Given two integers N and K, the task is to generate a permutation of N numbers (Every number from 1 to N occurs exactly once) such that the number of indices where a[i]>a[i+1] is exactly K. Print “Not possible” if no such permutation is possible.
Examples:
Input: N = 5, K = 3 Output: 5 4 3 1 2 Starting 3 indices satisfying the condition a[i] > a[i]+1 Input: N = 7, k = 4 Output: 7 6 5 4 1 2 3
Approach: Since the permutation should be lexicographically smallest with the condition satisfied for k indices i.e. a[i] > a[i+1]. So for that starting K+1 digits should be in decreasing order and the remaining should be in increasing order. So just print the K numbers from N to 1. Then print numbers from 1 to N-K.
For example: N = 6, K = 4
Print K numbers from N to 1 i.e. 6, 5, 4, 3
Print N-K numbers from 1 to N-K i.e. 1, 2
Permutation will be 654312 i.e. Starting 4 indices satisfy a[i] > a[i+1] for i = 1 to k.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
void printPermutation( int n, int k)
{ int i, mx = n;
for (i = 1; i <= k; i++) // Decreasing part
{
cout << mx << " " ;
mx--;
}
for (i = 1; i <= mx; i++) // Increasing part
cout << i << " " ;
} // Driver Code int main()
{ int N = 5, K = 3;
if (K >= N - 1)
cout << "Not Possible" ;
else
printPermutation(N, K);
return 0;
} |
// Java implementation of the above approach import java.io.*;
class GFG {
static void printPermutation( int n, int k)
{ int i, mx = n;
for (i = 1 ; i <= k; i++) // Decreasing part
{
System.out.print( mx + " " );
mx--;
}
for (i = 1 ; i <= mx; i++) // Increasing part
System.out.print( i + " " );
} // Driver Code public static void main (String[] args) {
int N = 5 , K = 3 ;
if (K >= N - 1 )
System.out.print( "Not Possible" );
else
printPermutation(N, K);
}
} // This code is contributed by inder_verma.. |
# Python3 implementation of the # above approach def printPermutation(n, k):
mx = n
for i in range ( 1 , k + 1 ): # Decreasing part
print (mx, end = " " )
mx - = 1
for i in range ( 1 , mx + 1 ): # Increasing part
print (i, end = " " )
# Driver Code if __name__ = = "__main__" :
N, K = 5 , 3
if K > = N - 1 :
print ( "Not Possible" )
else :
printPermutation(N, K)
# This code is contributed # by Rituraj Jain |
// C# implementation of the above approach using System;
class GFG {
static void printPermutation( int n, int k)
{ int i, mx = n;
for (i = 1; i <= k; i++) // Decreasing part
{
Console.Write( mx + " " );
mx--;
}
for (i = 1; i <= mx; i++) // Increasing part
Console.Write( i + " " );
} // Driver Code public static void Main () {
int N = 5, K = 3;
if (K >= N - 1)
Console.WriteLine( "Not Possible" );
else
printPermutation(N, K);
}
} // This code is contributed by inder_verma.. |
<?php // PHP implementation of the above approach function printPermutation( $n , $k )
{ $i ; $mx = $n ;
for ( $i = 1; $i <= $k ; $i ++) // Decreasing part
{
echo $mx , " " ;
$mx --;
}
for ( $i = 1; $i <= $mx ; $i ++) // Increasing part
echo $i , " " ;
} // Driver Code $N = 5; $K = 3;
if ( $K >= $N - 1)
echo "Not Possible" ;
else
printPermutation( $N , $K );
// This code is contributed by inder_verma.. ?> |
<script> // javascript implementation of the above approach function printPermutation(n , k)
{ var i, mx = n;
for (i = 1; i <= k; i++) // Decreasing part
{
document.write( mx + " " );
mx--;
}
for (i = 1; i <= mx; i++) // Increasing part
document.write( i + " " );
} // Driver Code var N = 5, K = 3;
if (K >= N - 1)
document.write( "Not Possible" );
else printPermutation(N, K);
// This code is contributed by 29AjayKumar </script> |
5 4 3 1 2
Time Complexity: O(N)