Counting Subarrays!

 https://www.interviewbit.com/old/problems/counting-subarrays/

 

int Solution::solve(vector<int> &A, int B) {
int i=0, j=0,count=0;
for (i=0; i<A.size(); i++){
    int sum=0;
    for (j=i; j>=0; j--){
        sum+=A[j];
        if(sum<B) count++;
        else break;
    }
}
return count;
} 


int Solution::solve(vector<int> &A, int B) {
 int sum=0;
 int ans=0;
 int st=0;
 int end=0;
while(end<A.size())
{
sum=sum+A[end];
if(sum<B)
{
end++;
ans=ans+(end-st);
}
else
{
while(sum>=B)
{
sum=sum-A[st];
st++;
}
end++;
ans=ans+(end-st);
}
}
return ans;
}

Comments

Post a Comment

Popular posts from this blog

Perfect Peak of Array

  Given an integer array A of size N . You need to check that whether there exist a element which is strictly greater than all the elements on left of it and strictly smaller than all the elements on right of it. If it exists return 1 else return 0 .  https://www.interviewbit.com/old/problems/perfect-peak-of-array/ int Solution::perfectPeak(vector<int> &A) { int ans=-1; int l[A.size()]; int r[A.size()]; l[0]=A[0]; r[A.size()-1]=A[A.size()-1]; for(int i=1;i<A.size();i++){     l[i] = max(A[i],l[i-1]); } for(int i=A.size()-2;i>=0;i--){     r[i] = min(A[i],r[i+1]); } for(int i=1;i<A.size()-1;i++){     if(A[i]>l[i-1]&&A[i]<r[i+1]){         return 1;     } } if(ans==-1){     return 0; } }  
Read more

Is Rectangle?

 https://www.interviewbit.com/old/problems/is-rectangle/ int Solution::solve(int A, int B, int C, int D) {      if (((A==B)&&(C==D))||((A==C)&&(B==D))||((A==D)&&(B==C))){         return 1;     }else{         return 0;     } }  
Read more

Sort array with squares!

 https://www.interviewbit.com/old/problems/sort-array-with-squares/      vector<int> Solution::solve(vector<int> &arr) {         int mid=0;     for(int i=0;i<arr.size();i++)     {         if(arr[i]>=0)         {             mid=i;             break;         }     }     int i=mid-1,j=mid;     vector<int>res(arr.size());     int k=0;     while(i>=0 && j<arr.size())     {         if((arr[i]*arr[i])<(arr[j]*arr[j]))         {             res...
Read more