523. Continuous 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.

 

Approach #1: Brute force. [C++]

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        if (nums.size() < 2) return false;
        int l = nums.size();
        if (k == 0) {
            for (int i = 0; i < l-1; ++i)
                if (nums[i] + nums[i+1] == 0) return true;
            return false;
        }
        vector<int> sum(l+1, 0);
        for (int i = 0; i < l; ++i)
            sum[i+1] = sum[i] + nums[i];
        for (int i = 0; i < l-1; ++i) {
            for (int j = i+2; j <= l; ++j) {
                if ((sum[j] - sum[i]) % k == 0) return true;
            }
        }
        
        return false;
    }
};

　　

sum[i] : the sum of nums from 0 to i;

 

Approach #2: Set. [C++]

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int n = nums.size(), sum = 0, pre = 0;
        unordered_set<int> modk;
        for (int i = 0; i < n; ++i) {
            sum += nums[i];
            int mod = k == 0 ? sum : sum % k;
            if (modk.count(mod)) return true;
            modk.insert(pre);
            pre = mod;
        }
        return false;
    }
};