Add MedianOfTwoSortedArrays algorithm (#5554)

This commit is contained in:
Hardik Pawar
2024-10-09 11:26:08 +05:30
committed by GitHub
parent 403649d404
commit 5c79e5de5d
3 changed files with 99 additions and 0 deletions

View File

@ -0,0 +1,53 @@
package com.thealgorithms.divideandconquer;
public final class MedianOfTwoSortedArrays {
private MedianOfTwoSortedArrays() {
}
/**
* Finds the median of two sorted arrays in logarithmic time.
*
* @param nums1 the first sorted array
* @param nums2 the second sorted array
* @return the median of the combined sorted array
* @throws IllegalArgumentException if the input arrays are not sorted
*/
public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums1.length > nums2.length) {
return findMedianSortedArrays(nums2, nums1); // Ensure nums1 is the smaller array
}
int m = nums1.length;
int n = nums2.length;
int low = 0;
int high = m;
while (low <= high) {
int partition1 = (low + high) / 2; // Partition in the first array
int partition2 = (m + n + 1) / 2 - partition1; // Partition in the second array
int maxLeft1 = (partition1 == 0) ? Integer.MIN_VALUE : nums1[partition1 - 1];
int minRight1 = (partition1 == m) ? Integer.MAX_VALUE : nums1[partition1];
int maxLeft2 = (partition2 == 0) ? Integer.MIN_VALUE : nums2[partition2 - 1];
int minRight2 = (partition2 == n) ? Integer.MAX_VALUE : nums2[partition2];
// Check if partition is valid
if (maxLeft1 <= minRight2 && maxLeft2 <= minRight1) {
// If combined array length is odd
if (((m + n) & 1) == 1) {
return Math.max(maxLeft1, maxLeft2);
}
// If combined array length is even
else {
return (Math.max(maxLeft1, maxLeft2) + Math.min(minRight1, minRight2)) / 2.0;
}
} else if (maxLeft1 > minRight2) {
high = partition1 - 1;
} else {
low = partition1 + 1;
}
}
throw new IllegalArgumentException("Input arrays are not sorted");
}
}