Minimum Moves to Equal Array Elements - Practice Coding Problems

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: 10

💡 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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Understanding the Visualization

Input Array

Array with different element values

Key Insight

Incrementing n-1 elements = decrementing 1 element

Final State

All elements equal to minimum value

Key Takeaway

🎯 Key Insight: Instead of incrementing n-1 elements, think of it as decrementing 1 element to reach the minimum

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

Approach	Time	Space	Notes
✓ Simulation Approach	O(n × max_diff)	O(1)	Simulate the actual moves by repeatedly incrementing n-1 elements
Mathematical Formula	O(n)	O(1)	Use mathematical insight: sum of differences from minimum

Simulation Approach — Algorithm Steps

Find minimum element in array
Increment all other n-1 elements by 1
Repeat until all elements are equal

Visualization

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Step-by-Step Walkthrough

Find Min

Identify minimum element (don't increment it)

Increment Others

Add 1 to all other n-1 elements

Repeat

Continue until all elements are equal

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int moves = 0;
    while (1) {
        int minVal = INT_MAX, maxVal = INT_MIN;
        for (int i = 0; i < numsSize; i++) {
            if (nums[i] < minVal) minVal = nums[i];
            if (nums[i] > maxVal) maxVal = nums[i];
        }
        if (minVal == maxVal) break;
        for (int i = 0; i < numsSize; i++) {
            if (nums[i] != minVal) {
                nums[i]++;
            }
        }
        moves++;
    }
    return moves;
}

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 × max_diff)

In worst case, need max_diff iterations, each taking O(n) time to update array

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for counters

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Simulation Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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