Given an array arr of distinct integers, the task is to find the count of sub-arrays of size i having all elements from 1 to i, in other words, the sub-array is any permutation of elements from 1 to i, with 1 < = i <= N.
Examples:
Input: arr[] = {2, 3, 1, 5, 4}
Output: 3
Explanation:
we have {1}, {2, 3, 1} and {2, 3, 1, 5, 4} subarray for i=1, i=3, i=5 respectively.
Permutation of size 4 and size 2 can’t be made because 5 and 3 are in the way respectively.Input: arr[] = {1, 3, 5, 4, 2}
Output: 2
Explanation:
we have {1} and {1, 3, 5, 4, 2} subarray for i=1 and i=5 respectively.
A Naive approach is to start from each index and try to find the subarray of every size(i) and check whether all elements from 1 to i are present.
Time complexity: O(N2)
An Efficient approach can be given by checking if it is possible to create a subarray of size i for every value of i from 1 to N.
As we know, every subarray of size K must be a permutation of all elements from 1 to K, knowing that we can look at the index of the numbers from 1 to N in order and calculate the index of the minimum and maximum values at every step.
- If maximum_ind – minimum_ind + 1 = K, then we have a permutation of size K, else not.
- Update the value of minimum_ind and maximum_ind at every step.
Time complexity: O(n)
Illustration:
Given Arr = {2, 3, 1, 5, 4}, let’s start with min_ind = INF and max_ind = -1
- index of 1 is 2, so min_ind = min(min_ind, 2) = 2 and max_ind = max(max_ind, 2) = 2,
2-2+1 = 1 so we have a permutation of size 1- index of 2 is 0, so min_ind = min(min_ind, 0) = 0 and max_ind = max(max_ind, 0) = 2,
2-0+1 = 3 so we don’t have a permutation of size 2- index of 3 is 1, so min_ind = min(min_ind, 1) = 0 and max_ind = max(max_ind, 1) = 2,
2-0+1 = 3 so we have a permutation of size 3- index of 4 is 4, so min_ind = min(min_ind, 4) = 0 and max_ind = max(max_ind, 4) = 4,
4-0+1 = 5 so we don’t have a permutation of size 4- index of 5 is 3, so min_ind = min(min_ind, 3) = 0 and max_ind = max(max_ind, 4) = 4,
4-0+1 = 5 so we have a permutation of size 5So answer is 3
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <iostream>#include <unordered_map>#include <vector>using namespace std;int find_permutations(vector<int>& arr){ int cnt = 0; int max_ind = -1, min_ind = 10000000; int n = arr.size(); unordered_map<int, int> index_of; // Save index of numbers of the array for (int i = 0; i < n; i++) { index_of[arr[i]] = i + 1; } for (int i = 1; i <= n; i++) { // Update min and max index // with the current index // and check if it's a valid permutation max_ind = max(max_ind, index_of[i]); min_ind = min(min_ind, index_of[i]); if (max_ind - min_ind + 1 == i) cnt++; } return cnt;}// Driver codeint main(){ vector<int> nums; nums.push_back(2); nums.push_back(3); nums.push_back(1); nums.push_back(5); nums.push_back(4); cout << find_permutations(nums); return 0;} |
Java
// Java program to implement // the above approach import java.util.*;class GFG{ public static int find_permutations( Vector<Integer> arr) { int cnt = 0; int max_ind = -1, min_ind = 10000000; int n = arr.size(); HashMap<Integer, Integer> index_of = new HashMap<>(); // Save index of numbers of the array for(int i = 0; i < n; i++) { index_of.put(arr.get(i), i + 1); } for(int i = 1; i <= n; i++) { // Update min and max index with // the current index and check // if it's a valid permutation max_ind = Math.max(max_ind, index_of.get(i)); min_ind = Math.min(min_ind, index_of.get(i)); if (max_ind - min_ind + 1 == i) cnt++; } return cnt; } // Driver Codepublic static void main(String[] args){ Vector<Integer> nums = new Vector<Integer>(); nums.add(2); nums.add(3); nums.add(1); nums.add(5); nums.add(4); System.out.print(find_permutations(nums));}}// This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program to implement# the above approachdef find_permutations(arr): cnt = 0 max_ind = -1 min_ind = 10000000; n = len(arr) index_of = {} # Save index of numbers of the array for i in range(n): index_of[arr[i]] = i + 1 for i in range(1, n + 1): # Update min and max index with the # current index and check if it's a # valid permutation max_ind = max(max_ind, index_of[i]) min_ind = min(min_ind, index_of[i]) if (max_ind - min_ind + 1 == i): cnt += 1 return cnt# Driver codeif __name__ == "__main__": nums = [] nums.append(2) nums.append(3) nums.append(1) nums.append(5) nums.append(4) print(find_permutations(nums))# This code is contributed by chitranayal |
C#
// C# program to implement// the above approachusing System; using System.Collections; using System.Collections.Generic; class GFG{ static int find_permutations(ArrayList arr){ int cnt = 0; int max_ind = -1, min_ind = 10000000; int n = arr.Count; Dictionary<int, int> index_of = new Dictionary<int, int>(); // Save index of numbers of the array for(int i = 0; i < n; i++) { index_of[(int)arr[i]] = i + 1; } for(int i = 1; i <= n; i++) { // Update min and max index with // the current index and check // if it's a valid permutation max_ind = Math.Max(max_ind, index_of[i]); min_ind = Math.Min(min_ind, index_of[i]); if (max_ind - min_ind + 1 == i) cnt++; } return cnt;}// Driver Codepublic static void Main(string[] args){ ArrayList nums = new ArrayList(); nums.Add(2); nums.Add(3); nums.Add(1); nums.Add(5); nums.Add(4); Console.Write(find_permutations(nums));}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript implementationfunction find_permutations(arr){ var cnt = 0; var max_ind = -1, min_ind = 10000000; var n = arr.length; var index_of = new Map(); // Save index of numbers of the array for (var i = 0; i < n; i++) { index_of.set(arr[i], i + 1); } for (var i = 1; i <= n; i++) { // Update min and max index // with the current index // and check if it's a valid permutation max_ind = Math.max(max_ind, index_of.get(i)); min_ind = Math.min(min_ind, index_of.get(i)); if (max_ind - min_ind + 1 == i) cnt++; } return cnt;} var nums = [];nums.push(2);nums.push(3);nums.push(1);nums.push(5);nums.push(4);document.write(find_permutations(nums));// This code contributed by shubhamsingh10</script> |
Output:
3
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