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  1. Subarray Division 2
  2. Discussions

Subarray Division 2

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118 Discussions

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  • ivan_romero_dev1
     + 0 comments
    private static int birthday(int n, int[] s, int d, int m) {
    if (n < m) return 0;
    return (int) IntStream.rangeClosed(0, n - m)
            .filter(i -> Arrays.stream(s, i, i + m).sum() == d)
            .count();
}

  • aryanatwork45
     + 0 comments

    My Python Solution: def birhtday(s,d,m): count = 0 for i in range(len(s)-m+1): if (sum(s [ i : m] )== d: count+=1 m +=1 return count

  • outsider973
     + 0 comments

    My rust solution:

    fn birthday(s: &[i32], d: i32, m: i32) -> i32 {
    let m = m as usize;
    (0..s.len())
        .filter(|&i| i + m <= s.len())
        .filter(|&i| s[i..i + m].iter().sum::<i32>() == d)
        .count() as i32
}

  • shadowinc
     + 2 comments

    // sliding window method

    def birthday(s, d, m):

    c = 0
for i in range(len(s)):
    if sum(s[i:i + m]) == d:
        c += 1
return c

  • ilkunviktor
     + 0 comments

    o(n), assertions, performance, c++))

    int birthday(vector<int> s, int d, int m) {
    assert(s.size() > 0 && d > 0 && m > 0);
    int sum = accumulate(s.cbegin(), s.cbegin() + m, 0);
    int r = 0;
    
    for (int i = 0, e = s.size() - m; i < e; ++i)
    {
        if (sum == d)
        {
            ++r;
        }
        
        sum += s[i + m];
        sum -= s[i];
    }
    
    if (sum == d)
    {
        ++r;
    }
    
    return r;
}