23. Median of Two Sorted Arrays
hardAsked at SquarespaceCompute the median of two sorted arrays without merging them; Squarespace uses it to test binary-search-on-partition fluency.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given two sorted arrays nums1 and nums2 of sizes m and n, return the median of the two sorted arrays combined. The overall runtime should be O(log(min(m, n))).
Constraints
0 <= m, n <= 10001 <= m + n <= 2000-10^6 <= nums1[i], nums2[i] <= 10^6
Examples
Example 1
nums1=[1,3], nums2=[2]2.0Example 2
nums1=[1,2], nums2=[3,4]2.5Approaches
1. Merge then index
Merge into one sorted array and pick the middle element(s). Simple but O(m + n).
- Time
- O(m+n)
- Space
- O(m+n)
const a=[...nums1,...nums2].sort((x,y)=>x-y);
const k=a.length;
return k%2 ? a[(k-1)/2] : (a[k/2-1]+a[k/2])/2;Tradeoff:
2. Binary search the partition
Pick a partition of the shorter array so left halves sum to half the total. Verify left max <= right min on both sides; binary search until valid.
- Time
- O(log min(m,n))
- Space
- O(1)
function findMedianSortedArrays(a, b) {
if (a.length > b.length) [a, b] = [b, a];
const m = a.length, n = b.length, half = (m + n + 1) >> 1;
let lo = 0, hi = m;
while (lo <= hi) {
const i = (lo + hi) >> 1, j = half - i;
const aL = i === 0 ? -Infinity : a[i - 1];
const aR = i === m ? Infinity : a[i];
const bL = j === 0 ? -Infinity : b[j - 1];
const bR = j === n ? Infinity : b[j];
if (aL <= bR && bL <= aR) {
if ((m + n) % 2) return Math.max(aL, bL);
return (Math.max(aL, bL) + Math.min(aR, bR)) / 2;
} else if (aL > bR) hi = i - 1; else lo = i + 1;
}
}Tradeoff:
Squarespace-specific tips
Squarespace likes when you sketch the partition diagram first and tie the log-time win to their analytics-rollup p50 percentile probe over two pre-sorted time buckets.
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