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  1. Subarray Division 2
  2. Discussions

Subarray Division 2

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117 Discussions

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  • 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;
}

  • fennessyeoin
     + 0 comments

    O(n) prefix sum solution in Golang

    func birthday(s []int32, d int32, m int32) int32 {
    prefixSums := make([]int32, len(s)+1)
    for i, n := range s {
        prefixSums[i+1] = prefixSums[i] + n
    }
    
    var countOfSegments int32
    for i := m; i < int32(len(prefixSums)); i++ {
        if prefixSums[i] - prefixSums[i-m] == d {
            countOfSegments++
        }
    }
    return countOfSegments
}