0523. Continuous Subarray Sum

https://leetcode.com/problems/continuous-subarray-sum

Description

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 <= 105

• 0 <= nums[i] <= 109

• 0 <= sum(nums[i]) <= 231 - 1

• 1 <= k <= 231 - 1

ac

math problem and kind of trick

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        // edge cases
        if (nums == null || nums.length <= 1) return false;

        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);  // -1 is to determine subarray length > 1
        int currSum = 0;
        for (int i = 0; i < nums.length; i++) {
            currSum += nums[i];
            if (k != 0) currSum %= k;  // k cannot be 0 in mod, if k == 0, the elements need to be 0 to get result
            if (map.containsKey(currSum)) {
                if (i - map.get(currSum) > 1) return true;  // subarray length > 1
            } else {
                map.put(currSum, i);
            }
        }

        return false;
    }
}
/*
key:
1. remainder, (a + n * k) % k == a % k, so: [23, 2, 4, 6, 7] ->  [23,25,31,35,42] -> [5,1,1,5,0], between two 5, you must had add some n * k. (sum2 - sum1)%k == 0 -> sum2%k - Sum1%k == 0 -> find sum2 %k == sum1 % k. 
2. careful about k == 0, when mod
3. don't override value for the same key in the map
*/