2616. Minimize the Maximum Difference of Pairs

 

You are given a 0-indexed integer array nums and an integer p. Find p pairs of indices of nums such that the maximum difference amongst all the pairs is minimized. Also, ensure no index appears more than once amongst the p pairs.

Note that for a pair of elements at the index i and j, the difference of this pair is |nums[i] - nums[j]|, where |x| represents the absolute value of x.

Return the minimum maximum difference among all p pairs. We define the maximum of an empty set to be zero.

 

Example 1:

Input: nums = [10,1,2,7,1,3], p = 2
Output: 1
Explanation: The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. 
The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.

Example 2:

Input: nums = [4,2,1,2], p = 1
Output: 0
Explanation: Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.

 

class Solution {

private int countValid(int[] nums,int thresold){

int index = 0;

int count = 0;

while(index < nums.length - 1){

if(nums[index + 1] - nums[index] <= thresold){

index++;

count++;

}

index++;

}

return count;

}

public int minimizeMax(int[] nums, int p) {

Arrays.sort(nums);

int left = 0;

int n = nums.length;

int right = nums[n - 1] - nums[0];

while(left < right){

int mid = left + (right - left) / 2;

if(countValid(nums,mid) >= p){

right = mid;

}else{

left = mid + 1;

}

}

return left;

}

}