Kefa and First Steps - CodeForces | C++ Implementation

Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1≤i≤n) he makes ai money. Kefa loves progress, that’s why he wants to know the length of the maximum non-decreasing subsegment in sequence ai. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.

The second line contains n integers a1, a2, …, an

Print a single integer — the length of the maximum non-decreasing subsegment of sequence a.

Read full problem here : Kefa and First Steps

Let earnings is an integer vector of size n which stores amount earned on each day.

To find the length of the maximum increasing subsegment in a given sequence, you can use the following algorithm:

1. Initialize three integer variables: current_earning, final_count, and local_count. current_earning should be set to the first element of the earnings vector, final_count should be set to 1, and local_count should be set to 1.

   int current_earning = earnings[0]; // stores highest earning of non-decreasing subsegment
   int final_count = 1; // stores length of longest non-decreasing subsegment
   int local_count = 1; // stores length of current subsegment in process
   

   ``

2. Iterate over the earnings vector, starting at the second element (index 1). For each element earnings[i], do the following:

   a. If earnings[i] > current_earning, increment local_count by 1 and set current_earning to earnings[i].

   b. If earnings[i] <= current_earning, set local_count to 1 and set current_earning to earnings[i].

   c. If local_count > final_count, set final_count to local_count.

   for (int i = 1; i < n; i++) 
   {
   	if (earnings[i] > current_earning) 
   	{
     		// if current earning is greater than previous earning, increment local count by 1
     		local_count++;
     		current_earning = earnings[i];
   	} 
   	else 
   	{
     		// if current earning is not greater than previous earning, reset local count to 1
     		local_count = 1;
     		current_earning = earnings[i];
   	}
   
   	// update final count if local count is greater than final count
   	if (local_count > final_count) 
   	{
     		final_count = local_count;
   	}
   }
   

   ``

3. After the loop finishes, final_count will contain the length of the maximum increasing subsegment.

#include <iostream>
#include <vector>

int main()
{
	int n;
	std::cin >> n;

	std::vector<int> earnings(n);

	int current_earning = earnings[0]; // stores highest earning of non-decreasing subsegment
 	int final_count = 1; // stores length of longest non-decreasing subsegment
  	int local_count = 1; // stores length of current subsegment in process

	for (int i = 1; i < n; i++) 
	{
    	if (earnings[i] > current_earning) 
    	{
      		// if current earning is greater than previous earning, increment local count by 1
      		local_count++;
      		current_earning = earnings[i];
    	} 
    	else 
    	{
      		// if current earning is not greater than previous earning, reset local count to 1
      		local_count = 1;
      		current_earning = earnings[i];
    	}

    	// update final count if local count is greater than final count
    	if (local_count > final_count) 
    	{
      		final_count = local_count;
    	}
	}
	std::cout << final_count << "\n";
}

Check out this on Github

If you have a different solution for finding the length of the maximum increasing subsegment in a given sequence, please share it in the comments below.