for these messages. Note every key would be in the dictionary.

Simple solution 1 BST like AVL or RBT to hold the pairs. O(log N=5000) … 5000 keys require about 12 levels. Possibly faster than hashtable.

Simple solution 2 standard hashtable to hold the pairs. The hashcode for string is usually fast enough, with minimal collision if load factor is set sufficiently low. Even if zero collision, this requires allocating 5000 linked nodes in memory.

In this case since the key strings are known in advance, I believe we can design a perfect hashing algorithm without collision. Interviewer said perfect hashing is hard in this case.

Solution 3: perfect hashing. Group the key strings by length, to 10 groups

  • If the number of distinct 8-char keys is small, for instance, we can use switch-case
  • for 1-char string keys, we maintain a dedicated bucket array (“table”) indexed by the ascii code
  • for 2-char string keys I can have a nested 2D array. Given key string “px”, “p” identifies the level2 array and within it, “x” identifies the bucket for the payload.
  • for 3-char keys, ditto

For those 10-char string keys, a 10-level nested array would be too large. Here’s my strategy. Assuming a lot of key strings (say, 3000 out of total population of 5000) are 4-char long, let’s tackle this sub-problem first:

  • for 4-char keys, I would designate an index range like 1 to 6000 (6000 buckets) to hold 3000 keys in this group. I would compute an initial hashcode := (char1*31+char2)*31+char3… % 6000 for this group of 3000 key strings and check for collision within this group. If there’s collision, then adjust some of the prime numbers and retry. Since load factor is 0.5 I hope we can avoid collision within this group. If it’s hard to avoid collision, I can reduce load factor, or introduce my collision resolver[2].
  • for 5-char keys, I would designate an index range like 6001 to 7800 (1800 buckets) to hold 900 keys in this group. I would compute initial hashcode := (char1*31+char2)*31+char3… %1800 for this group of keys and check for collision within this group.
  • So far, we have designed with a big combined bucket array. As an alternative, if there are only 7 key groups [4-char, 5-char, .. 10-char] then we can use seven dedicated bucket arrays, but what if 7000 groups? 7000 bucket arrays? I think so. Memory footprint is same as a combined bucket array.

[2] Here’s a simple collision resolver. Suppose Strings A and B(and C,D..) all hash to the same initial-hashcode of 55. I need to find another bucket for B. After I compute all the initial-hashcode values for each of 3000 key strings, I can easily find an unused bucket like 77. I will simply add an if-block to use bucket 77 for key B. Now this scheme is probably faster than open-addressing and separate chaining since the calc logic can be manually optimized , held in instruction-cache, or burned into FPGA. In contrast,

  • Separate chaining requires runtime memory access to follow the linked nodes. We can only hope the linked nodes are cached
  • open addressing requires runtime probing. Our load factor of 0.5 is not very good so open addressing can become slow.

http://www.dre.vanderbilt.edu/~schmidt/PDF/gperf.pdf is a published perfect hashing solution.

SCB-FM stack-based FIFO in O(1)amortized

Q: given a hardware-based stack API consisting of 3 functions {pop/push/isEmpty}, please implement a queue api consisting of 3 functions {enqueue/dequeue/isEmpty}

https://leetcode.com/problems/implement-queue-using-stacks/description/ is similar


service dequeue from a hidden stack.

When hidden stack is empty, pop all nodes from visible stack to hidden stack. Amortized O(1) pop()

isEmpty() must add up two sizes.

[[python cookbook]] P658 implements this classic algo in 9 lines.

max-profit # easy SCB-FM

Q: maximize profit on a given a price series … you can buy, sell, buy, sell any number of times, each time one share. No buy-buy-sell-sell allowed, i.e. at the time of buying, you must have zero inventory.

Q: ok you said your solution is O(N), can you do it in one scan?

====my answer

If price keeps dropping, then no trade possible

If price keeps rising steadily, then only one buy-sell pair

if i get peak-trough-peak-trough… then i must discard the first peak since I can’t do anything with it.


celebrity search in O(N) #MS

Q (MS 2018): write an algorithm to search for celebrities in a collection of people with the help of a blackbox predicate that can tell whether one person knows the other (bool a.knows(b)) which is an asymmetric relation. Celebrities are those who know no one but are known by everyone else.

Aha : There can be 0 or 1 celebrity.

There are exactly N*(N-1) relationships in the system, so brute force is O(N*N) but I propose a 2-pass O(N) solution:

  1. First pass: shrink the collection of Nodes [A, B .. N]
    1. check a random pair. if A knows B, then remove A from list; else, remove B from list. So first step always shrinks list by one
    2. Now check any random pair of persons in remaining list. Check either direction. Will remove exactly one person.
    3. list will shrink to 1 person (Dan) after N-1 steps
  2. Second pass: check this Dan thoroughly. Dan must be known to everyone and must not know anyone. We end up with either no celebrity OR one celebrity in Dan.

Aha — better use A/B/C or Alice/Bob/Cody instead of #1, #2, #3.

generate all abbr starting from longest.. +! recursion

I won’t insist on relative ordering among the shortest.

Idea 1() — Start with longest abbreviation i.e. the original string S, assuming 5 characters.

  1. populate the smallHM with the original word
  2. copy every char except the first. save into bigHM, then print/process this abbrevation.
  3. copy every char except the 2nd and print
  4. ..
  5. copy every char except the last. Now we have 5 strings in bigHM (a Level-4 hashmap), each length S-1=4
  6. make smallHM point to bigHM object; point bigHM to an empty hm
  7. now take a string from smallHM (Level-4 collection) and generate 4 shorter strings and save them in bigHM (a Level-3 collection), each length S-2=3
  8. now take 2nd string from Level-4 …
  9. After we finish Level-4, we have generated 20 strings in Level-3, but there are only 10 distinct items! so we need a L3 hashmap.


airport gate #maximum people alive

https://careercup.com/question?id=5153263227764736 defines the problem

Q (Amazon): In a city, year of birth/death of people who where born and died between year 1900 to 2000 are given. Write an algorithm to find the year in which max people were alive. Note the years are not unique and not sorted


Q (FlexTrade): For an airport gate system, flight arrival/departure times are given for yesterday. What’s the maximum number of gates required at the busiest time?

Solution1: O(N logN) merge-sort all timestamps, then scan it in one pass. If an arrival, then increment counter; if a departure then decrement it.

??Solution2 (assuming arrival times are pre-sorted) Using hashtable, keyed by arrival time. Value is a count of flights arriving at that time. Every arrival creates or updates in the hashtable. Every departure deletes or decrements. Maintain a separate total count.

I think we still need sorting.

Solution3: O(N). Use array if all the years are small integers. (Regular timestamp is also small integers — 0 to 2355 in steps of 5.) Fill all arrival/departure events as +1/-1 in an array indexed by year.

TowerResearch IV: c++,algo,past prj..

Q1: how could you use enum to improve performance
AA: use it to replace strings
AA: enum is used in template programming to move computation from run time to compile time. See Q4 and SFINAE in github.
%%A: see clever use of enum(char?) in demanding c++ app

Q1b: you put in your change but it turns out to hurt performance. Any idea?

Q4: how do you compute Fib(N) like Fib(99999) at compile time using template programming?
A: See my Fibonacci code in github

Q: how would you use multicast to send a message to a group of registered users?
%%A: less data copying than TCP

Q: In O(1) space, Given an unsorted array of natural numbers, how do you find any pair to produce a target sum? What’s your best time complexity?
%%A: if I assume the integer size is 32-bit then a fixed-size O(1) space radix structure can sort the array in O(N). See also my blog post on [[locate a pair with targetSum=55 #bbg IV #Morris]]
%%A: if integer size is unlimited, then I think the best is O(NN)

Q: How is your muni bond pricing engine designed internally?
Q2b: what kind of pricing rules live in the cache?
Q2c: how many threads to react to a given event?

Q: how did you calculate FX fwd points?
%%A: use the interest rates in the two currencies and the spot FX rate.

Q: how is your implied volatility computed from an option price?
%%A: if there’s not a closed-form formula, then Newton’s method or something similar should be able to converge quickly.

Q3 (OO design): How would you design a game software with Animal objects, various birds, and fish?
Q3a: Say the system is running fine, but now you need to add ostrich class?

binary search in rotated sorted array

https://leetcode.com/problems/search-in-rotated-sorted-array/description/ has the requirement. I don’t want to look at other people’s solution, so I have reproduced the requirements below. I have not encountered this problem in any coding interview.

====Q: Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]). You are given a target value to search. If found in the array return its index, otherwise return -1. You may assume no duplicate exists in the array. Your algorithm’s runtime complexity must be in the order of O(log n).

https://github.com/tiger40490/repo1/blob/cpp1/cpp/array/binSearchRoatedArr.cpp is my solution

–Solution 2: clean two-pass algo

first run a binary search to locate the pivot point. This pre-processing is a one-time investment.

Then run a O(1) test to discard one of the 2 segments. We are left with the remaining segment as a regular sorted array. Run binary search on it.

===Q (leetcode 153): find the pivot point, given the original array is ascending.

Not sure if I already solved this.

Compare a[0] vs a[1]. If descending, then a[1] is the answer.
Compare a[0] vs last. if a[0] is lower, then a[0] is the answer.

Initialize le at a[0] and ri at last. check the mid point

If mid is above a[0] then shift le
if mid is below last, then shift ri

check array@0-N +! auxDS #Nsdq#contrived

— Q1: Given a size-5 array of integers. For every element x: 0<= x <=4. Array would be considered Good if all elements are unique, i.e. all the numbers 0 to 4 show up exactly once each. Please write a program to check if the array is good. If Bad, then print every number that’s showing more than once. O(N) time and O(1) space please.

We will denote array size as sz, so N == sz-1.  N=4 and sz=5 in the opening example.

— comments:

I feel this question (even without the bonus) is too hard to complete on the spot, unless you have done it before.

I made a mistake that would be unforgivable in west coast and Bloomberg interviews — failed to clarify requirements and assumed input array is immutable.

— Q1b (bonus): Also print every missing number. (Requires iterating entire input.)

Example A {1,2,0,2,0} has repeating 0 and 2, and missing 3 and 4. Example B {3,1,4,2,0} is Good.

— comments:

If you feel the O() complexity requirements are daunting, then work out a less efficient solution first, and improve on it.

I feel in real projects, we always have enough memory to build a separate array, so the O(1) space requirement is contrived. A O(N) time and O(N) space solution is easier. Therefore, I rate this coding question as contrived and relatively non-standard. Yet the required techniques are “universal” in many high-end interviews.


has my 99% solution. The unfinished part is trivial.



max-profit #Nsdq short-sell

Q1: given a time series of price points within a past day, there are many ways to trade the day — one buy-sell, five buy-sells, or do-nothing … Short-sell allowed, but you must start and end the day holding no shares. For example, if you sell 9 times then you must buy 9 times, each time 1 lot (100 shares) exactly. Write a function to trade the day and produce the highest profit.  Note you can analyze the price points with perfect hindsight.

Interviewer asked for real code. Very little time given, so I believe the right solution is short, and much simpler than the question on “array of 0-N” (where interviewer asked for pure algorithm).

https://github.com/tiger40490/repo1/blob/cpp1/cpp/array/oracleDayTrader_Nsdq.cpp is my buggy”greedy” solution.

https://github.com/tiger40490/repo1/blob/py1/py/array/maxProfit_Nsdq.py is my new solution.

Q2: now you are allowed to buy UP-TO 100 shares each time. All other rules remain. When if ever is it optimal to buy/sell fewer than 100 shares (a.k.a. an odd lot)?
%%A: never

Q3: same as Q1 but now your holding must always be long 1 lot, short 1 lot or zero holding.

–(broken?) brute force solution I gave on the spot:

Start with just four price points 1,2,3,4. Name every pair A=1>2; B=1>3; C=1>4; D=2>3; E=2>4; F=3>4. Each pair represents a buy/sell round trip, with a profit (ignore it if unprofitable).

How many “plays” i.e. how many ways to trade the day? 2^6 plays.

Just evaluate each play and find the best. Beware that ABD is same as B x 2 and should be scaled down by 2. Similarly, AEBFC == C x 3

convert-to-0 using4transformations #Trex

Q: given a natural number, write a function to "transform" it to zero in a minimum number of transformations. Four transformations allowed:A) add 1
B) subtract 1
C) divide by 2 if divisible
D) divide by 4 if divisible

My own idea:

1) attempt D for as long as possible. Repeat for C. Now we have an odd number K
2) if K can be written as 4N-1, then do A, otherwise B.
3) go back to step 1

https://github.com/tiger40490/repo1/blob/py1/py/trivial/4conversion_ez.py is my code

bbg: allocate half the players to NY^SF

Always good to be concrete, so without loss of generality, let’s say J=99 i.e. 198 players.

Q1: Given 2J players, we need to allocate exactly half of them to NY and the rest to San Francisco, to form two teams for a inter-city contest. We receive the NY fare and SF fare for each player, as 2J (i.e. 198) pairs of integers. How can we minimize the total fares?

Q2(bonus): what if 2J+1 players, and you can decide which city gets an extra player?

— Analysis: how many possible allocations? 2J-choose-J. Very large number of allocations. Something like O(J!) so brute force is impractical.

If for any player the two fares both go up by $9800, it doesn’t change our algorithm at all. Therefore, we only care about the fare difference (N-S) for each player.

— solution: I will assume most players live near SF so SF fares are lower. I tabulate and sort the 198 fare differences “NY fare – SF fare” and suppose indeed at least 99 (half) are positive[1]. Therefore, my “base” is SF i.e. my base allocation is everyone-allocated-to-SF. Next I must pick 99 (half) of them and allocate to NY.

I will avoid the higher values in the table.

I simply pick the lower 99 (half) in the table, because these are the 99 “extra” costs I will incur. I want to minimize the total of these 99 values, whether they are mostly positive or mostly negative.

  • Note the highest number in the table indicates the most expensive player for NY relative to SF. If this number is negative, then SF is more expensive than NY for All players so follow the rule in [1] but notice he is the least expensive players for SF relative to NY. Regardless of positive/negative sign, we want to keep this person in SF .
  • Note the lowest number in the table indicates the least expensive player for NY relative to SF. If this number is negative, then SF is more expensive than NY for this player — normal. Regardless, we want to allocate her to NY.

[1] if proven otherwise (by the data), then we could easily treat NY as base to keep the solution intuitive. Actually even if we don’t change the base, the algorithm still works, albeit unintuitively.

A2: Say J=99 so total 199 players We already picked 99 for NY. Do we pick one more for NY or keep remaining 100 in SF?

Just look at the next lowest value in the table. If positive, then NY is more expensive for him, so keep him in SF.

25-horse puzzle #Google

Q: A single racetrack. P=25 mechanical horses each with a unique speed. You can race up to 5 horses each time, and get a printout of the ranking, but no timing.

What’s the minimum number of races to identify fastest #1 #2 #3 horses.

Youtube and my green quant-IV book both have the ananlysis.


Quick elimination — each time I can eliminate minimum 2 horses but can we eliminate faster?

What if T is 1 and P is 9? Two races exactly.

Here’s my algo #1

  1. first pick 5 random horses,
  2. race them and eliminate slowest 2 (reduce population by 2). Name the top 3 as AA/BB/CC
  3. race CC with 4 untested random horses. Four scenarios:
    1. if 3 or more beat CC, then eliminate the two slowest including CC, and put the top 3 in a group with AA/BB. Go back to Step b)
    2. if 2 beat CC, then eliminate the three slowest including CC, and put top 2 and an untested horse in a group with AA/BB. Go back to b)
    3. if only one horse beat CC, then eliminate the four slowest including CC, and put the #1 and 2 untested horses in a group with AA/BB. Go back to step b)
    4. if CC is #1 (happy path), then eliminate the other four horse and go back to step c)

Worst case — first race actually had the slowest 5, then we hit case C1 every time. So we only eliminate 2 horses each race

best case — first race had 5 fastest. total 6 races [1] but this is statistically unlikely. If you are not the most lucky, then we will need more than 6 races.

[1] first race eliminates 2… next 5 races each eliminates 4 only if CC is always fastest in every race.

–algo #2:

  1. [5 races] Split into 5 groups and run 5 unrelated races. Discard slowest 2 in each group. We end up with 15 horses.
  2. [1 race] Now race the 5 number one’s. Name them ABCDE. The D and E groups are completely eliminated (-6) and the C group is left with C alone (-2), and B group’s #3 is eliminated (-1). We end up with 6 horses. I will name the seven as A/B/C and the less-tested A2 A3 B2. There are only three possibilities:
    1. ABC
    2. A A2 A3
    3. A B B2
  3. [1 race] so I will race 5 of the 6 horses, sparing A since it’s already the final champion.
  4. — total 7 races best or worst case.


top N active stocks #bbg#cache,VO

Requirement: implement top_n(int N): top N most active stocks

Requirement 2: add a top_n_last_x_minute(int N, int X): top N that are last traded in the last X millisec

This program also demonstrates how to avoid storing the same string twice, as map key and also as field of the map value

#include <unordered_map> //requires -std=c++0x
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <chrono>
#include <assert.h>
using namespace std;

struct Rec{
  string const * strPtr; //a best-effort to avoid storing duplicate strings
  int cumvol; //cum vol
  long lastUpd; //millsec since epoch

  Rec(int vol): strPtr(NULL), cumvol(vol),
    lastUpd( (chrono::system_clock::now().time_since_epoch()).count()) { }

  bool operator<(Rec const & other) const{
        return this->cumvol < other.cumvol;
  friend ostream & operator<<(ostream & os, Rec const & r){
        os<<r.lastUpd<<"-"<<*(r.strPtr)<<"  : "<<r.cumvol;
        return os;

typedef unordered_map<string, Rec>::iterator ITR;

class App{
  unordered_map<string, Rec> lookup;
  vector<Rec> v;
  bool dirty;
  App(): dirty(true){}
  void update(string const & symbol, int vol){
    this->dirty = true;
    ITR it = lookup.find(symbol);
    if (it == lookup.end()){
        pair<ITR, bool> pair = lookup.insert(make_pair(symbol, Rec(vol)));
        assert(pair.first->second.strPtr == NULL);
        pair.first->second.strPtr = //second is the valye in the pair
          & pair.first->first; //the key of the new pair
        assert(it->second.strPtr != NULL);
        it->second.cumvol += vol;
        it->second.lastUpd = (std::chrono::system_clock::now().time_since_epoch()).count();
  void top_n(int N, unsigned long X=-1){ //unsigned -1 gives the largest positive value!
    size_t sz=lookup.size();
    if (dirty){
      cout<<"resorting due to updates to database ..."<<endl;
      for(ITR i=lookup.begin(); i!= lookup.end(); ++i){
      sort(v.begin(), v.end());
      dirty = false;
    long now = (std::chrono::system_clock::now().time_since_epoch()).count();
    //cout<<now<<" is now"<<endl;
    for(int k=sz-1, count=0; k>=0; --k){
        if (now - v[k].lastUpd > X) continue;
        if (++count >= N) break;
int main(){
  App app;
  app.update("ibm", 1000);
  app.update("gs", 700);
  app.update("gs", 500);
  app.update("goog", 600);
  app.update("msft", 800);
  app.update("ge", 500);
  app.update("ibm", 160);
  app.update("c", 400);

GS c++/dataStruct IV

Q1: what’s the problem to be solved by virtual inheritance
A: diamond..

Q2: consider such a diamond with A as grandparent, B/C virtually deriving from A, then grandchild D. Suppose class C has a foo() method that uses a field A.a. Now, within C, the offset of A.a depends on the inheritance hierarchy, so how does the compiler compile C.foo()?

A: I do see a problem here. In this ABCD case, the offset of A.a is one value, depending on the size of B. However, in a ABJKCD case (J/K virtually subclass A), the offset of A.a will be a different value, depending on the sizes of B/J/K. So the C.foo() method will compile to different object code! Yet the compiler has a single version of C.foo().

The trick is the offset of the A sub-object within C. This offset value can be saved in the vtable of C.

Note if there’s no virtual inheritance, then simpler. Within C, the A sub-object’s offset is a constant. No need to look up vtable.

Q3: appending a million integers to a list vs a vector, which is faster and by how much?
%%A: vector requires reallocation, roughly 20 times, assuming doubling each time, much fewer than 1M allocations for the list.  The reallocation entails element-by-element copy, but that’s cheaper than allocations. If I must estimate how much faster, I need to estimate how many allocations and how many copying:

  • for list, 1M allocations + 1M copying
  • for vector, 20 allocations + about 2M copying

%%A: I still think allocation is a hotspot and much more expensive than copying, perhaps by 10 or 100 times.

%%A: of course, you can pre-allocate for vector, but for this problem we ignore that feature.

Q4: deque — known for fast append/prepend, and random access. How?
%%A: I think it’s implemented as segments of vectors, with the front and back segments not fully occupied and growable. Once one of them reaches a certain size, a new segment is created. RandomAccess is implemented by identifying the host segment. To optimize, segments should have equal length.

Q4b: what if we want to support fast midstream insertion?
%%A: this is a vector limitation inherited by deque. If we need to support it, we may need to allow unequal segment lengths. RandomAccess is now slightly slower. For a given subscript 16, we need to maintain a step function like

  • 5-10 in Segment 2
  • 11-13 in Segment 3
  • 14-20 in Segment 4
  • 21-30 in Segment 5
  • 31-32 in Segment 6
  • … so each time any segment expands, the step function needs an update
  • Once we have the step function, a binary search will do, at log(S) where S = Segment count

[17]deque with uneven segments #GS

Update — Now I think unrolled linked list for small payloads is very similar.

Q: (2017 GS interview question) We know std::deque offers efficient insert at both ends. But now we want efficient insert mid-stream, and willing to relax O(1) lookup.

I proposed a solution using uneven segments. Each segment has a max capacity. When reached, a new segment would be allocated and linked in. Lookup using a staircase function.

Segment sizes (not capacity) is a mutable int array. From this array, we lazily derive another int array, the “threshold array” or staircase array. Every lookup requires a binary search in this threshold array to locate the target segment. If we keep a hard limit (like 1024) on segment count, then the threshold array would be at most 1024 long, so within 10 hits we always locate the target segment — better than O(log N)

The two int arrays should be L1-cached.

min-stack #bbg

My friend Abhinav (not his real name, to protect his privacy) got this question at Bloomberg internship interview. I added some details to make it clearer:

Q: Design a stack that supports push, pop, top, and retrieving the minimum element all with O(1) time complexity in the worst case.

There exists a function compare(Item A, Item B) that returns 1 if A is greater, 0 if equal, and -1 if A is smaller.

  • getMin() — Retrieve the minimum element in the stack.
  • push(x) — Push element x onto stack.
  • pop() — Removes the element on top of the stack.
  • top() — Get the top element.

==== analysis =====

The most efficient getMin() data structure is the binary heap, but insert/delete runs in O(logN). Therefore I felt the requirements here are computationally impossible. But in this context, we only need to support deleting the last added item 🙂

Key insight — popping is a constrained form of deletion. It’s hard to hit O(1) while supporting unconstrained deletions, BUT with a constraint on deletions, all operations can be O(1).

I need a regular stack + a helper data structure. A linked list or vector can support the stack operations

— The helper — a naturally sorted stack (or vector or deque) to hold record-breakers and record-matchers.

IFF a new minimum (record breaker) or another item matching the existing minimum (record-matcher) is added, we push it to the sorted stack.

After every pop(), we check the popped item. If equal to top of sorted stack, then pop the sorted stack.

At any time, top of the sorted stack is the current minimum.

Vector and deque are actually fine. Interviewer may feel they are inferior to a stack, but with a modified requirement, they may become very useful.

— Here’s a design to beat binary_heap, based on finite-sized (32 or 64-bit int) keys

Assuming the comparison is based on 32-bit integer key (string or floats can also use radix sort). I will use a radix array structure. Perhaps 4 arrays of 256-elements.  Or perhaps a 4×256 matrix, Ultimately the payload stored in the data structure are pointers to stack nodes. This saves memory since a stack node may be 99MB.

Every time we push a node, we derive the new key value in O(1), use the key value to locate its ‘home’ in the radix structure and store the new node’s address in a linked list therein. After the push(), we can compare and update (in constant time) a global variable pointing to the minimum node.

Each stack node also has a back-pointer to the iterator into the list, so before pop() we can use the iterator to locate the object in the radix structure and delete it from the host linked list. We will also update the global variable.

factorize a natural number #AQR

My friend Dilip received this question in a 2017 AQR on-site.

Q: given a natural number (like 8), write a function to output every factorization such as (2,4) (2,2,2). You can ignore or include the trivial factorization (1,8). You can use recursion if you want.
— (incomplete) Analysis

  1. I would first generate all the prime numbers up to sqrt(N)
  2. among them, i would find all the factors. Once I find a prime factor x, keep dividing by x so I know in total I have 3 x’s, 2 y’s and 1 z, or (x,x,x,y,y,z). I call them 6 non-distinct “prime factors”.

From there, I might be able to write a (recursive?) function to output the factorization formulas. The ideal algo automatically avoids duplicate factorizations but Here’s my non-ideal design: generate all 2-way “splits”, all 3-way splits… If I keep all my “splits” in a hashtable, I can detect duplicates. So just treat the 6 factors as 6 distinct factors. Now the problem is well-defined — next split@N boys .

— trie solution based on generate combinationSum compositions #backtrack up] trie+tree

Candidates are the non-distinct prime factors and their products, each a factor of the big number.

— recursive solution by CSY

  • no prime number needed! A major drawback — if the target number is odd, we would still keep testing 2, 4, 6, 8 as possible divisors!

https://github.com/tiger40490/repo1/blob/cpp1/cpp/algo_comboPermu/factorize_AQR.cpp is very short solution by CSY. Here’s my analysis —

  • I believe every time the factorize(60) function finds a small factor like 2, it pushes the factor onto a global stack, then run factorize() on the quotient i.e. 30 — wherein every factorization formula on 30 is “decorated” with the stack.

https://github.com/tiger40490/repo1/blob/py1/py/algo_combo_perm/factorize_AQR.py is my modified/improved python solution

  • I replaced the global vector with a local immutable list on each call stack. It helps me reason. This is also thread-friendly, if the target number is large.
  • It’s highly instructive to work out the expected output from the recursive loops, as in my comments.
  • Just like the continuousSentence problem, the recursive solution is clever-looking but not scalable.

locate a pair with targetSum==55 #bbg IV #Morris


Q: any O(1) space sort?

Q: From an unsorted array of positive integers, is it possible to find a pair of integers that sum up to a given sum?

Constraints: This should be done in O(n) and in-place without any external storage like arrays, hash-maps, but you can use extra variables/pointers.

If this is not possible, can there be a proof given for the same?

—–Initial analysis—-
I wish I were allowed to use a hash table of “wanted ” values. (iterate once and build hashtable. For Each new value encountered, check if it is in the “wanted” list…)

I feel this is typical of west coast algorithm quiz.

I feel it’s technically impossible, but proof?  I don’t know the standard procedure to prove O(n) is impossible. Here’s my feeble attempt:

Worst case — the pair happens to be the 1st and last in the array. Without external storage, how do we remember the 1st element?  We can only use a small number of variables, even if the array size is 99999999. As we iterate the array, I guess there would be no clue  that 1st element is worth remembering. Obviously if we forget the 1st element, then when we see the last element we won’t recognize they are the pair.
—–2nd analysis—-
If we can find a O(n) in-place sort then problem is solvable [1]. Let’s look at Radix sort, one of the most promising candidates. https://stackoverflow.com/questions/463105/in-place-radix-sort has an implementation for DNA strings.

Assumption 1: the integers all have a maximum “size” in terms of digits. Let’s say 32-bit. then yes radix is O(n) but not sure about space complexity. Now, with any big-O analysis we impose no limit on the sample size. For example we could have 999888777666555444333 integers. Now, 32-bit gives about 4 billion distinct “pigeon-holes”, so by the pigeon-hole principle most integers in our sample have to be duplicates!

Therefore, Assumption 1 is questionable. In fact, some programming languages impose no limit on integer size. One integer, be it 32 thousand bits or 32 billion bits, could use up as much memory as there is in the system. Therefore, Assumption 1 is actually superfluous.

Without Assumption 1, and if we allow our sample to be freely distributed, we must assume nothing about the maximum number of digits. I would simply use

Assumption 2: maximum number of digits is about log(n). In that case radix sort is O(n log(n)), not linear time:(

—– new analysis as of Jun 2017 —-
Can Morris algo sort an array (external storage)? If yes,  then use the 2-moving-pointer algo in locate a pair whose diff=55

However, Morris needs a tree not an array.

[1] locate a pair with targetSum=55 : pre-sorted #O(N) 1-pass

dynamic glass-drop #BNP

http://en.wikipedia.org/wiki/Dynamic_programming#Egg_dropping_puzzle is a similar but more ambiguous problem.

If we toss a wine glass out of a 3rd floor window it may break, but it may not if tossed from a 2nd floor window.

Q: with exactly two wine glasses taken from a (Japanese:) production line, you want to identify starting at exactly which level of a 11-floor tower this brand of glass will break if tossed out of the window. Equivalently, determine the highest safe floor. Optimization goal — minimize worst-case cost. Top floor is known to break, so answer might be 1st floor, 2nd floor,… or top floor — 11 possible answers, but 10 floors to test.

Needless to say, if the glass breaks at 6th floor, it will surely break at a higher level.

—- Analysis —-
With just a single glass, you have no strategy. You have to try from 1st floor up. If your classmate starts on 2nd floor and breaks her only glass there, she is disqualified because with no more glass to use, it’s now impossible to know whether it would survive1st floor.

I feel this is NOT a probability problem, so each toss is bound to pass or fail with 100% certainty but we just don’t know it.

We don’t want to toss first glass @2nd, @4th, @6th… because too many first-glass tests. Neither do we want to toss first glass @5th because if broken, then 2nd glass must try @1@2@3@4. Therefore first glass toss should be somewhere between ( @2, @5 ) exclusive.

—-%% solution, with heavy use of symbols —–
I feel this problem is tractable by dynamic programming. Let
* L denote the highest _safe__ floor known so far, and
* U denote the lowest _unsafe_ floor known so far. U is initially at top floor, and L is initially 0, not 1
* u denote U-1

u >= L is an invariant. The Upper and Lower bounds. (u-L) is the maximum number of floors to test. When we get u-L = 0 and final Answer is the current value of U.

Here’s a typical example
6 —- U known unsafe
5 —- u, so 3 floors to test (u-L)
2 —- L known safe

z8 denotes the worst case # tests required to locate the Answer with both glasses when u-L = 8 i.e. 8 floors to test. Could be z(L=0, u=8) or z(L=1,  u=9). Same thing.

For example, z(1,3) means worst case # tests when 1st and 4th storeys need no testing. Note z(1,3) == z(2,4) == z2

z1 = 1 i.e. one toss needed
z2 = 2
z3 = 2 // B! C || BA. The untested floors are named (upward) ABCDEFG (if 7 floors to try), and trailing ! means broken.

z4 = 3 // B! CD || BA
z5 = 3 // C! +2 || C then z2 // if C broken, then must test last glass at D and E
z6 = 3 // C! +2 || C z3  // ABCDEF
z7 = 4 // (D! + 3 || D z3) or (C! + 2 || C z4)
z8 = 4 // (D! + 3||D z4) or the smarter (C! + 2||C z5)
z9 = 4 // (D! + 3||D z5) or (C! + 2 || C z6)
z10 = 4 // (D!+3||D z6). Can’t use (C! + 2 || C z7) — for a 11-storey tower, we need 4 tosses.

I think there are cleverer implementations. This is not brute force. It builds up useful results with lower z() values and use them to work out higher z() values — dynamic programming.

This solution not only finds where to start the test and the worst-case # tests. It also works out the exact decision tree.

–Kyle’s insight

For a build of 100, the max cost is $14 i.e. 14 drops, so the first floor to try is 14th. I didn’t believe him, so he showed

  • if glass AA breaks, then subsequent max cost is $13 as BB must try floor 1,2,3…13. Total worst cost = $14
  • else AA would try 14+13=27th floor. If AA breaks, then subsequent max cost is $12 as BB must try #15,#16,..#26. Total cost is still $14
  • else AA would try 27+12=39th floor. If AA breaks, then subsequent max cost is $11. Total cost is still $14..
  • …AA’s 10th try is #95, at a $10 cost
  • On the 11th try, AA will try #99. If AA breaks, then subsequent max cost is $3 i.e. #96,97,98. Total cost is $14
  • else on the 12th try, AA will try #100. Total cost is $12

https://brilliant.org/wiki/egg-dropping/ explains the math x(x+1)/2=100 so x = 13.65, so firs floor to try is #14.



16bit page-counter to Estimate when a webpage is hit the 1,000,000th time

We don’t need to estimate the hit Frequency which could be time-varying.

If we can use a 2 counters, we can precisely determine the 1,000,000th hit. One slow-counter one regular fast counter. But that means we use 32 bits for our counters, like a 32-bit counter!

void increment(){ // was named count()?
  static int count16bit;

  long unsigned int now = system_time_in_nanos();
  if (now%16 == 0){ //just check last 4 bits of “now”
     //now check count16bit
     if  (count16bit <= 0){cout<<"got it"<<endl; }

https://en.wikipedia.org/wiki/Approximate_counting_algorithm#Algorithm is the standard implementation of a similar idea.

data structure to hold spreadsheet content

Q: Assume string content in each cell. As a start let’s target 1,000,000 rows by 1,000,000 columns. Goals
– memory
– random access
– growable

%%A: Intermediate solution (good habit): a 3-column DB table with a composite key (row,col). Now how do we represent this in a data structure. First attempt would be a hashmap whose key is a customized big integer. Higher 64 bits represent row numbr, while lower half col number. Value is a reference with Copy-on-write. Hash access with a given (row,col) is …. considered random access with O(1).

Now I think a second try would be a hashmap where
main key -> row number
value -> sub-hashmap, where
sub-key -> col number

The other of the twin hashmap keeps col numbers as main keys. Every update to the spreadsheet requires updates to both.

estimate cubic root #GS IV

Q: Given any double, how do you create an algorithm to estimate its /cube root/ to 3 decimal places.
%%A: first multiply it by 8 until we hit the limit of a 64bit long integer. We can safely drop the decimal part since it’s guaranteed to be negligible. Now we have a large (at least 61 bit) INTEGER to work with, which is easier than a float/double.

Original question was designed to ask: how would you as a human estimate that, without a computer.