#4 Median of Two Sorted Arrays Leetcode Java Solutions | leetcode problems and solutions java
4. Median of Two Sorted Arrays
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
Solution:-
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m = nums1.length;
int n = nums2.length;
if (m > n) {
int[] temp = nums1; nums1 = nums2; nums2 = temp;
int tmp = m; m = n; n = tmp;
}
int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
while (iMin <= iMax) {
int i = (iMin + iMax) / 2;
int j = halfLen - i;
if (i < iMax && nums2[j-1] > nums1[i]){
iMin = i + 1;
}
else if (i > iMin && nums1[i-1] > nums2[j]) {
iMax = i - 1;
}
else {
int maxLeft = 0;
if (i == 0) { maxLeft = nums2[j-1]; }
else if (j == 0) { maxLeft = nums1[i-1]; }
else { maxLeft = Math.max(nums1[i-1], nums2[j-1]); }
if ( (m + n) % 2 == 1 ) { return maxLeft; }
int minRight = 0;
if (i == m) { minRight = nums2[j]; }
else if (j == n) { minRight = nums1[i]; }
else { minRight = Math.min(nums2[j], nums1[i]); }
return (maxLeft + minRight) / 2.0;
}
}
return 0.0;
}
}
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Quetions :-
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