# Number of non-decreasing sub-arrays of length greater than or equal to K

• Last Updated : 08 Mar, 2022

Given an array arr[] of N elements and an integer K, the task is to find the number of non-decreasing sub-arrays of length greater than or equal to K.
Examples:

Input: arr[] = {1, 2, 3}, K = 2
Output:
{1, 2}, {2, 3} and {1, 2, 3} are the valid subarrays.
Input: arr[] = {3, 2, 1}, K = 1
Output:

Naive approach: A simple approach is to generate all the sub-arrays of length greater than or equal to K and then check whether the sub-array satisfies the condition. Thus, the time complexity of the approach will be O(N3).
Efficient approach: A better approach will be using the two-pointer technique

• For any index i, find the largest index j such that the sub-array arr[i…j] is non-decreasing. This can be achieved by simply increasing the value of j, starting from i + 1 and checking whether arr[j] is greater than arr[j – 1].
• Lets say that the length of the sub-array found in the previous step is L. Calculate X = max(0, L – K + 1) and (X * (X + 1)) / 2 will be added to final answer. This is because for an array of length L, the number of sub-arrays with length â‰¥ K
• Number of such sub-arrays starting from the first element = L – K + 1 = X.
• Number of such sub-arrays starting from the second element = L – K = X – 1.
• Number of such sub-arrays starting from the third element = L – K – 1 = X – 2.
• And so on until 0 i.e. 1 + 2 + 3 + .. + X = (X * (X + 1)) / 2.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function to return the required countint findCnt(int* arr, int n, int k){    // To store the final result    int ret = 0;     // Two pointer loop    int i = 0;    while (i < n) {         // Initialising j        int j = i + 1;         // Looping till the subarray increases        while (j < n and arr[j] >= arr[j - 1])            j++;        int x = max(0, j - i - k + 1);         // Update ret        ret += (x * (x + 1)) / 2;         // Update i        i = j;    }     // Return ret    return ret;} // Driver codeint main(){    int arr[] = { 5, 4, 3, 2, 1 };    int n = sizeof(arr) / sizeof(int);    int k = 2;     cout << findCnt(arr, n, k);     return 0;}

## Java

 // Java implementation of the approachclass GFG{ // Function to return the required countstatic int findCnt(int []arr, int n, int k){    // To store the final result    int ret = 0;     // Two pointer loop    int i = 0;    while (i < n)    {         // Initialising j        int j = i + 1;         // Looping till the subarray increases        while (j < n && arr[j] >= arr[j - 1])            j++;        int x = Math.max(0, j - i - k + 1);         // Update ret        ret += (x * (x + 1)) / 2;         // Update i        i = j;    }     // Return ret    return ret;} // Driver codepublic static void main(String []args){    int arr[] = { 5, 4, 3, 2, 1 };    int n = arr.length;    int k = 2;     System.out.println(findCnt(arr, n, k));}} // This code is contributed by Rajput-Ji

## Python3

 # Python3 implementation of the approach # Function to return the required countdef findCnt(arr, n, k) :     # To store the final result    ret = 0;     # Two pointer loop    i = 0;    while (i < n) :         # Initialising j        j = i + 1;         # Looping till the subarray increases        while (j < n and arr[j] >= arr[j - 1]) :            j += 1;                     x = max(0, j - i - k);         # Update ret        ret += (x * (x + 1)) / 2;         # Update i        i = j;     # Return ret    return ret; # Driver codeif __name__ == "__main__" :     arr = [ 5, 4, 3, 2, 1 ];    n = len(arr);    k = 2;     print(findCnt(arr, n, k)); # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approachusing System; class GFG{ // Function to return the required countstatic int findCnt(int []arr, int n, int k){    // To store the final result    int ret = 0;     // Two pointer loop    int i = 0;    while (i < n)    {         // Initialising j        int j = i + 1;         // Looping till the subarray increases        while (j < n && arr[j] >= arr[j - 1])            j++;        int x = Math.Max(0, j - i - k + 1);         // Update ret        ret += (x * (x + 1)) / 2;         // Update i        i = j;    }     // Return ret    return ret;} // Driver codepublic static void Main(String []args){    int []arr = { 5, 4, 3, 2, 1 };    int n = arr.Length;    int k = 2;     Console.WriteLine(findCnt(arr, n, k));}} // This code is contributed by PrinciRaj1992

## Javascript


Output:
0

Time Complexity: O(n)

Auxiliary Space: O(1)

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