Lexicographically smallest permutation of a string that contains all substrings of another string
Last Updated :
21 Jun, 2021
Given two strings A and B, the task is to find lexicographically the smallest permutation of string B such that it contains every substring from the string A as its substring. Print “-1” if no such valid arrangement is possible.
Examples:
Input: A = “aa”, B = “ababab”
Output: aaabbb
Explanation:
All possible substrings of A are (‘a’, ‘a’, ‘aa’)
Rearrange string B to “aaabb”.
Now “aaabb” is the lexicographically the smallest arrangement of B which contains all the substrings of A.
Input: A = “aaa”, B = “ramialsadaka”
Output: aaaaadiklmrs
Explanation:
All possible substrings of A are (‘a’, ‘aa’, ‘aaa’)
Rearrange string B to “aaaaadiklmrs”.
Now “aaaaadiklmrs” is the lexicographically smallest arrangement of B which contains all the substrings of A.
Naive Approach: The simplest approach is to generate all possible permutations of the string B and then from all these permutations, find lexicographically the smallest permutation which contains all substrings of A.
Time Complexity: O(N!)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the main observation is that the smallest string that contains all the substrings of A is the string A itself. Therefore, for string B to be reordered and contain all substring of A, it must contain A as a substring. The reordered string B can contain A as its substring only if the frequency of each character in string B is greater than or equal to its frequency in A. Below are the steps:
- Count the frequency of each character in string B in an array freq[] and then subtract from it the frequencies of the corresponding characters in string A.
- In order to form lexicographically the smallest string, initialize an empty string result and then append to it, all the leftover characters which are less in value than the first character of string A.
- Before appending all the characters equal to the first character A to result, check if there is any character which is less than the first character in string A. If so, then append A to result first and then all the remaining characters equal to the first character of A to make the reordered string lexicographically smallest.
- Otherwise, append all remaining occurrences of A[0] and then append A.
- At last, append the remaining characters to the result.
- After the above steps, print the string result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
string reorderString(string A, string B)
{
int size_a = A.length();
int size_b = B.length();
int freq[300] = { 0 };
for ( int i = 0; i < size_b; i++)
freq[B[i]]++;
for ( int i = 0; i < size_a; i++)
freq[A[i]]--;
for ( int j = 'a' ; j <= 'z' ; j++) {
if (freq[j] < 0)
return "-1" ;
}
string answer;
for ( int j = 'a' ; j < A[0]; j++)
while (freq[j] > 0) {
answer.push_back(j);
freq[j]--;
}
int first = A[0];
for ( int j = 0; j < size_a; j++) {
if (A[j] > A[0])
break ;
if (A[j] < A[0]) {
answer += A;
A.clear();
break ;
}
}
while (freq[first] > 0) {
answer.push_back(first);
--freq[first];
}
answer += A;
for ( int j = 'a' ; j <= 'z' ; j++)
while (freq[j]--)
answer.push_back(j);
return answer;
}
int main()
{
string A = "aa" ;
string B = "ababab" ;
cout << reorderString(A, B);
return 0;
}
|
Java
class GFG{
static String reorderString( char []A,
char []B)
{
int size_a = A.length;
int size_b = B.length;
int freq[] = new int [ 300 ];
for ( int i = 0 ; i < size_b; i++)
freq[B[i]]++;
for ( int i = 0 ; i < size_a; i++)
freq[A[i]]--;
for ( int j = 'a' ; j <= 'z' ; j++)
{
if (freq[j] < 0 )
return "-1" ;
}
String answer = "" ;
for ( int j = 'a' ; j < A[ 0 ]; j++)
while (freq[j] > 0 )
{
answer+=j;
freq[j]--;
}
int first = A[ 0 ];
for ( int j = 0 ; j < size_a; j++)
{
if (A[j] > A[ 0 ])
break ;
if (A[j] < A[ 0 ])
{
answer += String.valueOf(A);
A = new char [A.length];
break ;
}
}
while (freq[first] > 0 )
{
answer += String.valueOf(( char )first);
--freq[first];
}
answer += String.valueOf(A);
for ( int j = 'a' ; j <= 'z' ; j++)
while (freq[j]-- > 0 )
answer += (( char )j);
return answer;
}
public static void main(String[] args)
{
String A = "aa" ;
String B = "ababab" ;
System.out.print(reorderString(A.toCharArray(),
B.toCharArray()));
}
}
|
Python3
def reorderString(A, B):
size_a = len (A)
size_b = len (B)
freq = [ 0 ] * 300
for i in range (size_b):
freq[ ord (B[i])] + = 1
for i in range (size_a):
freq[ ord (A[i])] - = 1
for j in range ( ord ( 'a' ), ord ( 'z' ) + 1 ):
if (freq[j] < 0 ):
return "-1"
answer = []
for j in range ( ord ( 'a' ), ord (A[ 0 ])):
while (freq[j] > 0 ):
answer.append(j)
freq[j] - = 1
first = A[ 0 ]
for j in range (size_a):
if (A[j] > A[ 0 ]):
break
if (A[j] < A[ 0 ]):
answer + = A
A = ""
break
while (freq[ ord (first)] > 0 ):
answer.append(first)
freq[ ord (first)] - = 1
answer + = A
for j in range ( ord ( 'a' ), ord ( 'z' ) + 1 ):
while (freq[j]):
answer.append( chr (j))
freq[j] - = 1
return "".join(answer)
if __name__ = = '__main__' :
A = "aa"
B = "ababab"
print (reorderString(A, B))
|
C#
using System;
class GFG{
static String reorderString( char []A,
char []B)
{
int size_a = A.Length;
int size_b = B.Length;
int []freq = new int [300];
for ( int i = 0; i < size_b; i++)
freq[B[i]]++;
for ( int i = 0; i < size_a; i++)
freq[A[i]]--;
for ( int j = 'a' ; j <= 'z' ; j++)
{
if (freq[j] < 0)
return "-1" ;
}
String answer = "" ;
for ( int j = 'a' ; j < A[0]; j++)
while (freq[j] > 0)
{
answer+=j;
freq[j]--;
}
int first = A[0];
for ( int j = 0; j < size_a; j++)
{
if (A[j] > A[0])
break ;
if (A[j] < A[0])
{
answer += String.Join( "" , A);
A = new char [A.Length];
break ;
}
}
while (freq[first] > 0)
{
answer += String.Join( "" , ( char )first);
--freq[first];
}
answer += String.Join( "" , A);
for ( int j = 'a' ; j <= 'z' ; j++)
while (freq[j]-- > 0)
answer += (( char )j);
return answer;
}
public static void Main(String[] args)
{
String A = "aa" ;
String B = "ababab" ;
Console.Write(reorderString(A.ToCharArray(),
B.ToCharArray()));
}
}
|
Javascript
<script>
function reorderString(A,B)
{
let size_a = A.length;
let size_b = B.length;
let freq = new Array(300);
for (let i=0;i<300;i++)
{
freq[i]=0;
}
for (let i = 0; i < size_b; i++)
freq[B[i].charCodeAt(0)]++;
for (let i = 0; i < size_a; i++)
freq[A[i].charCodeAt(0)]--;
for (let j = 'a' .charCodeAt(0); j <= 'z' .charCodeAt(0); j++)
{
if (freq[j] < 0)
return "-1" ;
}
let answer = "" ;
for (let j = 'a' .charCodeAt(0); j < A[0].charCodeAt(0); j++)
while (freq[j] > 0)
{
answer+=j;
freq[j]--;
}
let first = A[0];
for (let j = 0; j < size_a; j++)
{
if (A[j] > A[0])
break ;
if (A[j] < A[0])
{
answer += (A).join( "" );
A = new Array(A.length);
break ;
}
}
while (freq[first] > 0)
{
answer += (String.fromCharCode(first));
--freq[first];
}
answer += (A).join( "" );
for (let j = 'a' .charCodeAt(0); j <= 'z' .charCodeAt(0); j++)
while (freq[j]-- > 0)
answer += String.fromCharCode(j);
return answer;
}
let A = "aa" ;
let B = "ababab" ;
document.write(reorderString(A.split( "" ),
B.split( "" )));
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)
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