Discussion on Subarray Division Challenge

Sort by

recency

|

3047 Discussions

|

3 hours ago+ 0 comments

The "Subarray Division" problem typically involves finding the number of contiguous subarrays of a given array that sum up to a specific value. A common variation of this problem is where you are given an array of integers, a target sum, and a required subarray length. One effective approach to solving this problem is using Content Creation Optimization the sliding window technique if the subarray length is fixed.

1 day ago+ 0 comments

The Subarray Division algorithm is commonly used to find contiguous subarrays that satisfy specific conditions, such as a target sum or length. ?and how can we implement this in code exploring Cookout barbecue options ?

5 days ago+ 1 comment

In C#

public static int birthday(List<int> s, int d, int m)
    {
        int count = 0;
        if (m > s.Count) return 0;
        for (int i = 0; i <= s.Count - m; i++)
        {
            int segmentSum = 0;
            for (int j = 0; j < m; j++)
                segmentSum += s[i + j];
            if (segmentSum == d) count++;
        }
        return count;
    }

5 hours ago+ 0 comments

Subarray division is a common problem in computer science and competitive programming that involves finding contiguous subarrays within a given array that meet specific conditions. One common variation is determining the number of subarrays that sum up to a given target value. This problem is often encountered in algorithmic challenges and can be solved using techniques such as brute force, prefix sums, or hash maps to optimize performance of espaço invisível 2025. Efficient solutions, such as those using the sliding window or two-pointer approach, help in reducing time complexity, making the problem manageable for large datasets. Understanding subarray division is crucial for improving problem-solving skills in array manipulation and dynamic programming.

6 days ago+ 0 comments

Worst solution ever with typescript

function birthday(s, d, m) {
  var count = 0;
  var beggining = 0;
  var success = 0;
  for (var i = 0; i < s.length; i++) {
    count = 0;
    beggining = 0;
    if (s.length == 1 && s.length == m && s[i] == d) {
      success++;
      break;
    }
    for (var j = i + 1; j <= s.length; j++) {
      count++;
      beggining = beggining + s[j - 1];
      if (count == m && beggining == d) {
        success++;
        break;
      }
    }
  }
  return success;
}

3 weeks ago+ 0 comments

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;
 }

Sort by

|

3047 Discussions

|

Cookie support is required to access HackerRank

Please signup or login in order to view this challenge