Minimum Moves to Equal Array Elements

Minimum Moves to Equal Array Elements - Problem

Array Math Medium

Given an integer array nums of size n, return the minimum number of moves required to make all array elements equal.

In one move, you can increment n - 1 elements of the array by 1.

Note: Incrementing n - 1 elements is equivalent to decrementing one element by 1.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3]

› Output: 3

💡 Note: Need 3 moves: increment [1,2] by 1 (→[2,3,3]), increment [2,3] by 1 (→[3,4,3]), increment [3,4] by 1 (→[4,4,4]). Total: 3 moves.

Example 2 — All Equal

$ Input: nums = [1,1,1]

› Output: 0

💡 Note: All elements are already equal, so no moves needed.

Example 3 — Larger Differences

$ Input: nums = [1,2,9]

› Output: 9

💡 Note: Minimum is 1. Moves needed: (2-1) + (9-1) = 1 + 8 = 9 moves to equalize all to 1.

Constraints

2 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 45 a Amazon 35 M Microsoft 25 f Facebook 20

The key insight is that incrementing n-1 elements equals decrementing 1 element. Best approach is the mathematical formula: find minimum and sum differences. Time: O(n), Space: O(1)

Common Approaches

✓ Mathematical Formula

⏱️ Time: O(n) Space: O(1)

The key insight is that incrementing n-1 elements is equivalent to decrementing 1 element. So we want to bring all elements down to the minimum value. The answer is the sum of differences between each element and the minimum.

Simulation Approach

⏱️ Time: O(n × max_diff) Space: O(1)

Keep track of moves while repeatedly finding the minimum element and incrementing all other elements until all are equal. This directly simulates the problem description.

Mathematical Formula — Algorithm Steps

Find the minimum element in the array
Calculate difference between each element and minimum
Sum all the differences to get total moves

Visualization

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Step-by-Step Walkthrough

Find Minimum

Identify the smallest element

Calculate Differences

Find how much each element exceeds minimum

Sum Differences

Total moves equals sum of all differences

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int solution(int* nums, int numsSize) {
    int minVal = INT_MAX;
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] < minVal) {
            minVal = nums[i];
        }
    }
    int moves = 0;
    for (int i = 0; i < numsSize; i++) {
        moves += nums[i] - minVal;
    }
    return moves;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' || token[0] == '-') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to find minimum, then single pass to calculate sum of differences

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for minimum and sum

✓ Linear Space

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Input & Output

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Common Approaches

Mathematical Formula — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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