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## Problem

Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.

An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.


Example 2:

Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.


Example 3:

Input: nums = [23,2,6,4,7], k = 13
Output: false


Constraints:

• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^9
• 0 <= sum(nums[i]) <= 2^31 - 1
• 1 <= k <= 2^31 - 1

## Code

560. Subarray Sum Equals K

class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
HashMap<Integer, Integer> map = new HashMap<>();

int sum = nums[0] % k;
map.put(sum, 0);

for(int i = 1; i < nums.length; i++) {
int num = nums[i];

sum = (sum + num) % k;

// [23,2,4,6,6], 7 -> [2,4,1,0,6]
if(sum == 0) return true;
if(map.containsKey(sum) && i - map.get(sum) > 1) return true;

// [1,3,0,6], 6
if(!map.containsKey(sum)) {
map.put(sum, i);
}

}

return false;
}
}