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++ 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 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 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# 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 |
<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> |
3
Time Complexity: O(n)
Auxiliary Space: O(1)