4 - Median of Two Sorted Arrays

 JAVA

• class Solution {
      private int m;
      private int n;
      private int[] nums1;
      private int[] nums2;
  
      public double findMedianSortedArrays(int[] nums1, int[] nums2) {
          m = nums1.length;
          n = nums2.length;
          this.nums1 = nums1;
          this.nums2 = nums2;
          int a = f(0, 0, (m + n + 1) / 2);
          int b = f(0, 0, (m + n + 2) / 2);
          return (a + b) / 2.0;
      }
  
      private int f(int i, int j, int k) {
          if (i >= m) {
              return nums2[j + k - 1];
          }
          if (j >= n) {
              return nums1[i + k - 1];
          }
          if (k == 1) {
              return Math.min(nums1[i], nums2[j]);
          }
          int p = k / 2;
          int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
          int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
          return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
      }
  }

C++

• class Solution {
  public:
      double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
          int m = nums1.size(), n = nums2.size();
          function<int(int, int, int)> f = [&](int i, int j, int k) {
              if (i >= m) {
                  return nums2[j + k - 1];
              }
              if (j >= n) {
                  return nums1[i + k - 1];
              }
              if (k == 1) {
                  return min(nums1[i], nums2[j]);
              }
              int p = k / 2;
              int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
              int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
              return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
          };
          int a = f(0, 0, (m + n + 1) / 2);
          int b = f(0, 0, (m + n + 2) / 2);
          return (a + b) / 2.0;
      }
  };

JS

var findMedianSortedArrays = function (nums1, nums2) {
    const m = nums1.length;
    const n = nums2.length;
    const f = (i, j, k) => {
        if (i >= m) {
            return nums2[j + k - 1];
        }
        if (j >= n) {
            return nums1[i + k - 1];
        }
        if (k == 1) {
            return Math.min(nums1[i], nums2[j]);
        }
        const p = Math.floor(k / 2);
        const x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
        const y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
        return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
    };
    const a = f(0, 0, Math.floor((m + n + 1) / 2));
    const b = f(0, 0, Math.floor((m + n + 2) / 2));
    return (a + b) / 2;
};