LeetCode-in-All

4. Median of Two Sorted Arrays

Hard

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)).

Example 1:

Input: nums1 = [1,3], nums2 = [2]

Output: 2.00000

Explanation: merged array = [1,2,3] and median is 2.

Example 2:

Input: nums1 = [1,2], nums2 = [3,4]

Output: 2.50000

Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.

Example 3:

Input: nums1 = [0,0], nums2 = [0,0]

Output: 0.00000

Example 4:

Input: nums1 = [], nums2 = [1]

Output: 1.00000

Example 5:

Input: nums1 = [2], nums2 = []

Output: 2.00000

Constraints:

• nums1.length == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= m + n <= 2000
• -106 <= nums1[i], nums2[i] <= 106

Solution

import 'dart:math';

class Solution {
  double findMedianSortedArrays(List<int> nums1, List<int> nums2) {
    if (nums2.length < nums1.length) {
      return findMedianSortedArrays(nums2, nums1);
    }

    int n1 = nums1.length;
    int n2 = nums2.length;
    int low = 0;
    int high = n1;

    // substitute for Integer.MIN_VALUE
    int minValue = -pow(2, 31).toInt();
    // substitute for Integer.MAX_VALUE
    int maxValue = pow(2, 31).toInt() - 1;

    while (low <= high) {
      int cut1 = (low + high) ~/ 2;
      int cut2 = ((n1 + n2 + 1) ~/ 2) - cut1;

      int l1 = cut1 == 0 ? minValue : nums1[cut1 - 1];
      int l2 = cut2 == 0 ? minValue : nums2[cut2 - 1];
      int r1 = cut1 == n1 ? maxValue : nums1[cut1];
      int r2 = cut2 == n2 ? maxValue : nums2[cut2];

      if (l1 <= r2 && l2 <= r1) {
        if ((n1 + n2) % 2 == 0) {
          return (max(l1, l2) + min(r1, r2)) / 2.0;
        }
        return max(l1, l2).toDouble();
      } else if (l1 > r2) {
        high = cut1 - 1;
      } else {
        low = cut1 + 1;
      }
    }
    return 0.0;
  }
}

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