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

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 less than or equal to K.

Examples:

Input: arr[] = {1, 2, 3}, K = 2
Output: 5
{1}, {2}, {3}, {1, 2} and {2, 3} are the valid subarrays.



Input: arr[] = {3, 2, 1}, K = 1
Output: 3

Naive approach: A simple approach is to generate all the sub-arrays of length less 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) and (L * (L + 1)) / 2 – (X * (X + 1)) / 2 will be added to the final answer. This is because for an array of length L, the number of sub-arrays with length ≥ K.
    • Number of such sub-arrays starting from the first element = L – K = X.
    • Number of such sub-arrays starting from the second element = L – K – 1 = X – 1.
    • Number of such sub-arrays starting from the third element = L – K – 2 = X – 2.
    • And so on until 0 i.e. 1 + 2 + 3 + .. + X = (X * (X + 1)) / 2. If this value is subtracted from the total increasing subarrays then the result will be the count of increasing subarrays of length less than or equal to K

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the required count
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 and arr[j] >= arr[j - 1])
            j++;
        int x = max(0, j - i - k);
  
        // Update ret
        ret += ((j - i) * (j - i + 1)) / 2 - (x * (x + 1)) / 2;
  
        // Update i
        i = j;
    }
  
    // Return ret
    return ret;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(int);
    int k = 2;
  
    cout << findCnt(arr, n, k);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
  
// Function to return the required count
static 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);
  
        // Update ret
        ret += ((j - i) * (j - i + 1)) / 2
                          (x * (x + 1)) / 2;
  
        // Update i
        i = j;
    }
  
    // Return ret
    return ret;
}
  
// Driver code
public static void main(String []args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
    int k = 2;
  
    System.out.println(findCnt(arr, n, k));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach 
  
# Function to return the required count 
def 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 += ((j - i) * (j - i + 1)) // 2 - \
                          (x * (x + 1)) / 2
  
        # Update i 
        i = j; 
  
    # Return ret 
    return ret; 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 2, 3 ]; 
    n = len(arr); 
    k = 2
  
    print(findCnt(arr, n, k)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
                      
class GFG
{
  
// Function to return the required count
static 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);
  
        // Update ret
        ret += ((j - i) * (j - i + 1)) / 2 - 
                        (x * (x + 1)) / 2;
  
        // Update i
        i = j;
    }
  
    // Return ret
    return ret;
}
  
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
    int k = 2;
  
    Console.WriteLine(findCnt(arr, n, k));
}
}
  
// This code is contributed by Rajput-Ji

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Output:

5


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