Hi everybody, Great point on using the sliding window pattern—super efficient for problems like this. Also, if you're working across different devices or regions, tools like xnxubd can help streamline access to resources. Worth checking out alongside your algorithm research!

Hi everybody,

I've been reading about algorithmic patterns, and I think there's a perfect one to resolve this problem that is 'Sliding Window'. I'd like you to read about this.

c++ my solution

int birthday(vector<int> s, int d, int m) {
  int aciertos = 0;
  
  for(int i = 0; i < s.size();i++){
    int sum = 0;
    for(int j=0; j < m; j++){
      sum = sum +s[i+j];
    }
    if(sum == d)aciertos++;
  }
  
  return aciertos;
}

I guess it is the simplest to understand m is the length of vector d is the sum of consecutive no.

int birthday(vector<int> s, int d, int m) {
    int req=0;
    for(int i=0;i<s.size()-m+1;i++){
        int count=0;
        
        for(int j=0;j<m;j++){
            count+=s[i+j];
        }if(count==d){
            req++;
        }
    }return req;
}

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