O(n): function birthday(s, d, m) { // Write your code here let count = 0; for (let i=0; i 0 ? s[i-1]: 0); if (i == m -1 && s[i] == d) { count += 1 } if (i > m - 1 && s[i] - s[i-m] == d) { count += 1 } } return count; }

Here is my solution, using JavaScript

function birthday(s, d, m) {
    // Write your code here
    let count = 0;
    while (s.length >= m){
      let sum = 0;
      // sum terms
      for(let i=0; i<m; i++){
        sum += s[i];
    }
    if(sum===d) count++;    
    s.shift();
    } 
    return count;
}

//javascript function birthday(s, d, m) { // Write your code here

let contador = 0;

for (let i = 0; i <= s.length - m; i++) {

const bloque = s.slice(i, i + m);

const suma = bloque.reduce((acumulador, valorActual) => acumulador + valorActual, 0);

if (suma === d) {
  contador++;
}

}

return contador;

}

Here is my own java solution. Please rate if you don't mind. :)

public static int birthday(List s, int d, int m) {

    int count = 0;

    for(int i = 0; i < s.size(); i++){                      
        int squaresSum = 0;

        for(int j = 0; j < m; j++){                
            if(j+i < s.size()){                    
                squaresSum += s.get(j+i);
            }               
        }

        if(squaresSum == d){
            count++;
        }
    }

    return count;   
}

}

def birthday(s, d, m):
# Write your code here
count = 0


if isinstance(s, list):
    if m == 0:
        return "0"
    for i in range(len(s) - m + 1):
        if sum(s[i:i + m]) == d:
            count += 1

    elif isinstance(s, int):
    if m == 1 and s == d:
        count = 1

return f"{count}"
elif isinstance(s, int):
    if m == 1 and s == d:
        count = 1

return f"{count}"