23. Median of Two Sorted Arrays
hardAsked at AdyenFind the median of two sorted arrays in O(log(min(m, n))) time.
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 combined sorted array. The overall run time complexity 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 fully, then pick the median.
- Time
- O(m + n)
- Space
- O(m + n)
const all = [...nums1, ...nums2].sort((a,b)=>a-b);
const n = all.length;
return n % 2 ? all[(n-1)/2] : (all[n/2 - 1] + all[n/2]) / 2;Tradeoff:
2. Binary search on partition
Binary-search the smaller array for the partition that splits both arrays into equal-size halves.
- 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 ? a[i-1] : -Infinity, aR = i < m ? a[i] : Infinity;
const bL = j ? b[j-1] : -Infinity, bR = j < n ? b[j] : Infinity;
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:
Adyen-specific tips
Adyen interviewers grade this for partition reasoning — they reuse the half-split logic when blending dual fraud-signal score streams and want the log-time discipline.
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