Build lexicographically smallest String from two given Strings

Given two strings X and Y of lowercase letters, of length N and M respectively, the task is to build another string Z by performing two types of operations:

• Choose any character from the string X, remove it from X, and add it to the end of Z.
• Choose any character from the string Y, remove it from Y, and add it to the end of Z.

Note: You can only K consecutive operations in a single string. Perform the operations until either X or Y becomes empty.

Examples:

Input: X = "aaaa", Y ="bbbb", K = 2
Output: aabaa
Explanation: The smallest lexicographically string possible for K = 2 is "aabaa".
Select "aa" from X. Then select "b" from Y and "aa" from X again.

Input: X = "ccaaa", Y ="bbeedd", K = 3
Output: aaabbcc

Approach: To solve the problem follow the below idea:

We solve this problem by using greedy method. 

Sort the strings X and Y and take the characters as follows: 

• Keep selecting consecutive characters from the string whose characters are lexicographically smaller till K consecutive characters are selected, or all the characters of that string are selected or the character of the other string becomes lexicographically smaller.
• If K consecutive characters are selected, then do the above operation again on the other string. Repeat these two processes till at least one of the string becomes empty.

Follow the below steps to solve the problem:

• Sort the strings in ascending order.
• Initialize the pointers i, j, p, q to 0.
• Run the loop while any of the strings is non-empty.
  • If (X[i] < Y[j] and p < k) or q == k, then,
    • Append X[i] to the end of Z and increment i and p by 1 and set q = 0.
  • Otherwise, do the following:
    • Append Y[j] to the end of Z and increment j and q by 1 and also set p to 0.
• Return the resultant string.

Below is the implementation of the above approach.

// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;

#define ll long long

// Function to build smallest lexicographically
// string possible for given K
string buildString(string a, string b, int k)
{
  
    // Calculate the sizes
    ll n = a.size();
    ll m = b.size();

    // Sort the strings
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());

    // Initialize the pointers
    ll i = 0, j = 0, p = 0, q = 0;

    string c = "";

    // While any of string is non-empty
    while (i < n && j < m) {
        if ((a[i] < b[j] && p < k) || q == k) {
            c += a[i];
            p++;
            i++;
            q = 0;
        }
        else {
            c += b[j];
            q++;
            j++;
            p = 0;
        }
    }

    // Return resultant string
    return c;
}

// Driver Code
int main()
{

    string X = "aaaa";
    string Y = "bbbb";
    int K = 2;

    // Function call
    cout << buildString(X, Y, K) << endl;

    return 0;
}

// JAVA code to implement the above approach
import java.util.Arrays;
import java.util.Scanner;
import java.io.*;

class GFG
{
  
    // Function to build smallest lexicographically
    // string possible for given k 
    static String buildstring(String a, String b, int k)
    {
        // Calculate the sizes
        int n = a.length();
        int m = b.length();
        
        //Sort the strings
        char c1[] = a.toCharArray();
        Arrays.sort(c1);
        a = new String(c1);
        char c2[] = b.toCharArray();
        Arrays.sort(c2);
        b = new String(c2);
        
        // Initialize the pointers
        int i = 0, j = 0, p = 0, q = 0;
         String c = "";
  
    // While any of string is non-empty
       while (i < n && j < m) 
       {
        if ((a.charAt(i) < b.charAt(j) && p < k) || q == k) 
        {
            c += a.charAt(i);
            p++;
            i++;
            q = 0;
        }
        else 
        {
            c += b.charAt(j);
            q++;
            j++;
            p = 0;
        }
       }
  
     // Return resultant string
     return c;
    }
    
    // driver code
    public static void main (String[] args) 
    {
        String x = "aaaa";
        String y = "bbbb";
        int k = 2;
        
        // function call
        System.out.println(buildstring(x,y,k));
    }
}

// This code is written by Ujjwal Kumar Bhardwaj

# Python3 code to implement the above approach

# Function to build smallest lexicographically
# string possible for given K
def buildString(a, b, k) :
  
    # Calculate the sizes
    n = len(a);
    m = len(b);

    a = list(a)
    b = list(b)
    
    # Sort the strings
    a.sort();
    b.sort();

    # Initialize the pointers
    i = 0; j = 0; p = 0; q = 0;

    c = "";

    # While any of string is non-empty
    while (i < n and j < m) :
        if ((a[i] < b[j] and p < k) or q == k) :
            c += a[i];
            p += 1;
            i += 1;
            q = 0;

        else :
            c += b[j];
            q += 1;
            j += 1;
            p = 0;
    
    c = "".join(c)
    
    # Return resultant string
    return c;

# Driver Code
if __name__ ==  "__main__" :


    X = "aaaa";
    Y = "bbbb";
    K = 2;

    # Function call
    print(buildString(X, Y, K));

    # This code is contributed by AnkThon

// C# code to implement the above approach

using System;
using System.Collections;

public class GFG {

    // Function to build smallest lexicographically
    // string possible for given k
    static String buildstring(String a, String b, int k)
    {
        // Calculate the sizes
        int n = a.Length;
        int m = b.Length;

        // Sort the strings
        char[] c1 = a.ToCharArray();
        Array.Sort(c1);
        a = new String(c1);
        char[] c2 = b.ToCharArray();
        Array.Sort(c2);
        b = new String(c2);

        // Initialize the pointers
        int i = 0, j = 0, p = 0, q = 0;
        String c = "";

        // While any of string is non-empty
        while (i < n && j < m) {
            if ((a[i] < b[j] && p < k) || q == k) {
                c += a[i];
                p++;
                i++;
                q = 0;
            }
            else {
                c += b[j];
                q++;
                j++;
                p = 0;
            }
        }

        // Return resultant string
        return c;
    }

    static public void Main()
    {

        // Code
        String x = "aaaa";
        String y = "bbbb";
        int k = 2;

        // function call
        Console.WriteLine(buildstring(x, y, k));
    }
}

// This code is written by lokeshmvs21.

<script>
//  Javascript code to implement the above approach

// Function to build smallest lexicographically
// string possible for given K
function buildString( a,  b,  k)
{
  
    // Calculate the sizes
    let n = a.length;
    let m = b.length;

    // Sort the strings
    a.sort;
    b.sort;

    // Initialize the pointers
    let i = 0, j = 0, p = 0, q = 0;

    let c = "";

    // While any of string is non-empty
    while (i < n && j < m) {
        if ((a[i] < b[j] && p < k) || q == k) {
            c += a[i];
            p++;
            i++;
            q = 0;
        }
        else {
            c += b[j];
            q++;
            j++;
            p = 0;
        }
    }

    // Return resultant string
    return c;
}

// Driver Code
    let X = "aaaa";
    let Y = "bbbb";
    let K = 2;

    // Function call
    document.write(buildString(X, Y, K));

// This code is contributed by hrithikgarg03188.
</script>

aabaa

Time Complexity: O(N*logN +M*logM)
Auxiliary Space: O(N+M)