Find Minimum Operations to Make All Elements Divisible by Three

Find Minimum Operations to Make All Elements Divisible by Three - Problem

Array Math Easy

You are given an integer array nums. In one operation, you can add or subtract 1 from any element of nums.

Return the minimum number of operations to make all elements of nums divisible by 3.

Input & Output

Example 1 — Mixed Numbers

$ Input: nums = [3,1,2,4,0]

› Output: 3

💡 Note: 3 is divisible (0 ops), 1→0 (1 op), 2→3 (1 op), 4→3 (1 op), 0 is divisible (0 ops). Total: 3 operations

Example 2 — All Divisible

$ Input: nums = [0,3,6,9]

› Output: 0

💡 Note: All numbers are already divisible by 3, so no operations needed

Example 3 — All Need Adjustment

$ Input: nums = [1,2,4,5]

› Output: 4

💡 Note: 1→0 (1 op), 2→3 (1 op), 4→3 (1 op), 5→6 (1 op). Total: 4 operations

Constraints

1 ≤ nums.length ≤ 50
0 ≤ nums[i] ≤ 50

Visualization

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Understanding the Visualization

Input Array

Array with mixed numbers

Calculate Operations

Check each number's distance to nearest multiple of 3

Sum Total

Return total operations needed

Key Takeaway

🎯 Key Insight: Any integer is at most 1 step away from a multiple of 3

Asked in

G Google 15 a Amazon 12

The key insight is that any number is at most 1 operation away from being divisible by 3. Use modulo arithmetic: if remainder is 1 or 2, we need exactly 1 operation. Best approach is direct calculation with Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Direct Calculation	O(n)	O(1)	Check each number's remainder when divided by 3
Mathematical Formula	O(n)	O(1)	Use modulo arithmetic directly

Direct Calculation — Algorithm Steps

Iterate through each number in the array
Calculate remainder when divided by 3
If remainder is 1, need 1 operation (subtract 1)
If remainder is 2, need 1 operation (add 1)
Sum all operations needed

Visualization

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Step-by-Step Walkthrough

Check Remainders

Calculate num % 3 for each element

Count Operations

Remainder 1 or 2 needs 1 operation, remainder 0 needs 0

Sum Total

Add up all operations needed

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int operations = 0;
    for (int i = 0; i < numsSize; i++) {
        int remainder = nums[i] % 3;
        if (remainder == 1) {
            operations += 1;  // subtract 1
        } else if (remainder == 2) {
            operations += 1;  // add 1
        }
    }
    return operations;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for [1,2,3] format
    int nums[1000];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] != '\n' && token[0] != ' ') {
            nums[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array of n elements

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable

✓ Linear Space

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Medium Frequency

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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