Continuous Subarray Sum - LeetCode solutions

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

Solution (Java)

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        map.put(0, -1);
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            int remainder = sum % k;
            if (map.containsKey(remainder)) {
                if (map.get(remainder) + 1 < i) {
                    return true;
                }
            } else {
                map.put(remainder, i);
            }
        }
        return false;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).