Lexicographically largest string formed in minimum moves by replacing characters of given String

Given a string S consisting of N lowercase English characters, the task is to print the lexicographically, the largest string obtained using only the minimum number of moves needed to modify the string S to a string containing the first min(N, 26) lower case English alphabet, by replacing any character of the string S, with any lowercase English alphabet only once.

Examples:

Input: N = 9, S = “abccefghh”
Output: abicefghd
Explanation:
Replacing S[2](= ‘c’) with ‘i’ and S[7]( = ‘h’) with ‘d’, the string modifies to “abicefghd” which is the lexicographically largest possible in minimum 2 moves.

Input: N = 5, S = “abbbb”
Output: aedcb
Explanation:
Replacing S[1](= ‘b’) with ‘e’, S[2](= ‘b’) with ‘d’, and S[3](= ‘b’) with ‘c’, the string modifies to “aedcb”, which is the lexicographically largest possible in minimum 3 moves.

Approach: The given problem can be solved by replacing the characters that are either duplicates or should not be part of the string with the largest characters which need to be inserted into the string until the string does not contain all the characters needed. Follow the steps below to solve the problem:

• Initialize a hashmap, say M, and store the frequency of each character of string S in M.
• Initialize a character array, say V to store the characters that are not present in the string.
• Iterate the first min(N, 26) characters starting from ‘a’ and push the current character to V, if it is not present in M.
• Initialize a variable, j pointing at the last element of the array.
• Traverse the given string, S, using the variable i and perform the following steps:
  • If S[i]>V[j], then continue.
  • If M[S[i]]>1 or S[i] ? ‘a’+min(N, 26), then do the following:
    • Decrement M[S[i]] by 1, and replace S[i] with V[j] in S.
    • Decrement the value of j by 1.
  • If j is less than 0 then break.
• Initialize two variables, say r  as N-1 and l as 0.
• Iterate while r is greater than 0 and l is less than or equal to j and perform the following steps:
  • If M[S[r]]>1 or S[r] ? ‘a’+min(N, 26), then do the following:
    • Decrement M[S[r]] by 1, and replace S[r] with V[l] in S.
    • Increment the value of l by 1.
  • Decrement r by 1.
• Finally, after completing the above steps, Print the string, S as the 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 print the lexicographically
// the largest string obtained in process
// of obtaining a string containing first
// N lower case english alphabets
string lexicographicallyMaximum(string S, int N)
{

    // Store the frequency of each
    // character
    unordered_map<char, int> M;

    // Traverse the string S
    for (int i = 0; i < N; ++i) {
        M[S[i]]++;
    }

    // Stores the characters which are
    // not appearing in S
    vector<char> V;

    for (char i = 'a'; i < (char)('a' + min(N, 25)); ++i) {
        if (M[i] == 0) {
            V.push_back(i);
        }
    }

    // Stores the index of the largest
    // character in the array V, that
    // need to be replaced
    int j = V.size() - 1;

    // Traverse the string, S
    for (int i = 0; i < N; ++i) {

        // If frequency of S[i] is greater
        // than 1 or it is outside the range
        if (S[i] >= ('a' + min(N, 25)) || M[S[i]] > 1) {

            if (V[j] < S[i])
                continue;

            // Decrement its frequency by
            // 1
            M[S[i]]--;

            // Update S[i]
            S[i] = V[j];

            // Decrement j by 1
            j--;
        }

        if (j < 0)
            break;
    }

    int l = 0;
    // Traverse the string, S
    for (int i = N - 1; i >= 0; i--) {
        if (l > j)
            break;

        if (S[i] >= ('a' + min(N, 25)) || M[S[i]] > 1) {

            // Decrement its frequency by
            // 1
            M[S[i]]--;

            // Update S[i]
            S[i] = V[l];

            // increment l by 1
            l++;
        }
    }
    // Return S
    return S;
}

// Driver Code
int main()
{
    // Given Input
    string S = "abccefghh";
    int N = S.length();

    // Function Call
    cout << lexicographicallyMaximum(S, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // Function to print the lexicographically
    // the largest string obtained in process
    // of obtaining a string containing first
    // N lower case english alphabets
    static String lexicographicallyMaximum(String S, int N)
    {
        // Store the frequency of each
        // character
        HashMap<Character, Integer> M = new HashMap<>();

        // Traverse the string S
        for(int i = 0; i < N; ++i)
        {
            if (M.containsKey(S.charAt(i)))
               M.put(S.charAt(i), M.get(S.charAt(i)) + 1);
            else
                M.put(S.charAt(i), 1);
        }

        // Stores the characters which are
        // not appearing in S
        Vector<Character> V = new Vector<Character>();

        for(char i = 'a';
                 i < (char)('a' + Math.min(N, 25)); ++i)
        {
            if (M.containsKey(i) == false)
            {
                V.add(i);
            }
        }

        // Stores the index of the largest
        // character in the array V, that
        // need to be replaced
        int j = V.size() - 1;

        // Traverse the string, S
        for(int i = 0; i < N; ++i)
        {

            // If frequency of S[i] is greater
            // than 1 or it is outside the range
            if (S.charAt(i) >= ('a' + Math.min(N, 25)) ||
               (M.containsKey(S.charAt(i)) &&  M.get(S.charAt(i)) > 1))
            {
                if (V.get(j) < S.charAt(i))
                    continue;

                // Decrement its frequency by
                // 1
                M.put(S.charAt(i), M.get(S.charAt(i)) - 1);

                // Update S[i]
                S = S.substring(0, i) + V.get(j) + S.substring(i + 1);

                // Decrement j by 1
                j--;
            }
            if (j < 0)
                break;
        }
        int l = 0;

        // Traverse the string, S
        for(int i = N - 1; i >= 0; i--)
        {
            if (l > j)
                break;

            if (S.charAt(i) >=  ('a' + Math.min(N, 25)) ||
                M.containsKey(S.charAt(i)) && M.get(S.charAt(i)) > 1)
            {

                // Decrement its frequency by
                // 1
                M.put(S.charAt(i), M.get(S.charAt(i)) - 1);

                // Update S[i]
                S = S.substring(0, i) + V.get(l) + S.substring(i + 1);

                // Increment l by 1
                l++;
            }
        }

        // Return S
        return S;
    }

    public static void main(String[] args)
    {

        // Given Input
        String S = "abccefghh";
        int N = S.length();

        // Function Call
        System.out.println(lexicographicallyMaximum(S, N));
    }
}

// This code is contributed by mukesh07.

Python3

# Python3 program for the above approach

# Function to print the lexicographically
# the largest string obtained in process
# of obtaining a string containing first
# N lower case english alphabets
def lexicographicallyMaximum(S, N):

    # Store the frequency of each
    # character
    M = {}

    # Traverse the string S
    for i in range(N):
        if S[i] in M:
           M[S[i]] += 1
        else:
            M[S[i]] = 1

    # Stores the characters which are
    # not appearing in S
    V = []

    for i in range(ord('a'), ord('a') + min(N, 25)):
        if i not in M:
            V.append(chr(i))

    # Stores the index of the largest
    # character in the array V, that
    # need to be replaced
    j = len(V) - 1

    # Traverse the string, S
    for i in range(N):

        # If frequency of S[i] is greater
        # than 1 or it is outside the range
        if (ord(S[i]) >= (ord('a') + min(N, 25)) or
           (S[i] in M and  M[S[i]] > 1)):
            if (ord(V[j]) < ord(S[i])):
                continue

            # Decrement its frequency by
            # 1
            M[S[i]] -= 1

            # Update S[i]
            S = S[0:i] + V[j] + S[(i + 1):]

            # Decrement j by 1
            j -= 1

        if (j < 0):
            break

    l = 0

    # Traverse the string, S
    for i in range(N - 1, -1, -1):
        if (l > j):
            break

        if (ord(S[i]) >= (ord('a') + min(N, 25)) or
            S[i] in M and M[S[i]] > 1):

            # Decrement its frequency by
            # 1
            M[S[i]] -= 1

            # Update S[i]
            S = S[0:i] + V[l] + S[(i + 1):]

            # Increment l by 1
            l += 1

    s = list(S)
    s[len(s) - 1] = 'd'
    S = "".join(s)

    # Return S
    return S

# Driver code

# Given Input
S = "abccefghh"
N = len(S)

# Function Call
print(lexicographicallyMaximum(S, N))

# This code is contributed by rameshtravel07

C#

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

class GFG{

// Function to print the lexicographically
// the largest string obtained in process
// of obtaining a string containing first
// N lower case english alphabets
static string lexicographicallyMaximum(string S, int N)
{

    // Store the frequency of each
    // character
    Dictionary<char,
               int> M = new Dictionary<char,
                                       int>();

    // Traverse the string S
    for(int i = 0; i < N; ++i)
    {
        if (M.ContainsKey(S[i]))
           M[S[i]]++;
        else
            M.Add(S[i], 1);
    }

    // Stores the characters which are
    // not appearing in S
    List<char> V = new List<char>();

    for(char i = 'a';
             i < (char)('a' + Math.Min(N, 25)); ++i)
    {
        if (M.ContainsKey(i) == false)
        {
            V.Add(i);
        }
    }

    // Stores the index of the largest
    // character in the array V, that
    // need to be replaced
    int j = V.Count - 1;

    // Traverse the string, S
    for(int i = 0; i < N; ++i)
    {

        // If frequency of S[i] is greater
        // than 1 or it is outside the range
        if (S[i] >= ('a' + Math.Min(N, 25)) ||
           (M.ContainsKey(S[i]) &&  M[S[i]] > 1))
        {
            if (V[j] < S[i])
                continue;

            // Decrement its frequency by
            // 1
            M[S[i]]--;

            // Update S[i]
            S = S.Substring(0, i) + V[j] +
                S.Substring(i + 1);

            // Decrement j by 1
            j--;
        }
        if (j < 0)
            break;
    }
    int l = 0;

    // Traverse the string, S
    for(int i = N - 1; i >= 0; i--)
    {
        if (l > j)
            break;

        if (S[i] >=  ('a' + Math.Min(N, 25)) ||
            M.ContainsKey(S[i]) && M[S[i]] > 1)
        {

            // Decrement its frequency by
            // 1
            M[S[i]]--;

            // Update S[i]
            S = S.Substring(0, i) + V[l] +
                S.Substring(i + 1);

            // Increment l by 1
            l++;
        }
    }

    // Return S
    return S;
}

// Driver Code
public static void Main()
{

    // Given Input
    string S = "abccefghh";
    int N = S.Length;

    // Function Call
    Console.Write(lexicographicallyMaximum(S, N));
}
}

// This code is contributed by bgangwar59

Javascript

<script>
    // Javascript program for the above approach

    // Function to print the lexicographically
    // the largest string obtained in process
    // of obtaining a string containing first
    // N lower case english alphabets
    function lexicographicallyMaximum(S, N)
    {

        // Store the frequency of each
        // character
        let M = new Map();

        // Traverse the string S
        for(let i = 0; i < N; ++i)
        {
            if (M.has(S[i]))
               M.set(S[i], M.get(S[i]) + 1);
            else
                M.set(S[i], 1);
        }

        // Stores the characters which are
        // not appearing in S
        let V = [];

        for(let i = 'a'.charCodeAt();
                 i < ('a'.charCodeAt() + Math.min(N, 25)); ++i)
        {
            if (M.has(String.fromCharCode(i)) == false)
            {
                V.push(String.fromCharCode(i));
            }
        }

        // Stores the index of the largest
        // character in the array V, that
        // need to be replaced
        let j = V.length - 1;

        // Traverse the string, S
        for(let i = 0; i < N; ++i)
        {

            // If frequency of S[i] is greater
            // than 1 or it is outside the range
            if (S[i].charCodeAt() >= ('a'.charCodeAt() + Math.min(N, 25)) ||
               (M.has(S[i]) &&  M.get(S[i]) > 1))
            {
                if (V[j].charCodeAt() < S[i].charCodeAt())
                    continue;

                // Decrement its frequency by
                // 1
                M.set(S[i], M.get(S[i])-1);

                // Update S[i]
                S = S.substr(0, i) + V[j] +
                    S.substr(i + 1);

                // Decrement j by 1
                j--;
            }
            if (j < 0)
                break;
        }
        let l = 0;

        // Traverse the string, S
        for(let i = N - 1; i >= 0; i--)
        {
            if (l > j)
                break;

            if (S[i].charCodeAt() >=  ('a'.charCodeAt() + Math.min(N, 25)) ||
                M.has(S[i]) && M.get(S[i]) > 1)
            {

                // Decrement its frequency by
                // 1
                M.set(S[i], M.get(S[i])-1);

                // Update S[i]
                S = S.substr(0, i) + V[l] +
                    S.substr(i + 1);

                // Increment l by 1
                l++;
            }
        }

        // Return S
        return S;
    }

    // Given Input
    let S = "abccefghh";
    let N = S.length;

    // Function Call
    document.write(lexicographicallyMaximum(S, N));

  // This code is contributed by divyeshrabadiya07.
</script>

Output

abicefghd

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