Leetcode 453. Minimum Moves to Equal Array Elements
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.
Q: Can the array contain negative numbers? A: Yes, the array can contain both positive and negative integers as well as zero.
Q: Is the array allowed to contain duplicate numbers? A: Yes, the array can contain duplicate numbers.
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.
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:
n - 1 elements by 1 is equivalent to decrementing one element by 1 when considering the offset to balance all elements.#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;
}
n is the number of elements in the array.
This efficiency ensures the solution is scalable and handles the upper constraint limits gracefully.
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