Count subarrays consisting of first K natural numbers in descending order
Given an array arr[] of size N and an integer K, the task is to count the number of subarrays which consists of first K natural numbers in descending order.
Examples:
Input: arr[] = {1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1}, K = 3
Output: 2
Explanation: The subarray {3, 2, 1} occurs twice in the array.Input: arr = {100, 7, 6, 5, 4, 3, 2, 1, 100}, K = 6
Output: 1
Naive Approach
The idea is to find all subarrays and then find those subarrays whose length is equal to k. After that from those subarrays find the number of subarrays that consist of the first K natural number in decreasing order.
Steps to implement-
- Initialize a variable ans with value 0 to store the final answer
- Run two for loops to find all subarray
- Simultaneously find the length of the subarray
- If any subarray has a length of K
- Then check whether it contains the first K natural numbers in descending order
- If Yes then increment the ans by 1
Code-
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count subarray having // the decreasing sequence K to 1 int CountSubarray( int arr[], int n, int k) { //To store answer int ans=0; //Find all subarray for ( int i=0;i<n;i++){ //To store length int length=0; for ( int j=i;j<n;j++){ //Increment the length length++; //when length is equal to k if (length==k){ int count=k; int m=i; while (m<=j){ if (arr[m]==count){count--;} else { break ;} m++; } //when subarray consist of first //K natural numbers in descending order if (count==0){ans++;} } } } return ans; } // Driver Code int main() { int arr[] = { 1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1 }; int N = sizeof (arr) / sizeof (arr[0]); int K = 3; // Function Call cout << CountSubarray(arr, N, K); return 0; } |
Java
// Java program for the above approach import java.util.*; public class GFG { // Function to count subarrays having the decreasing sequence K to 1 static int countSubarray( int [] arr, int n, int k) { // To store the answer int ans = 0 ; // Find all subarrays for ( int i = 0 ; i < n; i++) { // To store the length int length = 0 ; for ( int j = i; j < n; j++) { // Increment the length length++; // When length is equal to k if (length == k) { int count = k; int m = i; while (m <= j) { if (arr[m] == count) { count--; } else { break ; } m++; } // When subarray consists of the first // K natural numbers in descending order if (count == 0 ) { ans++; } } } } return ans; } // Driver Code public static void main(String[] args) { int [] arr = { 1 , 2 , 3 , 7 , 9 , 3 , 2 , 1 , 8 , 3 , 2 , 1 }; int N = arr.length; int K = 3 ; // Function Call System.out.println(countSubarray(arr, N, K)); } } |
Python3
# Python3 program for the above approach # Function to count subarrays having the decreasing sequence K to 1 def count_subarray(arr, n, k): # To store the answer ans = 0 # Find all subarrays for i in range (n): # To store length length = 0 for j in range (i, n): # Increment the length length + = 1 # When length is equal to k if length = = k: count = k m = i while m < = j: if arr[m] = = count: count - = 1 else : break m + = 1 # When subarray consists of the first K natural numbers # in descending order if count = = 0 : ans + = 1 return ans # Driver Code if __name__ = = "__main__" : arr = [ 1 , 2 , 3 , 7 , 9 , 3 , 2 , 1 , 8 , 3 , 2 , 1 ] N = len (arr) K = 3 # Function Call print (count_subarray(arr, N, K)) |
C#
// C# program for the above approach using System; class GFG { // Function to count subarray having // the decreasing sequence K to 1 static int CountSubarray( int [] arr, int n, int k) { // To store answer int ans = 0; // Find all subarray for ( int i = 0; i < n; i++) { // To store length int length = 0; for ( int j = i; j < n; j++) { // Increment the length length++; // when length is equal to k if (length == k) { int count = k; int m = i; while (m <= j) { if (arr[m] == count) { count--; } else { break ; } m++; } // when subarray consist of first // K natural numbers in descending order if (count == 0) { ans++; } } } } return ans; } // Driver code static public void Main() { int [] arr = { 1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1 }; int N = arr.Length; int K = 3; // Function Call Console.Write(CountSubarray(arr, N, K)); } } |
Javascript
// Javascript program for the above approach // Function to count subarrays having a decreasing sequence of length K to 1 function CountSubarray(arr, n, k) { let ans = 0; // To store the count of valid subarrays // Loop through the array to find subarrays for (let i = 0; i < n; i++) { let length = 0; // To store the current subarray length for (let j = i; j < n; j++) { length++; // Increment the subarray length // Check if the subarray length is equal to K if (length === k) { let count = k; // Counter to track the decreasing sequence from K to 1 let m = i; // Index to traverse the subarray // Check if the subarray follows the decreasing sequence while (m <= j) { if (arr[m] === count) { count--; } else { break ; } m++; } // If the counter reaches 0, it means the subarray // consists of the required sequence if (count === 0) { ans++; // Increment the count of valid subarrays } } } } return ans; // Return the total count of valid subarrays } // Driver Code const arr = [1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1]; const N = arr.length; // Get the length of the array const K = 3; // Define the required sequence length // Function Call and Output console.log(CountSubarray(arr, N, K)); |
2
Time Complexity: O(N3), because of two loops to find all subarray and a third loop to choose subarray which consists of first K natural numbers in descending order
Auxiliary Space: O(1), because no extra space has been used
Approach: The idea is to traverse the array and check if the required decreasing sequence is present starting from the current index or not. Follow the steps below to solve the problem:
- Initialize two variables, temp to K, that checks the pattern, and count with 0, to store the count of total subarray matched.
- Traverse the array arr[] using the variable i and do the following:
- If arr[i] is equal to temp and the value of temp is 1, then increment the count by 1 and update temp as K. Else decrement temp by 1.
- Otherwise, update temp as temp = K and if arr[i] is equal to K, decrement i by 1.
- After the above steps, print the value of count as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count subarray having // the decreasing sequence K to 1 int CountSubarray( int arr[], int n, int k) { int temp = k, count = 0; // Traverse the array for ( int i = 0; i < n; i++) { // Check if required sequence // is present or not if (arr[i] == temp) { if (temp == 1) { count++; temp = k; } else temp--; } // Reset temp to k else { temp = k; if (arr[i] == k) i--; } } // Return the count return count; } // Driver Code int main() { int arr[] = { 1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1 }; int N = sizeof (arr) / sizeof (arr[0]); int K = 3; // Function Call cout << CountSubarray(arr, N, K); return 0; } // This code is contributed by Dharanendra L V |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to count subarray having // the decreasing sequence K to 1 static int CountSubarray( int arr[], int n, int k) { int temp = k, count = 0 ; // Traverse the array for ( int i = 0 ; i < n; i++) { // Check if required sequence // is present or not if (arr[i] == temp) { if (temp == 1 ) { count++; temp = k; } else temp--; } // Reset temp to k else { temp = k; if (arr[i] == k) i--; } } // Return the count return count; } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 3 , 7 , 9 , 3 , 2 , 1 , 8 , 3 , 2 , 1 }; int N = arr.length; int K = 3 ; // Function Call System.out.println(CountSubarray(arr, N, K)); } } // This code is contributed by shivanisinghss2110 |
Python3
# Python3 program for the above approach # Function to count subarray having # the decreasing sequence K to 1 def CountSubarray(arr, n, k): temp = k count = 0 # Traverse the array for i in range (n): # Check if required sequence # is present or not if (arr[i] = = temp): if (temp = = 1 ): count + = 1 temp = k else : temp - = 1 # Reset temp to k else : temp = k if (arr[i] = = k): i - = 1 # Return the count return count # Driver Code if __name__ = = "__main__" : arr = [ 1 , 2 , 3 , 7 , 9 , 3 , 2 , 1 , 8 , 3 , 2 , 1 ] N = len (arr) K = 3 # Function Call print (CountSubarray(arr, N, K)) # This code is contributed by chitranayal |
C#
// C# program for the above approach using System; class GFG{ // Function to count subarray having // the decreasing sequence K to 1 static int CountSubarray( int [] arr, int n, int k) { int temp = k, count = 0; // Traverse the array for ( int i = 0; i < n; i++) { // Check if required sequence // is present or not if (arr[i] == temp) { if (temp == 1) { count++; temp = k; } else temp--; } // Reset temp to k else { temp = k; if (arr[i] == k) i--; } } // Return the count return count; } // Driver code static public void Main() { int [] arr = { 1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1 }; int N = arr.Length; int K = 3; // Function Call Console.Write(CountSubarray(arr, N, K)); } } // This code is contributed by Dharanendra L V |
Javascript
<script> // JavaScript program for the above approach // Function to count subarray having // the decreasing sequence K to 1 function CountSubarray(arr, n, k) { var temp = k, count = 0; // Traverse the array for ( var i = 0; i < n; i++) { // Check if required sequence // is present or not if (arr[i] == temp) { if (temp == 1) { count++; temp = k; } else temp--; } // Reset temp to k else { temp = k; if (arr[i] == k) i--; } } // Return the count return count; } // Driver Code var arr = [1, 2, 3, 7, 9, 3, 2, 1, 8, 3, 2, 1]; var N = arr.length; var K = 3; // Function Call document.write(CountSubarray(arr, N, K)); </script> |
2
Time Complexity: O(N)
Auxiliary Space: O(1)