// 2616. Minimize the Maximum Difference of Pairs #include "leetcode.h" /* * 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 * minimised. also, ensure no index appears more than once amongst that '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. */ class Solution { bool can_form_pair(vector &nums, int p, int mid) { int cnt = 0; for (int i = 0; i < nums.size() - 1 && cnt < p;) { if (nums[i + 1] - nums[i] <= mid) { cnt++; i += 2; } else i++; } return cnt >= p; } public: int minimizeMax(vector &nums, int p) { sort(nums.begin(), nums.end()); int left = 0, right = nums.end() - nums.begin(); while (left < right) { int mid = (left + right) / 2; if (can_form_pair(nums, p, mid)) right = mid; else left = mid + 1; } return left; } }; int main() { Solution obj; vector n1 = {10, 1, 2, 7, 1, 3}, n2 = {4, 2, 1, 2}; printf("%d\n", obj.minimizeMax(n1, 2)); // expect: 1 printf("%d\n", obj.minimizeMax(n2, 1)); // expect: 0 }