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23. Median of Two Sorted Arrays

hardAsked at Yelp

Find the median of two sorted arrays in O(log(min(m,n))) — Yelp uses this binary-search-on-partition trick to test whether candidates can find log-time medians in merged review-score streams.

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 overall runtime must be O(log(min(m, n))).

Constraints

  • 0 <= m, n <= 1000
  • 1 <= m + n <= 2000
  • -10^6 <= nums1[i], nums2[i] <= 10^6

Examples

Example 1

Input
nums1 = [1,3], nums2 = [2]
Output
2.0

Example 2

Input
nums1 = [1,2], nums2 = [3,4]
Output
2.5

Approaches

1. Merge then index

Merge into a sorted array and grab the middle element(s).

Time
O(m + n)
Space
O(m + n)
const m = [...nums1, ...nums2].sort((a, b) => a - b);
const n = m.length;
return n % 2 ? m[(n-1)/2] : (m[n/2 - 1] + m[n/2]) / 2;

Tradeoff:

2. Binary search on partition

Binary-search the shorter array for a cut so that left-half size equals the median split and the four boundary values satisfy 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, 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) hi = i - 1;
    else if (bL > aR) lo = i + 1;
    else {
      if ((m + n) % 2) return Math.max(aL, bL);
      return (Math.max(aL, bL) + Math.min(aR, bR)) / 2;
    }
  }
}

Tradeoff:

Yelp-specific tips

Yelp will pivot to review ranking — be ready to discuss how a log-time median lets them recompute a city's review-score median across two sharded indexes without a full merge.

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Output

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