Continuous Subarray Sum

#1

question–>


code–>

 class Solution {
    public:
        bool checkSubarraySum(vector<int>& nums, int k) {
            unordered_map<int,int> m;
            m[0]=-1;
           
            int* csum=new int[nums.size()];
            csum[0]=nums[0];
            for(int i=1;i<nums.size();i++){
                csum[i]=nums[i]+csum[i-1];
            }
            for(int i=0;i<nums.size();i++){
                if(k!=0){
                if(m.count(csum[i]%k)){
                    if(i-m[csum[i]%k]>1){
                        return true;
                    }
                }
                m[csum[i]%k]=i;}
                else if(k==0 and nums[i]==0 and i<nums.size()-1 and nums[i+1]==0 ){
                    return true;
                }
            }
            return false;
        }
    };

test case not passing
[0,0]
-1

#2

@tusharaggarwal272 are you there?

#3

hello @sheikhhaji18 yes what is your doubt?

#4

the code above is not passing all testcases help

#5

@sheikhhaji18 where your are submitting on leetcode?

#6

yes iam submitting in leetcode @tusharaggarwal272

#7

Go to the details of the test cases and you will got to know which test cases are not getting passes and what can you do and i am also trying.
Happy Learning!!

#8

yes i am trying @tuasharaggarwal

#10

@sheikhhaji18i think your logic is right but it should be framed agaoin to handle corner test cases.
class Solution {
public:
bool checkSubarraySum(vector& nums, int k) {

    int N = nums.size();
    
    for (int i = 0; i < N; i++)
    {
        int sum = nums[i];
        
        for (int j = i+1; j < N; j++)
        {
            sum += nums[j];
            
            if (sum == k) return true;
            
            if (k != 0 && sum % k == 0) return true;
        }
    }
    
    return false;
}

};
see this code it is passing every test case:
if you have any doubt you can ask here:
Happy Learnin!!

#11

okay @tusharaggarwal272 i didnot see the constraint i just saw the video and implemented it ,it will work fine for n^2

#12

@sheikhhaji18 yes its okay.
you can mark this doubt as resolved.
Happy Learning!!

#13

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