# Count the number of strictly increasing subarrays in an array

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Vinay Khatri
Last updated on September 20, 2022

## Problem

Given an array, count the total number of strictly increasing subarrays in it.

#### Sample Input

`[1, 2, 3, 1]`

`3`

#### Explanation

Let’s say the endpoints of a strictly increasing subarray are ``` start ``` and ``` end ``` . Then, the subarray ``` arr[start, end+1] ``` will be strictly increasing if the element at ``` end+1 ``` is greater than the element at ``` end ``` . The same thing goes for elements ``` end+1, end+2… ``` and so on.

### C++ Programming

```#include <iostream>
using namespace std;

int solve(int arr[], int n)
{

int ans = 0;

for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
if (arr[j - 1] >= arr[j])
{
break;
}
ans++;
}
}

return ans;
}

int main()
{
int arr[] = { 1, 2, 3, 2 };
int n = sizeof(arr) / sizeof(arr[0]);

cout << solve(arr, n);

return 0;
}```

`3`

### C Programming

```#include <stdio.h>

int solve(int arr[], int n)
{

int ans = 0;

for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{

if (arr[j - 1] >= arr[j])
{

break;
}
ans++;
}
}

return ans;
}

int main()
{
int arr[] = { 1, 2, 3, 2 };
int n = sizeof(arr) / sizeof(arr[0]);

printf("%d",solve(arr, n));

return 0;
}```

`3`

### Python Programming

```def solve(arr):

ans = 0

for i in range(len(arr)):
for j in range(i + 1, len(arr)):

if arr[j - 1] >= arr[j]:

break

ans = ans + 1

return ans

arr = [1, 2, 3, 2]

print(solve(arr))```

```3
```