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LeetCode 523: Continuous Subarray Sum Solution

Master LeetCode problem 523 (Continuous Subarray Sum), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

523. Continuous Subarray Sum

Medium
ArrayHash TableMathPrefix Sum

Problem Explanation

Explanation:

  1. Algorithmic Idea:

    • We can iterate through the array and keep track of the running sum.
    • At each step, we calculate the running sum modulo k.
    • If we encounter the same modulo value again, it means the sum of elements between those two indexes is a multiple of k.
    • We also need to consider cases where the running sum itself is a multiple of k.
    • To handle the case where the subarray size is at least 2, we store the running sum modulo k in a HashMap with initial key 0 and value -1.
    • If the difference between the current index and the index stored in the HashMap corresponding to the same modulo value is greater than 1, we have found a good subarray.

  2. Time Complexity: O(n) where n is the number of elements in the input array.

  3. Space Complexity: O(min(n, k)) where n is the number of elements in the input array.

:

Solution Code

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;
    }
}

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