https://leetcode.com/problems/median-of-two-sorted-arrays/description/ is similar except X and Y can be unequal length. My solution solves the harder, generalized problem.

This “coding” question is really math problem. Once you work out the math techniques, the coding is simple.

Designate arr1 as the shorter array. compare med(arr1) vs med(arr2)

Suppose former is lower, i can discard lower half of arr1 (s items). Can i discard highest s items in arr2? I think so because upper half of arr2 cannot have that median element, so any subset of it can be discarded

repeat until arr1 is completely discarded or left to a single element .. might be the final median. Now answer is close to the med of the remaining arr2.

–For the equal-length problem, My own idea on the spot — find the median of X and median of Y. If med(X) < med(Y) then discard the lower portion of X i.e. the “XB group”, and higher portion of Y (“YA group”). Then repeat.

- Note len(XB) == len(YA) == min(len(X), len(Y))/2 := K. So every iteration would shrink the shorter array by half (i.e. K), and shrink the longer array by K. K would drop in value in next iteration.
- loop exit — When the shorter of the two (say it’s X) shrinks to length 1, we are lucky — find the numbers around median(Y) and adjust the answer based on X[0].

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Insight — Why can’t the final “winner”be somewhere in XB group? Because XA + YA already constitute half the population, and all of them are higher.

I always like concrete examples. So Suppose there are 512 items in the lower portion “XB group”, and the higher portion “XA” has 512 items. Suppose there are 128 items each in YB and YA groups. So in this iteration, we discard YA and the lowest 128 items in XB.

Definition of lower portion —

* all lower items up to but not including med(X) If len(X) is odd

* exactly the lower half of X if len(X) is even