LeetCode Challenge Day 75 — 3512.Minimum Operations to Make Array Sum Divisible by K

Hey folks 👋

This is Day 75 of my LeetCode streak 🚀
Today's problem is Minimum Operations to Make Array Sum Divisible by K — one of the cleanest math-based problems in the series.

📌 Problem Statement

You are given an array nums and an integer k.
Your task: find the minimum number of operations needed to make the sum of the array divisible by k.

Each operation allows you to increment any element by 1.

💡 Intuition

To make the total sum divisible by k, we only need to check:

remainder = sum(nums) % k

If remainder == 0, the sum is already divisible → 0 operations.
Otherwise, we need exactly remainder increments to reach the next multiple of k.

This works because every increment increases the total sum by 1.

🔑 Approach

1. Compute the total sum of the array.

2. Return sum % k — the remainder directly represents the number of operations needed.

This solution is optimal because increments are the only allowed operation.

⏱️ Complexity Analysis

ComplexityValue TimeO(n) SpaceO(1)

🧑‍💻 Code (JavaScript)

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var minOperations = function(nums, k) {
    const sum = nums.reduce((a, b) => a + b, 0);
    return sum % k;
};

🎯 Reflection

This is a perfect example of how simple modulo arithmetic can turn a problem into a one-line solution.

• ✔ No complex logic

• ✔ Pure math

• ✔ Runs in linear time with constant space

That's it for Day 75 of my LeetCode challenge 💪
See you tomorrow!

Happy Coding 👨‍💻