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23. Median of Two Sorted Arrays

hardAsked at SoFi

Find the median of two sorted arrays in O(log(m+n)) time — SoFi uses this hard problem to filter for candidates who can do binary search on answer space, the same pattern used in portfolio-risk percentile computations.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays. The overall run time complexity should be O(log(m+n)).

Constraints

  • nums1.length == m
  • nums2.length == n
  • 0 <= m, n <= 1000
  • 1 <= m + n <= 2000

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 + index

Merge both arrays in linear time, then pick the middle.

Time
O(m+n)
Space
O(m+n)
function findMedianSortedArrays(a, b) {
  const m = [...a, ...b].sort((x,y) => x-y);
  const mid = m.length / 2;
  return m.length % 2
    ? m[Math.floor(mid)]
    : (m[mid - 1] + m[mid]) / 2;
}

Tradeoff:

2. Binary search partition

Binary search on the smaller array for a partition i such that left halves of both arrays are <= right halves. The median is derived from the boundary values at this partition.

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:

SoFi-specific tips

SoFi probes whether you can binary-search on a derived property (partition index) — VaR (value-at-risk) tail computations on portfolio return distributions use the same trick to avoid scanning the full series.

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