Count of subarrays which forms a permutation from given Array elements
Given an array A[] consisting of integers [1, N], the task is to count the total number of subarrays of all possible lengths x (1 ? x ? N), consisting of a permutation of integers [1, x] from the given array.
Examples:
Input: A[] = {3, 1, 2, 5, 4} Output: 4
Explanation:
Subarrays forming a permutation are {1}, {1, 2}, {3, 1, 2} and {3, 1, 2, 5, 4}.
Input: A[] = {4, 5, 1, 3, 2, 6} Output: 4
Explanation:
Subarrays forming a permutation are {1}, {1, 3, 2}, {4, 5, 1, 3, 2} and {4, 5, 1, 3, 2, 6}.
Naive Approach:
Follow the steps below to solve the problem:
- The simplest approach to solve the problem is to generate all possible subarrays.
- For each subarray, check if it is a permutation of elements in the range [1, length of subarray].
- For every such subarray found, increase count. Finally, print the count.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach:
To optimize the above approach, follow the steps below:
- For every element from i = [1, N], check the maximum and minimum index, at which the elements of the permutation [1, i] are present.
- If the difference between the maximum and minimum index is equal to i, then it means there is a valid contiguous permutation for i.
- For every such permutation, increase the count. Finally, print the count.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function returns the required countint PermuteTheArray(int A[], int n){ int arr[n]; // Store the indices of the // elements present in A[]. for (int i = 0; i < n; i++) { arr[A[i] - 1] = i; } // Store the maximum and // minimum index of the // elements from 1 to i. int mini = n, maxi = 0; int count = 0; for (int i = 0; i < n; i++) { // Update maxi and mini, to // store minimum and maximum // index for permutation // of elements from 1 to i+1 mini = min(mini, arr[i]); maxi = max(maxi, arr[i]); // If difference between maxi // and mini is equal to i if (maxi - mini == i) // Increase count count++; } // Return final count return count;}// Driver Codeint main(){ int A[] = { 4, 5, 1, 3, 2, 6 }; cout << PermuteTheArray(A, 6); return 0;} |
Java
Python3
C#
Javascript
4
Time Complexity: O(N)
Auxiliary Space: O(N)
