Given a string
- If S[i] == “D”, then A[i] > A[i+1]
- If S[i] == “I”, then A[i] < A[i+1].
Note that output must contain distinct elements.
Examples:
Input: S = “DDI”
Output: [3, 2, 0, 1]Input: S = “IDID”
Output: [0, 4, 1, 3, 2]
Approach:
If S[0] == “I”, then choose
Below is the implementation of the above approach:
//C++ Implementation of above approach #include<bits/stdc++.h> using namespace std;
// function to find minimum required permutation
void StringMatch(string s)
{
int lo=0, hi = s.length(), len=s.length();
vector< int > ans;
for ( int x=0;x<len;x++)
{
if (s[x] == 'I' )
{
ans.push_back(lo) ;
lo += 1;
}
else
{
ans.push_back(hi) ;
hi -= 1;
}
}
ans.push_back(lo) ;
cout<< "[" ;
for ( int i=0;i<ans.size();i++)
{
cout<<ans[i];
if (i!=ans.size()-1)
cout<< "," ;
}
cout<< "]" ;
} // Driver code int main()
{ string S = "IDID" ;
StringMatch(S); return 0;
} //contributed by Arnab Kundu |
// Java Implementation of above approach import java.util.*;
class GFG
{ // function to find minimum required permutation static void StringMatch(String s)
{ int lo= 0 , hi = s.length(), len=s.length();
Vector<Integer> ans = new Vector<>();
for ( int x = 0 ; x < len; x++)
{
if (s.charAt(x) == 'I' )
{
ans.add(lo) ;
lo += 1 ;
}
else
{
ans.add(hi) ;
hi -= 1 ;
}
}
ans.add(lo) ;
System.out.print( "[" );
for ( int i = 0 ; i < ans.size(); i++)
{
System.out.print(ans.get(i));
if (i != ans.size()- 1 )
System.out.print( "," );
}
System.out.print( "]" );
} // Driver code public static void main(String[] args)
{ String S = "IDID" ;
StringMatch(S);
} } // This code is contributed by Rajput-Ji |
# Python Implementation of above approach # function to find minimum required permutation def StringMatch(S):
lo, hi = 0 , len (S)
ans = []
for x in S:
if x = = 'I' :
ans.append(lo)
lo + = 1
else :
ans.append(hi)
hi - = 1
return ans + [lo]
# Driver code S = "IDID"
print (StringMatch(S))
|
// C# Implementation of above approach using System;
using System.Collections.Generic;
class GFG
{ // function to find minimum required permutation static void StringMatch(String s)
{ int lo=0, hi = s.Length, len=s.Length;
List< int > ans = new List< int >();
for ( int x = 0; x < len; x++)
{
if (s[x] == 'I' )
{
ans.Add(lo) ;
lo += 1;
}
else
{
ans.Add(hi) ;
hi -= 1;
}
}
ans.Add(lo) ;
Console.Write( "[" );
for ( int i = 0; i < ans.Count; i++)
{
Console.Write(ans[i]);
if (i != ans.Count-1)
Console.Write( "," );
}
Console.Write( "]" );
} // Driver code public static void Main(String[] args)
{ String S = "IDID" ;
StringMatch(S);
} } // This code is contributed by 29AjayKumar |
<script> // Javascript implementation of above approach // function to find minimum required permutation function StringMatch(s)
{
var lo=0, hi = s.length, len=s.length;
var ans=[];
for ( var x=0;x<len;x++)
{
if (s[x] == 'I' )
{
ans.push(lo) ;
lo += 1;
}
else
{
ans.push(hi) ;
hi -= 1;
}
}
ans.push(lo) ;
document.write( "[" );
for ( var i=0;i<ans.length;i++)
{
document.write(ans[i]);
if (i!=ans.length -1)
document.write( ", " );
}
document.write( "]" );
} var S = "IDID" ;
StringMatch(S); // This code is contributed by SoumikMondal </script> |
Output
[0,4,1,3,2]
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(n), where n is the length of the given string.