Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
You are given an integer array nums and an integer k. You can perform the following operation any number of times:
i and replace nums[i] with nums[i] - 1.Return the minimum number of operations required to make the sum of the array divisible by k.
Example 1:
Input: nums = [3,9,7], k = 5
Output: 4
Explanation:
nums[1] = 9. Now, nums = [3, 5, 7].Example 2:
Input: nums = [4,1,3], k = 4
Output: 0
Explanation:
Example 3:
Input: nums = [3,2], k = 6
Output: 5
Explanation:
nums[0] = 3 and 2 operations on nums[1] = 2. Now, nums = [0, 0].Constraints:
1 <= nums.length <= 10001 <= nums[i] <= 10001 <= k <= 100Problem summary: You are given an integer array nums and an integer k. You can perform the following operation any number of times: Select an index i and replace nums[i] with nums[i] - 1. Return the minimum number of operations required to make the sum of the array divisible by k.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math
[3,9,7] 5
[4,1,3] 4
[3,2] 6
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3512: Minimum Operations to Make Array Sum Divisible by K
class Solution {
public int minOperations(int[] nums, int k) {
return Arrays.stream(nums).sum() % k;
}
}
// Accepted solution for LeetCode #3512: Minimum Operations to Make Array Sum Divisible by K
func minOperations(nums []int, k int) (ans int) {
for _, x := range nums {
ans = (ans + x) % k
}
return
}
# Accepted solution for LeetCode #3512: Minimum Operations to Make Array Sum Divisible by K
class Solution:
def minOperations(self, nums: List[int], k: int) -> int:
return sum(nums) % k
// Accepted solution for LeetCode #3512: Minimum Operations to Make Array Sum Divisible by K
impl Solution {
pub fn min_operations(nums: Vec<i32>, k: i32) -> i32 {
nums.iter().sum::<i32>() % k
}
}
// Accepted solution for LeetCode #3512: Minimum Operations to Make Array Sum Divisible by K
function minOperations(nums: number[], k: number): number {
return nums.reduce((acc, x) => acc + x, 0) % k;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.