Kefa and First Steps  CodeForces  C++ Implementation
Table of Content
 Problem:  {:.} Input  {:.} Output
 Solution:  {:.} C++ Implementation
Problem:
Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the ith day (1≤i≤n) he makes a_{i} money. Kefa loves progress, that’s why he wants to know the length of the maximum nondecreasing subsegment in sequence a_{i}. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called nondecreasing if all numbers in it follow in the nondecreasing order.
Input
The first line contains integer n.
The second line contains n integers a_{1}, a_{2}, …, a_{n}
Output
Print a single integer — the length of the maximum nondecreasing subsegment of sequence a.
Read full problem here : Kefa and First Steps
Solution:
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:

Initialize three integer variables:
current_earning
,final_count
, andlocal_count
.current_earning
should be set to the first element of theearnings
vector,final_count
should be set to 1, andlocal_count
should be set to 1.1 2 3
int current_earning = earnings[0]; // stores highest earning of nondecreasing subsegment int final_count = 1; // stores length of longest nondecreasing subsegment int local_count = 1; // stores length of current subsegment in process

Iterate over the
earnings
vector, starting at the second element (index 1). For each elementearnings[i]
, do the following:a. If
earnings[i] > current_earning
, incrementlocal_count
by 1 and setcurrent_earning
toearnings[i]
.b. If
earnings[i] <= current_earning
, setlocal_count
to 1 and setcurrent_earning
toearnings[i]
.c. If
local_count > final_count
, setfinal_count
tolocal_count
.1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
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; } }

After the loop finishes, final_count will contain the length of the maximum increasing subsegment.
C++ Implementation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
#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 nondecreasing subsegment
int final_count = 1; // stores length of longest nondecreasing 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.