Skip to content
Related Articles

Related Articles

Improve Article
Number of non-decreasing sub-arrays of length greater than or equal to K
  • Last Updated : 14 May, 2021

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 ≥ 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 <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 + 1);
 
        // Update ret
        ret += (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
int 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 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 + 1);
 
        // Update ret
        ret += (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
public 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 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 += (x * (x + 1)) / 2;
 
        # Update i
        i = j;
 
    # Return ret
    return ret;
 
# Driver code
if __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 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 + 1);
 
        // Update ret
        ret += (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
public 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




<script>
 
// Javascript implementation of the approach
 
// Function to return the required count
function findCnt(arr, n, k)
{
    // To store the final result
    var ret = 0;
 
    // Two pointer loop
    var i = 0;
    while (i < n) {
 
        // Initialising j
        var j = i + 1;
 
        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
        var x = Math.max(0, j - i - k + 1);
 
        // Update ret
        ret += (x * (x + 1)) / 2;
 
        // Update i
        i = j;
    }
 
    // Return ret
    return ret;
}
 
// Driver code
var arr = [5, 4, 3, 2, 1];
var n = arr.length;
var k = 2;
document.write( findCnt(arr, n, k));
 
 
</script>
Output: 
0

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :