Number of non-decreasing sub-arrays of length K

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Given an array arr[] of length N, the task is to find the number of non-decreasing sub-arrays of length K.
Examples:

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

Naive approach Generate all the sub-arrays of length K and then check whether the sub-array satisfies the condition. Thus, the time complexity of the approach will be O(N * K).
Better approach: A better approach will be using two-pointer technique. Let’s say the current index is i.

Find the largest index j, such that the sub-array arr[i…j] is non-decreasing. This can be achieved by simply incrementing the value of j starting from i + 1 and checking whether arr[j] is greater than arr[j – 1].
Let’s say the length of the sub-array found in the previous step is L. The number of sub-arrays of length K contained in it will be max(L – K + 1, 0).
Now, update i = j and repeat the above steps while i is in the index range.

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 count of
// increasing subarrays of length k
int cntSubArrays(int* arr, int n, int k)
{
    // To store the final result
    int res = 0;
 
    int i = 0;
    // Two pointer loop
    while (i < n) {
 
        // Initialising j
        int j = i + 1;
 
        // Looping till the subarray increases
        while (j < n and arr[j] >= arr[j - 1])
            j++;
 
        // Updating the required count
        res += max(j - i - k + 1, 0);
 
        // Updating i
        i = j;
    }
 
    // Returning res
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 2, 5 };
    int n = sizeof(arr) / sizeof(int);
    int k = 2;
 
    cout << cntSubArrays(arr, n, k);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the count of
// increasing subarrays of length k
static int cntSubArrays(int []arr, int n, int k)
{
    // To store the final result
    int res = 0;
 
    int i = 0;
     
    // Two pointer loop
    while (i < n)
    {
 
        // Initialising j
        int j = i + 1;
 
        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
 
        // Updating the required count
        res += Math.max(j - i - k + 1, 0);
 
        // Updating i
        i = j;
    }
 
    // Returning res
    return res;
}
 
// Driver code
public static void main(String []args)
{
    int arr[] = { 1, 2, 3, 2, 5 };
    int n = arr.length;
    int k = 2;
 
    System.out.println(cntSubArrays(arr, n, k));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach 
 
# Function to return the count of 
# increasing subarrays of length k 
def cntSubArrays(arr, n, k) : 
 
    # To store the final result 
    res = 0; 
 
    i = 0; 
     
    # Two pointer loop 
    while (i < n) :
 
        # Initialising j 
        j = i + 1; 
 
        # Looping till the subarray increases 
        while (j < n and arr[j] >= arr[j - 1]) :
            j += 1; 
 
        # Updating the required count 
        res += max(j - i - k + 1, 0); 
 
        # Updating i 
        i = j; 
 
    # Returning res 
    return res; 
 
# Driver code 
if __name__ == "__main__" : 
     
    arr = [ 1, 2, 3, 2, 5 ]; 
    n = len(arr); 
    k = 2; 
 
    print(cntSubArrays(arr, n, k)); 
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
// Function to return the count of
// increasing subarrays of length k
static int cntSubArrays(int []arr, int n, int k)
{
    // To store the final result
    int res = 0;
 
    int i = 0;
     
    // Two pointer loop
    while (i < n)
    {
 
        // Initialising j
        int j = i + 1;
 
        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
 
        // Updating the required count
        res += Math.Max(j - i - k + 1, 0);
 
        // Updating i
        i = j;
    }
 
    // Returning res
    return res;
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 3, 2, 5 };
    int n = arr.Length;
    int k = 2;
 
    Console.WriteLine(cntSubArrays(arr, n, k));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to return the count of
// increasing subarrays of length k
function cntSubArrays(arr, n, k)
{
    // To store the final result
    var res = 0;
 
    var i = 0;
    // Two pointer loop
    while (i < n) {
 
        // Initialising j
        var j = i + 1;
 
        // Looping till the subarray increases
        while (j < n && arr[j] >= arr[j - 1])
            j++;
 
        // Updating the required count
        res += Math.max(j - i - k + 1, 0);
 
        // Updating i
        i = j;
    }
 
    // Returning res
    return res;
}
 
// Driver code
var arr = [ 1, 2, 3, 2, 5 ];
var n = arr.length;
var k = 2;
document.write( cntSubArrays(arr, n, k));
 
</script>

Output:

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

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