Consecutive numbers with sum equal to n

Given a positive integer n, print the consecutive numbers whose sum is equal to n. For example,

    n = 20
    Output: 2, 3, 4, 5, 6 (because 2+3+4+5+6+7 = 20)
    n = 29
    Output: 14, 15

Solution:
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The brute force solution to this problem is to use two nested loops and use all possible start and end points between 1 to n (both inclusive)

void findNumbersBruteForce(int n)
{
  for(int start = 1; start<=n; start++)
  {
    for(int end = start; end<=n; end++)
    {
      int sum = 0;
      for(int i=start; i<=end; i++)
        sum += i;
      if(sum == n)
      {
        // PRINT NUMBERS FROM start TO end
        for(int i=start; i<=end; i++)
          cout<<i<<" ";
        return;
      }
    }
  }
}

This function takes O(n²) time. A better solution is to use a sliding window and adjust the start and end as per the below logic:

start = 1
end = 1
sum = 1; // SUM OF ELEMENTS FROM start TO end (both including)
while(true)
{
    IF (sum == n)
       PRINT NUMBERS FROM start TO end AND RETURN
    ELSE IF (sum < n)
       end++;            // MOVE end FORWARD
       sum += end;
    ELSE
       sum -= start;    // MOVE start FORWARD
       start++;
}

The while loop is not an infinite loop because the solution exist for all numbers (in the worst case, start=end=n is the solution).

void findNumbers(unsigned int n)
{
  int start = 1;
  int end = 1;
  int sum = 1;
  while(1)
  {
    if(sum == n)
    {
      // PRINT NUMBERS FROM start TO end
      while(start <= end)
      {
        cout<< start<<" ";
        start++;
      }
      return;
    }
    else if(sum<n)
    {
      end++;
      sum += end;
    }
    else
    {
      sum -= start;
      start++;
    }
  }
}

This code takes O(n) time because in each iteration we are moving either start or end forward. In the worst case, the loop may iterate 2*n times before both start and end becomes more than n itself.