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

Subarray Division

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3013 Discussions

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  • alban_tyrex
     + 1 comment

    Here is my O(N) c++ solution, you can watch the explanation here : https://youtu.be/L2fkDuGrxiI

    int birthday(vector<int> s, int d, int m) {
    if(s.size() < m) return 0;
    int su = accumulate(s.begin(), s.begin() + m, 0), result = 0;
    if(su == d) result++;
    for(int i = m; i < s.size(); i++){
        su += s[i];
        su -= s[i-m];
        if(su == d) result++;
    }
    return result;
 }

  • joao_martinho_m1
     + 0 comments

    In java:

    public static int birthday(List s, int d, int m) { // Write your code here

    //m --> array lenght
//d --> total sum of lenght
//s --> array

Integer numberOfPossibleCombinations = 0;
Integer chocolateBarSize = s.size();

for (int i = 0; i < chocolateBarSize ; i++){  

    List<Integer> sumList = s.stream().skip(i).limit(m).collect(Collectors.toList());    
    if(m > sumList.size()){
        break;
    }

    System.out.print("sumList " + sumList);

    Integer sum = sumList.stream().mapToInt(Integer::intValue).sum();
    System.out.println("sum " + sum);

    if(sum == d){
        numberOfPossibleCombinations++;
    }
}

return numberOfPossibleCombinations;
}

  • rabearimananaha1
     + 0 comments
    > def birthday(s, d, m):
    # Write your code here
    left = 0
    right = m
    contiguous_sum = 0
    for i in range (len(s)-m+1):
        current_sum = 0
        for j in range(m):
            current_sum += s[i+j]

    sliding wondow technique
    if current_sum==d: contiguous_sum += 1 return contiguous_sum

  • maxedward419
     + 0 comments

    How can i inegrate this code on living room wall art prints ? can any one guide me I am using kadence theme in wordpress.

  • vknakul43
     + 0 comments

    Sliding window technique O(n)

    JAVA

        public static int birthday(List<Integer> s, int d, int m) {
        int n = s.size();
        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 < n; i++) {
            sum += s.get(i);
            sum -= s.get(i - m);
            if (sum == d)
            count++;
        }
        
        return count;
    }