[LeetCode] 523. Continous Subarray Sum

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.
Example 1:
Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.
Example 2:
Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.
Note:
  1. The length of the array won't exceed 10,000.
  2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

Thought process:
Obtain the prefix sum at each index and brute force the sum of every sub-array.

Solution 1 (prefix sum):
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class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums.length == 0) {
            return false;
        }
        
        int[] prefixSum = new int[nums.length + 1];
        prefixSum[0] = 0;
        for (int i = 1; i < prefixSum.length; i++) {
            prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
        }
        
        for (int i = 0; i < prefixSum.length; i++) {
            for (int j = i + 2; j < prefixSum.length; j++) {
                int sum = prefixSum[j] - prefixSum[i];
                if (sum == k || (k != 0 && sum % k == 0)) {
                    return true;
                }
            }
        }
        return false;
    }
}

Time complexity: O(n^2).

There is another solution using hash map. Iterate through the array. Keep track of prefix sum. Put sum % k -> index mappings into a hash map. At each index, check if the map already contains the remainder as a key. If so, check if the sub-array's length is larger than one. If so, return true. This is because if a sub-array sums to a multiple of k, then its calculated remainder will be remainder + sum % k = remainder + 0 = remainder.

Solution 2 (hash map):
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class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        int sum = 0;
        
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (k != 0) {
                sum %= k;
            }
            
            if (map.containsKey(sum)) {
                if (i - map.get(sum) > 1) {
                    return true;
                }
            } else {
                map.put(sum, i);
            }
        }
        return false;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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