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23. Median of Two Sorted Arrays

hardAsked at Coursera

Find the median of two sorted arrays in O(log(m+n)) time, a binary search hard problem Coursera uses to test candidates' ability to reason about partitioning for analytics pipelines.

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
  • -10^6 <= nums1[i], nums2[i] <= 10^6

Examples

Example 1

Input
nums1 = [1,3], nums2 = [2]
Output
2.00000

Example 2

Input
nums1 = [1,2], nums2 = [3,4]
Output
2.50000

Approaches

1. Merge and find median (O(m+n))

Merge both arrays into a sorted array, then take the middle element(s) — fails the O(log) requirement.

Time
O(m+n)
Space
O(m+n)
function findMedianSortedArrays(nums1, nums2) {
  const merged = [...nums1, ...nums2].sort((a,b) => a-b);
  const mid = Math.floor(merged.length / 2);
  return merged.length % 2 ? merged[mid] : (merged[mid-1]+merged[mid])/2;
}

Tradeoff:

2. Binary search on partition

Binary search on the smaller array to find the correct partition point such that all left-half elements are <= all right-half elements in both arrays combined. Compute median from the four boundary values.

Time
O(log(min(m,n)))
Space
O(1)
function findMedianSortedArrays(nums1, nums2) {
  if (nums1.length > nums2.length) return findMedianSortedArrays(nums2, nums1);
  const m = nums1.length, n = nums2.length;
  let lo = 0, hi = m;
  while (lo <= hi) {
    const i = Math.floor((lo + hi) / 2);
    const j = Math.floor((m + n + 1) / 2) - i;
    const maxL1 = i === 0 ? -Infinity : nums1[i-1];
    const minR1 = i === m ? Infinity : nums1[i];
    const maxL2 = j === 0 ? -Infinity : nums2[j-1];
    const minR2 = j === n ? Infinity : nums2[j];
    if (maxL1 <= minR2 && maxL2 <= minR1) {
      const maxL = Math.max(maxL1, maxL2);
      const minR = Math.min(minR1, minR2);
      return (m + n) % 2 ? maxL : (maxL + minR) / 2;
    } else if (maxL1 > minR2) { hi = i - 1; }
    else { lo = i + 1; }
  }
}

Tradeoff:

Coursera-specific tips

Coursera interviews emphasize algorithms for educational platforms, content recommendation systems, and scalable delivery pipelines. Medium-difficulty graph and DP problems are typical.

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