minimize_the_maximum_difference_of_pairs.dart

/*

-* 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.
 

Constraints:

1 <= nums.length <= 105
0 <= nums[i] <= 109
0 <= p <= (nums.length)/2

*/

class Solution {
  int minimizeMax(List<int> nums, int p) {
    nums.sort();
    int left = 0;
    int right = nums[nums.length - 1] - nums[0];

    while (left < right) {
      int mid =left + (right + left) ~/ 2;
      if (helper(nums, mid, p)) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }

    return left;
  }

  bool helper(List<int> nums, int mid, int p) {
    int count = 0;
    for (int i = 0; i < nums.length - 1 && count < p;) {
      if (nums[i + 1] - nums[i] <= mid) {
        count++;
        i += 2;
      } else {
        i++;
      }
    }
    return count >= p;
  }
}