2616

2616. Minimize the Maximum Difference of Pairs (Medium)

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

Companies: Amazon, Apple, Microsoft

Related Topics:
Array, Binary Search, Greedy

Similar Questions:

• Minimum Absolute Difference (Easy)
• Minimum Difference Between Largest and Smallest Value in Three Moves (Medium)

Solution 1. Binary Search

// OJ: https://leetcode.com/problems/minimize-the-maximum-difference-of-pairs
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int minimizeMax(vector<int>& A, int P) {
        sort(begin(A), end(A));
        int N = A.size(), L = 0, R = A.back() - A[0];
        auto valid = [&](int t) {
            int cnt = 0;
            for (int i = 1; i < N && cnt < P;) {
                if (A[i] - A[i - 1] <= t) ++cnt, i += 2;
                else ++i;
            }
            return cnt == P;
        };
        while (L <= R) {
            int M = (L + R) / 2;
            if (valid(M)) R = M - 1;
            else L = M + 1;
        }
        return L;
    }
};