LeetCode/MedianTwoArrays.java at main · JBecnel/LeetCode · GitHub

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/*
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.


Constraints:

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

class MedianTwoArrays {
    public static void main(String[] args) {
        int[] a = {1,3,9,10, 12,141};
        int[] b = {2,4,6,8, 100};
        System.out.println(medianSortedArrays(a,b));
        System.out.println(medianSortedArrays(b,a));


        int[] c = {1,3,9, 10, 12, 141};
        int[] d = {2,4,6, 8, 100, 200};
        System.out.println(medianSortedArrays(c,d));
        System.out.println(medianSortedArrays(d,c));

        int[] e = {1,3,5, 7, 9, 11};
        int[] f = {12,14,16, 18, 110, 200};
        System.out.println(medianSortedArrays(e,f));
        System.out.println(medianSortedArrays(f,e));
    }

    public static double medianSortedArrays(int[] array1, int[] array2) {
       int n = array1.length;
       int m = array2.length;

       int i = 0;
       int j = 0;

       int target = (n+m)/2;

      for (int k = 0; k < target; k++) {
            if (i < array1.length-1 && j < array2.length-1) {
                if (array1[i] < array2[j])
                    i++;
                else
                    j++;
            } else {
                if (i == array1.length-1)
                    j++;
                else
                    i++;
            }
       }

       int min1 = Math.min(array1[i], array2[j]);
       int min2 = Math.max(array1[i], array2[j]);
       if (i+1 < array1.length)
            min2 = Math.min(min2, array1[i+1]);
       if (j+1 < array2.length)
            min2 = Math.min(min2, array2[j+1]);

       System.out.println(min1 + " " + min2);

       if ( (n+m) % 2 == 0)
           return (min1+min2)/2.0;
       else
           return min1;
    }

    public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;

        int l1 = 0; int r1 = m-1;
        int l2 = 0; int r2 = n-1;

        int pos1, pos2 = 0;

        int b = Math.max(nums1[m-1], nums2[n-1]);
        int a = Math.min(nums1[0], nums2[0]);

        pos1 = (l1+r1)/2;
        pos2 = (l2+r2)/2;

        while (a < b) {
            System.out.println("pos 1,2  " + pos1 + " " + pos2);

            if (nums1[pos1] > nums2[pos2]) {
                r1 = pos1;
                l2 = pos2;
                a = Math.max(nums2[pos2], a);
                b = Math.min(nums1[pos1], b);

                pos1 = (l1+r1)/2 + ((l1+r1) % 2);
                pos2 = (l2+r2)/2;


            }
            else {
                r2 = pos2;
                l1 = pos1;
                a = Math.max(nums1[pos1], a);
                b = Math.min(nums2[pos2], b);

                pos1 = (l1+r1)/2;
                pos2 = (l2+r2)/2 + ((l2+r2) % 2);

            }
            System.out.println("pos:  1: " + "("+ l1 + ", " + r1 + ")" +  "       2: " + "("+ l2 + ", " + r2 + ")" +  "m,M: " + a + " " + b);


        }


        return nums2[pos2];

    }
}


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