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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0523_continuous_subarray_sum;

// #Medium #Array #Hash_Table #Math #Prefix_Sum
// #2022_07_28_Time_37_ms_(41.45%)_Space_109.7_MB_(5.07%)

import java.util.HashMap;
import java.util.Map;

/**
 * 523 - Continuous Subarray Sum\.
 *
 * Medium
 *
 * 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
**/
public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        Map 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;
    }
}