Class Solution
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public final class Solution3512 - Minimum Operations to Make Array Sum Divisible by K.
Easy
You are given an integer array
numsand an integerk. You can perform the following operation any number of times:Select an index
iand replacenums[i]withnums[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:
Perform 4 operations on
nums[1] = 9. Now,nums = [3, 5, 7].The sum is 15, which is divisible by 5.
Example 2:
Input: nums = 4,1,3, k = 4
Output: 0
Explanation:
The sum is 8, which is already divisible by 4. Hence, no operations are needed.
Example 3:
Input: nums = 3,2, k = 6
Output: 5
Explanation:
Perform 3 operations on
nums[0] = 3and 2 operations onnums[1] = 2. Now,nums = [0, 0].The sum is 0, which is divisible by 6.
Constraints:
1 <= nums.length <= 10001 <= nums[i] <= 10001 <= k <= 100
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final IntegerminOperations(IntArray nums, Integer k)-
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Method Detail
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minOperations
final Integer minOperations(IntArray nums, Integer k)
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