22. Median of Two Sorted Arrays
hardAsked at ByteDanceFind the median of two sorted arrays in logarithmic time — ByteDance uses it to test binary-search-on-answer reasoning that mirrors quantile estimation in their ranking pipeline.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given two sorted arrays nums1 and nums2 of size m and n, return the median of the combined sorted array. The expected overall run time is O(log(min(m, n))).
Constraints
0 <= m, n <= 10001 <= m + n <= 2000-10^6 <= nums[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 the two arrays into one sorted array; return the middle element(s).
- Time
- O(m + n)
- Space
- O(m + n)
const m=[...a,...b].sort((x,y)=>x-y); const k=m.length;
return k%2 ? m[(k-1)/2] : (m[k/2-1]+m[k/2])/2;Tradeoff:
2. Binary partition on smaller array
Pick a cut in the smaller array so left halves combined hold the lower half of all elements. Adjust until max(left) <= min(right).
- 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:
ByteDance-specific tips
ByteDance interviewers reward the explicit invariant that the left partition holds exactly half the elements — the same precondition their quantile-estimation jobs assert before publishing ranking thresholds.
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