Count of subarrays of size K which is a permutation of numbers from 1 to K
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 5
So answer is 3
Below is the implementation of the above approach:
C++
#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;
for ( int i = 0; i < n; i++) {
index_of[arr[i]] = i + 1;
}
for ( int i = 1; i <= n; i++) {
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;
}
int 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
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<>();
for ( int i = 0 ; i < n; i++)
{
index_of.put(arr.get(i), i + 1 );
}
for ( int i = 1 ; i <= n; i++)
{
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;
}
public 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));
}
}
|
Python3
def find_permutations(arr):
cnt = 0
max_ind = - 1
min_ind = 10000000 ;
n = len (arr)
index_of = {}
for i in range (n):
index_of[arr[i]] = i + 1
for i in range ( 1 , n + 1 ):
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
if __name__ = = "__main__" :
nums = []
nums.append( 2 )
nums.append( 3 )
nums.append( 1 )
nums.append( 5 )
nums.append( 4 )
print (find_permutations(nums))
|
C#
using 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 >();
for ( int i = 0; i < n; i++)
{
index_of[( int )arr[i]] = i + 1;
}
for ( int i = 1; i <= n; i++)
{
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;
}
public 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));
}
}
|
Javascript
<script>
function find_permutations(arr)
{
var cnt = 0;
var max_ind = -1, min_ind = 10000000;
var n = arr.length;
var index_of = new Map();
for ( var i = 0; i < n; i++) {
index_of.set(arr[i], i + 1);
}
for ( var i = 1; i <= n; i++) {
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));
</script>
|
Last Updated :
31 May, 2021
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