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+// 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<int> &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<int> &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<int> 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
+}