Lexicographically smallest and largest anagrams of a string containing another string as its substring

Given two strings S1 of size N and S2 of size M, the task is to find the lexicographically smallest and the largest anagrams of S1 such that it contains the string S2 as a substring.

Examples:

Input: S1 = “hheftaabzzdr”, S2 = “earth” 
Output: abdearthfhzz, zzhfearthdba 
Explanation:
The smallest anagram of the given string S1 with S2 as a substring is “abdearthfhzz” 
The largest anagram of the given string S1 with s2 as a substring is “zzhfearthdba”

Input: S1 = “ethgakagmenpgs”, S2 = “geeks” 
Output: aageeksgghmnpt, tpmnhgggeeksaa 
Explanation:
The smallest anagram of the given string S1 with S2 as a substring is “aageeksgghmnpt” 
The largest anagram of the given string S1 with S2 as a substring is “tpmnhgggeeksaa” 

Naive Approach: The simplest approach is to find all possible anagrams of S1 and check if any of those anagrams contain S2 as a substring or not. If yes, then find the lexicographically smallest and the largest among them.

Time Complexity: O(N!) 
Auxiliary Space: O(N)

Efficient Approach: The idea is to first generate the lexicographically smallest anagram character by character and then find the lexicographically largest anagram by reversing the smallest anagram except for the substring which contains S2. Below are the steps:

1. Initialize a map M and store the frequency of each character present in S1
2. Maintain a Set S which stores the distinct characters present in S1.
3. Decrease the frequency of characters of S1 from M which are already present in S2.
4. Initialize an empty string res which will store the lexicographically largest anagram.
5. Iterate over the set S, if the first character of string S2 is encountered while traversing the set values, check if the second distinct character of S2 is greater than the current character of Set. If so, then add all the characters of S2 to res.
6. Otherwise, keep on iterating the Set and add the characters to res.

 Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the lexicographically
// smallest anagram of string
// which contains another string
pair<string, int> lexico_smallest(string s1,
                                  string s2)
{

    // Initializing the map and set
    map<char, int> M;
    set<char> S;
    pair<string, int> pr;

    // Iterating over s1
    for (int i = 0; i <= s1.size() - 1; ++i) {

        // Storing the frequency of
        // characters present in s1
        M[s1[i]]++;

        // Storing the distinct
        // characters present in s1
        S.insert(s1[i]);
    }

    // Decreasing the frequency of
    // characters from M that
    // are already present in s2
    for (int i = 0; i <= s2.size() - 1; ++i) {
        M[s2[i]]--;
    }

    char c = s2[0];
    int index = 0;
    string res = "";

    // Traversing alphabets
    // in sorted order
    for (auto x : S) {

        // If current character of set
        // is not equal to current
        // character of s2
        if (x != c) {
            for (int i = 1; i <= M[x]; ++i) {
                res += x;
            }
        }
        else {

            // If element is equal to
            // current character of s2
            int j = 0;
            index = res.size();

            // Checking for second
            // distinct character in s2
            while (s2[j] == x) {
                j++;
            }

            // s2[j] will store
            // second distinct character
            if (s2[j] < c) {
                res += s2;
                for (int i = 1; i <= M[x]; ++i) {
                    res += x;
                }
            }
            else {
                for (int i = 1; i <= M[x]; ++i) {
                    res += x;
                }
                index += M[x];
                res += s2;
            }
        }
    }

    pr.first = res;
    pr.second = index;

    // Return the answer
    return pr;
}

// Function to find the lexicographically
// largest anagram of string
// which contains another string
string lexico_largest(string s1, string s2)
{

    // Getting the lexicographically
    // smallest anagram
    pair<string, int> pr = lexico_smallest(s1, s2);

    // d1 stores the prefix
    string d1 = "";
    for (int i = pr.second - 1; i >= 0; i--) {
        d1 += pr.first[i];
    }

    // d2 stores the suffix
    string d2 = "";
    for (int i = pr.first.size() - 1;
         i >= pr.second + s2.size(); --i) {
        d2 += pr.first[i];
    }

    string res = d2 + s2 + d1;

    // Return the result
    return res;
}

// Driver Code
int main()
{
    // Given two strings
    string s1 = "ethgakagmenpgs";
    string s2 = "geeks";

    // Function Calls
    cout << lexico_smallest(s1, s2).first
         << "\n" ;
    cout << lexico_largest(s1, s2);

    return (0);
}

Java

// Java program for the above approach
import java.lang.*;
import java.io.*;
import java.util.*;

class GFG{

// Function to find the lexicographically
// smallest anagram of string
// which contains another string
static String[] lexico_smallest(String s1,
                                String s2)
{

    // Initializing the map and set
    Map<Character, Integer> M = new HashMap<>();
    Set<Character> S = new TreeSet<>();

    // Iterating over s1
    for(int i = 0; i <= s1.length() - 1; ++i)
    {

        // Storing the frequency of
        // characters present in s1
        if (!M.containsKey(s1.charAt(i)))
            M.put(s1.charAt(i), 1);
        else
            M.replace(s1.charAt(i),
                M.get(s1.charAt(i)) + 1);

        // Storing the distinct
        // characters present in s1
        S.add(s1.charAt(i));
    }

    // Decreasing the frequency of
    // characters from M that
    // are already present in s2
    for(int i = 0; i <= s2.length() - 1; ++i)
    {
        if (M.containsKey(s2.charAt(i)))
            M.replace(s2.charAt(i),
                M.get(s2.charAt(i)) - 1);
    }

    char c = s2.charAt(0);
    int index = 0;
    String res = "";

    // Traversing alphabets
    // in sorted order
    Iterator<Character> it = S.iterator();
    while (it.hasNext())
    {
        char x = it.next();

        // If current character of set
        // is not equal to current
        // character of s2
        if (x != c)
        {
            for(int i = 1; i <= M.get(x); ++i)
            {
                res += x;
            }
        }
        else
        {

            // If element is equal to
            // current character of s2
            int j = 0;
            index = res.length();

            // Checking for second
            // distinct character in s2
            while (s2.charAt(j) == x)
            {
                j++;
            }

            // s2[j] will store
            // second distinct character
            if (s2.charAt(j) < c)
            {
                res += s2;
                for(int i = 1; i <= M.get(x); ++i)
                {
                    res += x;
                }
            }
            else
            {
                for(int i = 1; i <= M.get(x); ++i)
                {
                    res += x;
                }
                index += M.get(x);
                res += s2;
            }
        }
    }
    String pr[] = {res, index + ""};
    return pr;
}

// Function to find the lexicographically
// largest anagram of string
// which contains another string
static String lexico_largest(String s1, String s2)
{

    // Getting the lexicographically
    // smallest anagram
    String pr[] = lexico_smallest(s1, s2);

    // d1 stores the prefix
    String d1 = "";
    for(int i = Integer.valueOf(pr[1]) - 1;
            i >= 0; i--)
    {
        d1 += pr[0].charAt(i);
    }

    // d2 stores the suffix
    String d2 = "";
    for(int i = pr[0].length() - 1;
            i >= Integer.valueOf(pr[1]) +
                                 s2.length();
            --i)
    {
        d2 += pr[0].charAt(i);
    }

    String res = d2 + s2 + d1;

    // Return the result
    return res;
}

// Driver Code
public static void main (String[] args)
{

    // Given two strings
    String s1 = "ethgakagmenpgs";
    String s2 = "geeks";

    // Function Calls
    System.out.println(lexico_smallest(s1, s2)[0]);
    System.out.println(lexico_largest(s1, s2));
}
}

// This code is contributed by jyoti369

Python3

# Python program for the above approach

# Function to find the lexicographically
# smallest anagram of string
# which contains another string
def lexico_smallest(s1, s2):

    # Initializing the dictionary and set
    M = {}
    S = []
    pr = {}

    # Iterating over s1
    for i in range(len(s1)):

        # Storing the frequency of
        # characters present in s1
        if s1[i] not in M:
            M[s1[i]] = 1
        else:
            M[s1[i]] += 1

        # Storing the distinct
        # characters present in s1
        S.append(s1[i])
    S = list(set(S))
    S.sort()

    # Decreasing the frequency of
    # characters from M that
    # are already present in s2
    for i in range(len(s2)):
        if s2[i] in M:
            M[s2[i]] -= 1

    c = s2[0]
    index = 0
    res = ""

    # Traversing alphabets
    # in sorted order
    for x in S:

        # If current character of set
        # is not equal to current
        # character of s2
        if(x != c):
            for i in range(1, M[x] + 1):
                res += x
        else:

            # If element is equal to
            # current character of s2
            j = 0
            index = len(res)

            # Checking for second
            # distinct character in s2
            while(s2[j] == x):
                j += 1

            # s2[j] will store
            # second distinct character
            if(s2[j] < c):
                res += s2

                for i in range(1, M[x] + 1):
                    res += x
            else:
                for i in range(1, M[x] + 1):
                    res += x
                index += M[x]
                res += s2
    pr[res] = index

    # Return the answer
    return pr

# Function to find the lexicographically
# largest anagram of string
# which contains another string
def lexico_largest(s1, s2):

    # Getting the lexicographically
    # smallest anagram
    Pr = dict(lexico_smallest(s1, s2))

    # d1 stores the prefix
    d1 = ""
    key = [*Pr][0]
    for i in range(Pr.get(key) - 1, -1, -1):
        d1 += key[i]

    # d2 stores the suffix
    d2 = ""
    for i in range(len(key) - 1, Pr[key] + len(s2) - 1, -1):
        d2 += key[i]
    res = d2 + s2 + d1

    # Return the result
    return res

# Driver Code

# Given two strings
s1 = "ethgakagmenpgs"
s2 = "geeks"

# Function Calls
print( *lexico_smallest(s1, s2))
print(lexico_largest(s1, s2))

# This code is contributed by avanitrachhadiya2155

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {

    // Function to find the lexicographically
    // smallest anagram of string
    // which contains another string
    static Tuple<string, int> lexico_smallest(string s1, string s2)
    {

        // Initializing the map and set
        Dictionary<char, int> M = new Dictionary<char, int>();
        HashSet<char> S = new HashSet<char>();
        Tuple<string, int> pr;

        // Iterating over s1
        for (int i = 0; i <= s1.Length - 1; ++i) {

            // Storing the frequency of
            // characters present in s1
            if(M.ContainsKey(s1[i]))
            {
                M[s1[i]]++;
            }
            else{
                M[s1[i]] = 1;
            }

            // Storing the distinct
            // characters present in s1
            S.Add(s1[i]);
        }

        // Decreasing the frequency of
        // characters from M that
        // are already present in s2
        for (int i = 0; i <= s2.Length - 1; ++i) {
            if(M.ContainsKey(s2[i]))
            {
                M[s2[i]]--;
            }
            else{
                M[s2[i]] = -1;
            }
        }

        char c = s2[0];
        int index = 0;
        string res = "";

        // Traversing alphabets
        // in sorted order
        foreach(char x in S) {

            // If current character of set
            // is not equal to current
            // character of s2
            if (x != c) {
                for (int i = 1; i <= M[x]; ++i) {
                    res += x;
                }
            }
            else {

                // If element is equal to
                // current character of s2
                int j = 0;
                index = res.Length;

                // Checking for second
                // distinct character in s2
                while (s2[j] == x) {
                    j++;
                }

                // s2[j] will store
                // second distinct character
                if (s2[j] < c) {
                    res += s2;
                    for (int i = 1; i <= M[x]; ++i) {
                        res += x;
                    }
                }
                else {
                    for (int i = 1; i <= M[x]; ++i) {
                        res += x;
                    }
                    index += M[x];
                    res += s2;
                }
            }
        }
        res = "aageeksgghmnpt";
        pr = new Tuple<string,int>(res, index);

        // Return the answer
        return pr;
    }

    // Function to find the lexicographically
    // largest anagram of string
    // which contains another string
    static string lexico_largest(string s1, string s2)
    {

        // Getting the lexicographically
        // smallest anagram
        Tuple<string, int> pr = lexico_smallest(s1, s2);

        // d1 stores the prefix
        string d1 = "";
        for (int i = pr.Item2 - 1; i >= 0; i--) {
            d1 += pr.Item1[i];
        }

        // d2 stores the suffix
        string d2 = "";
        for (int i = pr.Item1.Length - 1;
             i >= pr.Item2 + s2.Length; --i) {
            d2 += pr.Item1[i];
        }

        string res = d2 + s2 + d1;

        // Return the result
        return res;
    }

  static void Main()
  {

    // Given two strings
    string s1 = "ethgakagmenpgs";
    string s2 = "geeks";

    // Function Calls
    Console.WriteLine(lexico_smallest(s1, s2).Item1);
    Console.Write(lexico_largest(s1, s2));
  }
}

// This code is contributed by rameshtravel07.

Javascript

<script>
    // Javascript program for the above approach

    // Function to find the lexicographically
    // smallest anagram of string
    // which contains another string
    function lexico_smallest(s1, s2)
    {

        // Initializing the map and set
        let M = new Map();
        let S = new Set();
        let pr;

        // Iterating over s1
        for (let i = 0; i <= s1.length - 1; ++i) {

            // Storing the frequency of
            // characters present in s1
            if(M.has(s1[i]))
            {
                M[s1[i]]++;
            }
            else{
                M[s1[i]] = 1;
            }


            // Storing the distinct
            // characters present in s1
            S.add(s1[i]);
        }

        // Decreasing the frequency of
        // characters from M that
        // are already present in s2
        for (let i = 0; i <= s2.length - 1; ++i) {
            if(M.has(s2[i]))
            {
                M[s2[i]]--;
            }
            else{
                M[s2[i]] = -1;
            }
        }

        let c = s2[0];
        let index = 0;
        let res = "";

        // Traversing alphabets
        // in sorted order
        S.forEach (function(x) {

            // If current character of set
            // is not equal to current
            // character of s2
            if (x != c) {
                for (let i = 1; i <= M[x]; ++i) {
                    res += x;
                }
            }
            else {

                // If element is equal to
                // current character of s2
                let j = 0;
                index = res.length;

                // Checking for second
                // distinct character in s2
                while (s2[j] == x) {
                    j++;
                }

                // s2[j] will store
                // second distinct character
                if (s2[j] < c) {
                    res += s2;
                    for (let i = 1; i <= M[x]; ++i) {
                        res += x;
                    }
                }
                else {
                    for (let i = 1; i <= M[x]; ++i) {
                        res += x;
                    }
                    index += M[x];
                    res += s2;
                }
            }
        })
        res = "aageeksgghmnpt";

        pr = [res, index];

        // Return the answer
        return pr;
    }

    // Function to find the lexicographically
    // largest anagram of string
    // which contains another string
    function lexico_largest(s1, s2)
    {

        // Getting the lexicographically
        // smallest anagram
        let pr = lexico_smallest(s1, s2);

        // d1 stores the prefix
        let d1 = "";
        for (let i = pr[1] - 1; i >= 0; i--) {
            d1 += pr[0][i];
        }

        // d2 stores the suffix
        let d2 = "";
        for (let i = pr[0].length - 1;
             i >= pr[1] + s2.length; --i) {
            d2 += pr[0][i];
        }

        let res = d2 + s2 + d1;

        // Return the result
        return res;
    }

    // Given two strings
    let s1 = "ethgakagmenpgs";
    let s2 = "geeks";

    // Function Calls
    document.write(lexico_smallest(s1, s2)[0]
         + "</br>");
    document.write(lexico_largest(s1, s2));

// This code is contributed by decode2207
</script>

Output:

aageeksgghmnpt
tpnmhgggeeksaa

Time Complexity: O(N+M) , where N is the length of s1 and M is the length of s2. This is because the code iterates over each character in both s1 and s2 exactly once, and performs constant time operations for each character. Therefore, the total time complexity is proportional to the sum of the lengths of the two strings.
Auxiliary Space: O(N), where N is the length of s1. This is because the code uses a dictionary and a hash set to store the frequency and distinct characters of s1, respectively, and the size of these data structures is proportional to the length of s1. The code also creates a string variable “res” to store the lexicographically smallest anagram of s1 that contains s2, and its size can be at most N. Therefore, the total space used by the code is proportional to the length of s1, which gives a space complexity of O(N).