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Leetcode 453. Minimum Moves to Equal Array Elements

Problem Statement:

Given an integer array nums of size n, you need to find the minimum number of moves required to make all elements equal. In one move, you can increment n - 1 elements by 1.

Clarifying Questions:

  1. Q: Can the array contain negative numbers? A: Yes, the array can contain both positive and negative integers as well as zero.

  2. Q: Is the array allowed to contain duplicate numbers? A: Yes, the array can contain duplicate numbers.

  3. Q: Is there a constraint on the size of the array? A: The typical constraints apply, however, we should presume that 1 <= nums.length <= 10^4 and -10^6 <= nums[i] <= 10^6.

Strategy:

To solve this problem, we need to efficiently determine the minimum moves required to make all the elements equal. Here’s the step-by-step strategy:

  1. Understanding the Operation:
    • Incrementing n - 1 elements by 1 is equivalent to decrementing one element by 1 when considering the offset to balance all elements.
    • Thus, making all elements equal is equivalent to reducing every element to the minimum element in the array.
  2. Mathematical Insight:
    • If we reduce all elements to the minimum element in the array, the number of moves required is the sum of the difference between each element and the minimum element.

Plan:

  1. Find the minimum element in the array.
  2. Compute the sum of the differences between each element and the minimum element.
  3. Return the computed sum.

Code:

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

int minMoves(std::vector<int>& nums) {
    int minElement = *std::min_element(nums.begin(), nums.end());
    int moves = 0;
    
    for (int num : nums) {
        moves += num - minElement;
    }
    
    return moves;
}

int main() {
    std::vector<int> nums = {1, 2, 3};
    std::cout << "Minimum moves required: " << minMoves(nums) << std::endl;
    return 0;
}

Time Complexity:

This efficiency ensures the solution is scalable and handles the upper constraint limits gracefully.

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