22. Median of Two Sorted Arrays
hardAsked at FreshworksFind the median across two sorted arrays in O(log min(m, n)) — Freshworks asks this to probe binary-search fluency over a partitioned domain.
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. The runtime must be O(log(min(m, n))).
Constraints
nums1.length == m, nums2.length == n0 <= m, n <= 10001 <= m + n <= 2000
Examples
Example 1
nums1 = [1,3], nums2 = [2]2.0Example 2
nums1 = [1,2], nums2 = [3,4]2.5Approaches
1. Brute force (merge then index)
Merge both arrays in O(m+n) and pick the middle index/indices.
- Time
- O(m+n)
- Space
- O(m+n)
const merged = [];
let i=0,j=0;
while(i<nums1.length||j<nums2.length){ /* push smaller */ }
const mid=(merged.length-1)/2;
return (merged[Math.floor(mid)] + merged[Math.ceil(mid)]) / 2;Tradeoff:
2. Binary search on smaller array's partition
Pick the smaller array. Binary search a partition i in [0, m]. Compute j = (m + n + 1) / 2 - i. Check four boundary values; tighten the search until left_max <= right_min on both sides.
- 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;
const 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:
Freshworks-specific tips
Freshworks scores hard on whether you binary-search the SMALLER array — say that out loud before coding because that's where the log min(m, n) bound comes from.
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