Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 9

Description

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

Solution

class Solution {
  int pairs(const vector<int> &nums, int diff) {
    int count = 0;
    for(int i = 0; i < nums.size() - 1; ++i) {
      if(nums[i + 1] - nums[i] <= diff) {
        count += 1;
        i += 1;
      }
    }
    return count;
  }
public:
  int minimizeMax(vector<int> &nums, int p) {
    sort(nums.begin(), nums.end());
    int low = 0;
    int high = nums.back() - nums.front();
    while(high > low) {
      int mid = (high + low) / 2;
      if(pairs(nums, mid) < p) {
        low = mid + 1;
      } else {
        high = mid;
      }
    }
    return low;
  }
};

// Accepted
// 103/103 cases passed (140 ms)
// Your runtime beats 88.53 % of cpp submissions
// Your memory usage beats 82.46 % of cpp submissions (81.08 MB)