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Minimum Operations to Make Array Sum Divisible by K

This problem asks you to find the minimum operations to make the sum of an array divisible by a given integer k.

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Easy · Array plus Math

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Answer-first summary

This problem asks you to find the minimum operations to make the sum of an array divisible by a given integer k.

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The task requires calculating the minimum number of operations needed to make the sum of an array divisible by k. The solution revolves around finding the remainder of the sum when divided by k and determining how many steps are needed to adjust it. The problem involves both array manipulation and mathematical properties, particularly modular arithmetic.

Problem Statement

You are given an integer array nums and an integer k. The task is to perform operations on the array such that the sum of its elements becomes divisible by k. Each operation involves adding or subtracting from any element in the array.

Return the minimum number of operations required to make the sum of the array divisible by k. You can perform the operation any number of times.

Examples

Example 1

Input: nums = [3,9,7], k = 5

Output: 4

Example 2

Input: nums = [4,1,3], k = 4

Output: 0

Example 3

Input: nums = [3,2], k = 6

Output: 5

Constraints

  • 1 <= nums.length <= 1000
  • 1 <= nums[i] <= 1000
  • 1 <= k <= 100

Solution Approach

Calculate the Remainder

The first step is to compute the sum of the elements in the array and find its remainder when divided by k. This can be done using the modulo operation, sum(nums) % k.

Find the Minimum Adjustment

Next, you will need to find the minimum number of operations required to adjust the remainder to 0. This will involve determining how much you need to add or subtract from the sum to make it divisible by k.

Iterate Through the Array

For each element in the array, determine how it can contribute to adjusting the sum to a multiple of k. Keep track of the number of operations needed to achieve divisibility.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time complexity of this problem depends on the approach you use to calculate the sum and adjust the remainder. If you're iterating through the array once to compute the sum and then adjusting it, the time complexity is O(n), where n is the length of the array. The space complexity depends on whether you're using extra space for tracking operations, but it can be considered O(1) for the most optimized approach.

What Interviewers Usually Probe

  • Tests understanding of modulo and array manipulation.
  • Evaluates ability to work with array sums and math-based optimizations.
  • Assesses problem-solving skills in handling operations within constraints.

Common Pitfalls or Variants

Common pitfalls

  • Forgetting to account for the case where the sum is already divisible by k (requires zero operations).
  • Overcomplicating the problem by trying to use more than one operation per element.
  • Not correctly adjusting the sum by the minimum necessary amount, leading to incorrect results.

Follow-up variants

  • What if we allowed negative numbers in the array?
  • What if the array elements were limited to powers of two?
  • What if we needed to make the sum divisible by a prime number instead of k?

FAQ

How do I solve the Minimum Operations to Make Array Sum Divisible by K?

To solve this, you need to calculate the remainder of the sum of the array when divided by k. Then find the minimum adjustments required to make the sum divisible by k.

What is the best approach for solving this problem?

The best approach is to first calculate the sum of the array, then find the remainder when divided by k, and adjust the sum using the least number of operations to make it divisible by k.

How does the modulo operation help in this problem?

The modulo operation helps in determining the remainder of the sum when divided by k, which is key to finding out how much needs to be added or subtracted to achieve divisibility.

What if the sum is already divisible by k?

If the sum is already divisible by k, then no operations are needed, and the answer is zero.

Can this problem be solved using dynamic programming?

While dynamic programming could be applied to some variants of the problem, a direct approach using modular arithmetic and greedy adjustments is typically the most efficient.

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Solution

Solution 1: Sum and Modulo

The problem essentially asks for the result of the sum of the array elements modulo $k$. Therefore, we only need to iterate through the array, calculate the sum of all elements, and then take the modulo $k$. Finally, return this result.

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class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        return sum(nums) % k
Minimum Operations to Make Array Sum Divisible by K Solution: Array plus Math | LeetCode #3512 Easy