# Count the number of non-increasing subarrays

• Difficulty Level : Basic
• Last Updated : 09 Jun, 2022

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), since there runs a loop from 0 to (n – 2).
Auxiliary Space: O(1), since no extra space has been taken.

My Personal Notes arrow_drop_up
Recommended Articles
Page :