# Count the number of non-increasing subarrays

• Difficulty Level : Basic
• Last Updated : 07 Jun, 2021

Given an array of N integers. The task is to count the number of subarrays (of size at least one) that are non-increasing.

Examples:

```Input : arr[] = {1, 4, 3}
Output : 4
The possible subarrays are {1}, {4}, {3}, {4, 3}.

Input :{4, 3, 2, 1}
Output : 10
The possible subarrays are:
{4}, {3}, {2}, {1}, {4, 3}, {3, 2}, {2, 1},
{4, 3, 2}, {3, 2, 1}, {4, 3, 2, 1}.```

Simple Solution: A simple solution is to generate all possible subarrays, and for every subarray check if the subarray is non increasing or not. The time complexity of this solution would be O(N3).

Efficient Solution: A Better Solution is to use the fact that if a subarray arr[i:j] is not non increasing, then subarrays arr[i:j+1], arr[i:j+2], .. arr[i:n-1] cannot be non increasing. So, start traversing the array and for current subarray keep incrementing it’s length until it is non-increasing and update the count. Once the subarray starts increasing reset the length.

Below is the implementation of the above idea:

## C++

 `// C++ program to count number of non``// increasing subarrays``#include ``using` `namespace` `std;` `int` `countNonIncreasing(``int` `arr[], ``int` `n)``{``    ``// Initialize result``    ``int` `cnt = 0;` `    ``// Initialize length of current non``    ``// increasing subarray``    ``int` `len = 1;` `    ``// Traverse through the array``    ``for` `(``int` `i = 0; i < n - 1; ++i) {` `        ``// If arr[i+1] is less than or equal to arr[i],``        ``// then increment length``        ``if` `(arr[i + 1] <= arr[i])``            ``len++;` `        ``// Else Update count and reset length``        ``else` `{``            ``cnt += (((len + 1) * len) / 2);``            ``len = 1;``        ``}``    ``}` `    ``// If last length is more than 1``    ``if` `(len > 1)``        ``cnt += (((len + 1) * len) / 2);` `    ``return` `cnt;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 2, 3, 7, 1, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << countNonIncreasing(arr, n);` `    ``return` `0;``}`

## Java

 `// Java program to count number of non``// increasing subarrays``class` `GFG``{``    ` `static` `int` `countNonIncreasing(``int` `arr[], ``int` `n)``{``    ``// Initialize result``    ``int` `cnt = ``0``;` `    ``// Initialize length of current non``    ``// increasing subarray``    ``int` `len = ``1``;` `    ``// Traverse through the array``    ``for` `(``int` `i = ``0``; i < n - ``1``; ++i)``    ``{` `        ``// If arr[i+1] is less than or equal to arr[i],``        ``// then increment length``        ``if` `(arr[i + ``1``] <= arr[i])``            ``len++;` `        ``// Else Update count and reset length``        ``else``        ``{``            ``cnt += (((len + ``1``) * len) / ``2``);``            ``len = ``1``;``        ``}``    ``}` `    ``// If last length is more than 1``    ``if` `(len > ``1``)``        ``cnt += (((len + ``1``) * len) / ``2``);` `    ``return` `cnt;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``5``, ``2``, ``3``, ``7``, ``1``, ``1` `};``    ``int` `n =arr.length;` `    ``System.out.println(countNonIncreasing(arr, n));``}``}` `// This code is contributed by Code_Mech`

## Python3

 `# Python3 program to count number of non``# increasing subarrays``def` `countNonIncreasing(arr, n):``    ` `    ``# Initialize result``    ``cnt ``=` `0``;` `    ``# Initialize length of current``    ``# non-increasing subarray``    ``len` `=` `1``;` `    ``# Traverse through the array``    ``for` `i ``in` `range``(``0``, n ``-` `1``):` `        ``# If arr[i+1] is less than``        ``# or equal to arr[i],``        ``# then increment length``        ``if` `(arr[i ``+` `1``] <``=` `arr[i]):``            ``len` `+``=` `1``;` `        ``# Else Update count and reset length``        ``else``:``            ``cnt ``+``=` `(((``len` `+` `1``) ``*` `len``) ``/` `2``);``            ``len` `=` `1``;``            ` `    ``# If last length is more than 1``    ``if` `(``len` `> ``1``):``        ``cnt ``+``=` `(((``len` `+` `1``) ``*` `len``) ``/` `2``);` `    ``return` `int``(cnt);` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``5``, ``2``, ``3``, ``7``, ``1``, ``1``];``    ``n ``=` `len``(arr);` `    ``print``(countNonIncreasing(arr, n));` `# This code contributed by PrinciRaj1992`

## C#

 `// C# program to count number of non``// increasing subarrays``using` `System;``    ` `class` `GFG``{``    ` `static` `int` `countNonIncreasing(``int` `[]arr, ``int` `n)``{``    ``// Initialize result``    ``int` `cnt = 0;` `    ``// Initialize length of current non``    ``// increasing subarray``    ``int` `len = 1;` `    ``// Traverse through the array``    ``for` `(``int` `i = 0; i < n - 1; ++i)``    ``{` `        ``// If arr[i+1] is less than or equal to arr[i],``        ``// then increment length``        ``if` `(arr[i + 1] <= arr[i])``            ``len++;` `        ``// Else Update count and reset length``        ``else``        ``{``            ``cnt += (((len + 1) * len) / 2);``            ``len = 1;``        ``}``    ``}` `    ``// If last length is more than 1``    ``if` `(len > 1)``        ``cnt += (((len + 1) * len) / 2);` `    ``return` `cnt;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 5, 2, 3, 7, 1, 1 };``    ``int` `n = arr.Length;` `    ``Console.Write(countNonIncreasing(arr, n));``}``}` `// This code has been contributed by 29AjayKumar`

## PHP

 ` 1)``        ``\$cnt` `+= ((``\$len` `+ 1) * ``\$len``) / 2;` `    ``return` `\$cnt``;``}` `// Driver code``\$arr` `= ``array``( 5, 2, 3, 7, 1, 1 );``\$n` `= sizeof(``\$arr``);` `echo` `countNonIncreasing(``\$arr``, ``\$n``);` `// This code is contributed by akt_mit``?>`

## Javascript

 ``
Output
`10`

Time Complexity: O(N)
Auxiliary Space: O(1)

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