Link: https://leetcode.cn/problems/continuous-subarray-sum/

Question

difficulty: mid
adj diff: 2

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

就是用一个 map 不断的记录就行了。

Code

public boolean checkSubarraySum(int[] nums, int k) {
    int len = nums.length;
    int[] prefixSum = new int[len + 1];
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    map.put(0, -1);

    for (int i = 1; i <= len; i++) {
        prefixSum[i] = (prefixSum[i - 1] + nums[i - 1]) % k;
        if (map.containsKey(prefixSum[i])) {
            if (i - map.get(prefixSum[i]) > 2)
            // check the subarray length (need to be at least 2)
                return true;
        } else {
            map.put(prefixSum[i], i - 1);
        }
    }
    return false;
}

523. Continuous Subarray Sum

Question

Code