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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Subarray Division
  5. Discussions

Subarray Division

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3092 Discussions

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  • aayuuhere
     + 0 comments

    Java Solution

    Using Sliding Window Protocal O(n)

     public static int birthday(List<Integer> s, int d, int m) {
    // Write your code here
        int count = 0 ;
        
        int sum = 0 ;
        for(int i = 0 ; i < m ; i++){
            sum += s.get(i);
        }
        if(sum == d){
            count++;
        }
        
        for(int i = m ; i < s.size() ; i ++){
            sum = sum - s.get(i-m) + s.get(i);
            if(sum == d){
                count++;
            }
        }
        return count;
    }

  • AndyScript
     + 1 comment

    TypeScript ugly one-liner:

    function birthday(s: number[], d: number, m: number): number {
  return s.flatMap((_, i) => [s.slice(i, m + i)]).filter((arr) => arr.reduce((a, b) => a + b, 0) === d).length;
}

  • nicokurtz01
     + 0 comments
    int birthday(vector<int> s, int d, int m) {
    std::vector<int>::iterator left = s.begin();
    std::vector<int>::iterator right = left + m; // points behind sliding window
    
    int sum = std::accumulate(left, right, 0);
    int numWays = sum == d ? 1 : 0;
    
    while (right != s.end()) {
        sum = sum - *left + *right; // update sum
        left++; right++; // and move the sliding window
        
        if (sum == d)
            numWays++;
    }
    
    return numWays;
}

  • liuliming_kz
     + 1 comment

    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; }

  • wmgarciaporcel
     + 0 comments

    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;
}