algoadvance

Leetcode 2974. Minimum Number Game

Problem Statement

You are given an integer array nums consisting of 2 * n integers. You need to divide these integers into exactly n pairs such that the maximum minimum pair sum is minimized. A pair sum is defined as the sum of a number from the left half of the list and a number from the right half of the list, where the left half consists of the first n integers and the right half consists of the last n integers.

Return the minimized maximum pair sum.

Clarifying Questions

  1. What constraints can we assume for the length of the input array?
    • You can assume that nums will always have an even number of elements with the given constraints 2 <= 2 * n <= 10^5.
  2. What range can we expect for the integers in the array?
    • The integer values in nums are between 1 and 10^6.
  3. Can the integers in the array be negative?
    • No, in this problem, the integers are always positive as per the standard constraints.
  4. Should we consider the input array to contain distinct integers or can they be duplicated?
    • Integers may be duplicated in the array.

Strategy

To minimize the maximum pair sum, a good approach is:

  1. Sort the Array: First, sort the array in ascending order.
  2. Pair the Elements: After sorting, the most efficient way to achieve our goal is to consider the weakest pairs with the strongest ones:
    • Pair the smallest element from the left half with the largest from the right half,
    • i.e., if the array is [a1, a2, ..., an, b1, b2, ..., bn], pair ai with bn-i+1 for 1 <= i <= n.
  3. Compute Pair Sums: Calculate the pair sums and find the maximum pair sum.

Code

Here is the C++ implementation of the above strategy:

#include <iostream>
#include <vector>
#include <algorithm>

int minimizeMaximumPairSum(std::vector<int>& nums) {
    // Sort the array
    std::sort(nums.begin(), nums.end());
    int n = nums.size() / 2;
    int max_pair_sum = 0;
    
    for (int i = 0; i < n; ++i) {
        // Calculate the pair sum
        int pair_sum = nums[i] + nums[nums.size() - 1 - i];
        // Update the maximum pair sum
        max_pair_sum = std::max(max_pair_sum, pair_sum);
    }
    
    return max_pair_sum;
}

int main() {
    std::vector<int> nums = {3, 5, 2, 3, 8, 7, 4, 6};
    std::cout << "Minimized Maximum Pair Sum is: " << minimizeMaximumPairSum(nums) << std::endl;
    return 0;
}

Explanation

  1. Sort the Array: Sorting ensures that you can efficiently pair the smallest and largest elements.
  2. Pair and Calculate Sum: Iterate through the first half of the sorted array (left half) and pair each element with the corresponding element from the second half (right half).
  3. Update Maximum Pair Sum: Track the maximum pair sum encountered during the iteration.

Time Complexity

Thus, the overall time complexity is (O(n \log n)).

Cut your prep time in half and DOMINATE your interview with AlgoAdvance AI