Lexicographically smallest permutation of a string that contains all substrings of another string
Last Updated :
15 Jul, 2025
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++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to reorder the string B
// to contain all the substrings of A
string reorderString(string A, string B)
{
// Find length of strings
int size_a = A.length();
int size_b = B.length();
// Initialize array to count the
// frequencies of the character
int freq[300] = { 0 };
// Counting frequencies of
// character in B
for (int i = 0; i < size_b; i++)
freq[B[i]]++;
// Find remaining character in B
for (int i = 0; i < size_a; i++)
freq[A[i]]--;
for (int j = 'a'; j <= 'z'; j++) {
if (freq[j] < 0)
return "-1";
}
// Declare the reordered string
string answer;
for (int j = 'a'; j < A[0]; j++)
// Loop until freq[j] > 0
while (freq[j] > 0) {
answer.push_back(j);
// Decrement the value
// from freq array
freq[j]--;
}
int first = A[0];
for (int j = 0; j < size_a; j++) {
// Check if A[j] > A[0]
if (A[j] > A[0])
break;
// Check if A[j] < A[0]
if (A[j] < A[0]) {
answer += A;
A.clear();
break;
}
}
// Append the remaining characters
// to the end of the result
while (freq[first] > 0) {
answer.push_back(first);
--freq[first];
}
answer += A;
for (int j = 'a'; j <= 'z'; j++)
// Push all the values from
// frequency array in the answer
while (freq[j]--)
answer.push_back(j);
// Return the answer
return answer;
}
// Driver Code
int main()
{
// Given strings A and B
string A = "aa";
string B = "ababab";
// Function Call
cout << reorderString(A, B);
return 0;
}
// Java program for
// the above approach
class GFG{
// Function to reorder the String B
// to contain all the subStrings of A
static String reorderString(char []A,
char []B)
{
// Find length of Strings
int size_a = A.length;
int size_b = B.length;
// Initialize array to count the
// frequencies of the character
int freq[] = new int[300];
// Counting frequencies of
// character in B
for (int i = 0; i < size_b; i++)
freq[B[i]]++;
// Find remaining character in B
for (int i = 0; i < size_a; i++)
freq[A[i]]--;
for (int j = 'a'; j <= 'z'; j++)
{
if (freq[j] < 0)
return "-1";
}
// Declare the reordered String
String answer = "";
for (int j = 'a'; j < A[0]; j++)
// Loop until freq[j] > 0
while (freq[j] > 0)
{
answer+=j;
// Decrement the value
// from freq array
freq[j]--;
}
int first = A[0];
for (int j = 0; j < size_a; j++)
{
// Check if A[j] > A[0]
if (A[j] > A[0])
break;
// Check if A[j] < A[0]
if (A[j] < A[0])
{
answer += String.valueOf(A);
A = new char[A.length];
break;
}
}
// Append the remaining characters
// to the end of the result
while (freq[first] > 0)
{
answer += String.valueOf((char)first);
--freq[first];
}
answer += String.valueOf(A);
for (int j = 'a'; j <= 'z'; j++)
// Push all the values from
// frequency array in the answer
while (freq[j]-- > 0)
answer += ((char)j);
// Return the answer
return answer;
}
// Driver Code
public static void main(String[] args)
{
// Given Strings A and B
String A = "aa";
String B = "ababab";
// Function Call
System.out.print(reorderString(A.toCharArray(),
B.toCharArray()));
}
}
// This code is contributed by 29AjayKumar
# Python3 program for the above approach
# Function to reorder the B
# to contain all the substrings of A
def reorderString(A, B):
# Find length of strings
size_a = len(A)
size_b = len(B)
# Initialize array to count the
# frequencies of the character
freq = [0] * 300
# Counting frequencies of
# character in B
for i in range(size_b):
freq[ord(B[i])] += 1
# Find remaining character in B
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"
# Declare the reordered string
answer = []
for j in range(ord('a'), ord(A[0])):
# Loop until freq[j] > 0
while (freq[j] > 0):
answer.append(j)
# Decrement the value
# from freq array
freq[j] -= 1
first = A[0]
for j in range(size_a):
# Check if A[j] > A[0]
if (A[j] > A[0]):
break
# Check if A[j] < A[0]
if (A[j] < A[0]):
answer += A
A = ""
break
# Append the remaining characters
# to the end of the result
while (freq[ord(first)] > 0):
answer.append(first)
freq[ord(first)] -= 1
answer += A
for j in range(ord('a'), ord('z') + 1):
# Push all the values from
# frequency array in the answer
while (freq[j]):
answer.append(chr(j))
freq[j] -= 1
# Return the answer
return "".join(answer)
# Driver Code
if __name__ == '__main__':
# Given strings A and B
A = "aa"
B = "ababab"
# Function call
print(reorderString(A, B))
# This code is contributed by mohit kumar 29
// C# program for
// the above approach
using System;
class GFG{
// Function to reorder the String B
// to contain all the subStrings of A
static String reorderString(char []A,
char []B)
{
// Find length of Strings
int size_a = A.Length;
int size_b = B.Length;
// Initialize array to count the
// frequencies of the character
int []freq = new int[300];
// Counting frequencies of
// character in B
for (int i = 0; i < size_b; i++)
freq[B[i]]++;
// Find remaining character in B
for (int i = 0; i < size_a; i++)
freq[A[i]]--;
for (int j = 'a'; j <= 'z'; j++)
{
if (freq[j] < 0)
return "-1";
}
// Declare the reordered String
String answer = "";
for (int j = 'a'; j < A[0]; j++)
// Loop until freq[j] > 0
while (freq[j] > 0)
{
answer+=j;
// Decrement the value
// from freq array
freq[j]--;
}
int first = A[0];
for (int j = 0; j < size_a; j++)
{
// Check if A[j] > A[0]
if (A[j] > A[0])
break;
// Check if A[j] < A[0]
if (A[j] < A[0])
{
answer += String.Join("", A);
A = new char[A.Length];
break;
}
}
// Append the remaining characters
// to the end of the result
while (freq[first] > 0)
{
answer += String.Join("", (char)first);
--freq[first];
}
answer += String.Join("", A);
for (int j = 'a'; j <= 'z'; j++)
// Push all the values from
// frequency array in the answer
while (freq[j]-- > 0)
answer += ((char)j);
// Return the answer
return answer;
}
// Driver Code
public static void Main(String[] args)
{
// Given Strings A and B
String A = "aa";
String B = "ababab";
// Function Call
Console.Write(reorderString(A.ToCharArray(),
B.ToCharArray()));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript program for
// the above approach
// Function to reorder the String B
// to contain all the subStrings of A
function reorderString(A,B)
{
// Find length of Strings
let size_a = A.length;
let size_b = B.length;
// Initialize array to count the
// frequencies of the character
let freq = new Array(300);
for(let i=0;i<300;i++)
{
freq[i]=0;
}
// Counting frequencies of
// character in B
for (let i = 0; i < size_b; i++)
freq[B[i].charCodeAt(0)]++;
// Find remaining character in B
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";
}
// Declare the reordered String
let answer = "";
for (let j = 'a'.charCodeAt(0); j < A[0].charCodeAt(0); j++)
// Loop until freq[j] > 0
while (freq[j] > 0)
{
answer+=j;
// Decrement the value
// from freq array
freq[j]--;
}
let first = A[0];
for (let j = 0; j < size_a; j++)
{
// Check if A[j] > A[0]
if (A[j] > A[0])
break;
// Check if A[j] < A[0]
if (A[j] < A[0])
{
answer += (A).join("");
A = new Array(A.length);
break;
}
}
// Append the remaining characters
// to the end of the result
while (freq[first] > 0)
{
answer += (String.fromCharCode(first));
--freq[first];
}
answer += (A).join("");
for (let j = 'a'.charCodeAt(0); j <= 'z'.charCodeAt(0); j++)
// Push all the values from
// frequency array in the answer
while (freq[j]-- > 0)
answer += String.fromCharCode(j);
// Return the answer
return answer;
}
// Driver Code
// Given Strings A and B
let A = "aa";
let B = "ababab";
// Function Call
document.write(reorderString(A.split(""),
B.split("")));
// This code is contributed by patel2127
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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