Continuous Subarray Sum| leetcode solution

 523. Continuous Subarray Sum

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

solution:

class Solution {

public:

    bool checkSubarraySum(vector<int>& nums, int k) {

        

         unordered_map<int, int>umap;

        int s = nums[0];

        for(int i = 1 ; i < nums.size(); i++ )

        {    

             s = s+ nums[i];

             s = s%k; 

            if(umap.find(s) != umap.end())

            {

                if((i - umap[s]) > 1)

                    return true;

            }

            else

                umap[s] = i;

        }

        return false;

        

    }

};