Lexicographically smallest string possible by merging two sorted strings

Last Updated : 01 Jun, 2021

Given two sorted strings S1 and S2 of lengths N and M respectively, the task is to construct lexicographically the smallest string possible by merging the two given strings and not changing the order of occurrence of characters.

Examples:

Input: S1 = “eefgkors”, S2 = “eegks”
Output: “eeeefggkkorss”
Explanation:
String “eeeefggkkorss” is lexicographically the smallest string that can be formed after merging the two given string S1 and S2.

Input: S1 = “aabcdtx”, S2 = “achilp”
Output: “aaabccdhilptx”

Naive Approach: The simplest approach is to concatenate the given two strings and sort the resultant string to get lexicographically the smallest string.

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

Efficient Approach: The above approach can be optimized by using the Two-Pointer technique by comparing characters of the given strings and then, construct the required string accordingly.
Follow the steps below to solve the problem:

Initialize two pointers, say ptr1 and ptr2, pointing towards the beginning of both the strings S1 and S2 respectively.
Initialize a string ans = “”, to store the resultant lexicographically the smallest string.
Iterate until ptr1 is less than N and ptr2 is less than M and perform the following steps:
- If S1[ptr1] is less than the S2[ptr2], then append the character S1[ptr1] to the string ans. Increment ptr1.
- Otherwise, append the character S2[ptr2] to the string ans. Increment ptr2.
After completing the above steps, one of the pointers doesn’t reach the end of the string.
Therefore, add all the remaining characters of the string to the end of the string ans.
Print ans as the resultant string.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find lexicographically
// smallest string possible by merging
// two sorted strings
void mergeStrings(string s1, string s2)
{
    // Stores length of string s1
    int len1 = s1.size();
 
    // Stores length of string s2
    int len2 = s2.size();
 
    // Pointer to beginning
    // of string1 i.e., s1
    int pntr1 = 0;
 
    // Pointer to beginning
    // of string2 i.e., s2
    int pntr2 = 0;
 
    // Stores the final string
    string ans = "";
 
    // Traverse the strings
    while (pntr1 < len1 && pntr2 < len2) {
 
        // Append the smaller of the
        // two current characters
        if (s1[pntr1] < s2[pntr2]) {
            ans += s1[pntr1];
            pntr1++;
        }
 
        else {
            ans += s2[pntr2];
            pntr2++;
        }
    }
 
    // Append the remaining characters
    // of any of the two strings
    if (pntr1 < len1) {
        ans += s1.substr(pntr1, len1);
    }
    if (pntr2 < len2) {
        ans += s2.substr(pntr2, len2);
    }
 
    // Print the final string
    cout << ans;
}
 
// Driver Code
int main()
{
    string S1 = "abdcdtx";
    string S2 = "achilp";
 
    // Function Call
    mergeStrings(S1, S2);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to find lexicographically
// smallest string possible by merging
// two sorted strings
static void mergeStrings(String s1, String s2)
{
 
    // Stores length of string s1
    int len1 = s1.length();
     
    // Stores length of string s2
    int len2 = s2.length();
     
    // Pointer to beginning
    // of string1 i.e., s1
    int pntr1 = 0;
     
    // Pointer to beginning
    // of string2 i.e., s2
    int pntr2 = 0;
     
    // Stores the final string
    String ans = "";
     
    // Traverse the strings
    while (pntr1 < len1 && pntr2 < len2) 
    {
     
        // Append the smaller of the
        // two current characters
        if (s1.charAt(pntr1) < s2.charAt(pntr2)) 
        {
            ans += s1.charAt(pntr1);
            pntr1++;
        }
         
        else
        {
            ans += s2.charAt(pntr2);
            pntr2++;
        }
    }
     
    // Append the remaining characters
    // of any of the two strings
    if (pntr1 < len1)
    {
        ans += s1.substring(pntr1, len1);
    }
    if (pntr2 < len2) 
    {
        ans += s2.substring(pntr2, len2);
    }
     
    // Print the final string
    System.out.println(ans);
}
 
// Driver Code
public static void main (String[] args) 
{
    String S1 = "abdcdtx";
    String S2 = "achilp";
 
    // Function Call
    mergeStrings(S1, S2);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to find lexicographically
# smallest possible by merging
# two sorted strings
def mergeStrings(s1, s2):
   
    # Stores length of s1
    len1 = len(s1)
 
    # Stores length of s2
    len2 = len(s2)
 
    # Pointer to beginning
    # of string1 i.e., s1
    pntr1 = 0
 
    # Pointer to beginning
    # of string2 i.e., s2
    pntr2 = 0
 
    # Stores the final string
    ans = ""
 
    # Traverse the strings
    while (pntr1 < len1 and pntr2 < len2):
 
        # Append the smaller of the
        # two current characters
        if (s1[pntr1] < s2[pntr2]):
            ans += s1[pntr1]
            pntr1 += 1
        else:
            ans += s2[pntr2]
            pntr2 += 1
 
    # Append the remaining characters
    # of any of the two strings
    if (pntr1 < len1):
        ans += s1[pntr1:pntr1 + len1]
 
    if (pntr2 < len2):
        ans += s2[pntr2:pntr2 + len2]
 
    # Print the final string
    print (ans)
 
# Driver Code
if __name__ == '__main__':
    S1 = "abdcdtx"
    S2 = "achilp"
 
    # Function Call
    mergeStrings(S1, S2)
 
# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG 
{
   
  // Function to find lexicographically
  // smallest string possible by merging
  // two sorted strings
  static void mergeStrings(string s1, string s2)
  {
     
    // Stores length of string s1
    int len1 = s1.Length;
 
    // Stores length of string s2
    int len2 = s2.Length;
 
    // Pointer to beginning
    // of string1 i.e., s1
    int pntr1 = 0;
 
    // Pointer to beginning
    // of string2 i.e., s2
    int pntr2 = 0;
 
    // Stores the final string
    string ans = "";
 
    // Traverse the strings
    while (pntr1 < len1 && pntr2 < len2) {
 
      // Append the smaller of the
      // two current characters
      if (s1[pntr1] < s2[pntr2]) {
        ans += s1[pntr1];
        pntr1++;
      }
 
      else {
        ans += s2[pntr2];
        pntr2++;
      }
    }
 
    // Append the remaining characters
    // of any of the two strings
    if (pntr1 < len1) {
      ans += s1.Substring(pntr1, len1 - pntr1);
    }
    if (pntr2 < len2) {
      ans += s2.Substring(pntr2, len2 - pntr2);
    }
 
    // Print the final string
    Console.WriteLine(ans);
  }
 
  // Driver Code
  public static void Main()
  {
    string S1 = "abdcdtx";
    string S2 = "achilp";
 
    // Function Call
    mergeStrings(S1, S2);
  }
}
 
// This code is contributed by ukasp.

Javascript

<script>
     
// Javascript program for the above approach
 
// Function to find lexicographically
// smallest string possible by merging
// two sorted strings
function mergeStrings( s1,  s2)
{
    // Stores length of string s1
    var len1 = s1.length;
 
    // Stores length of string s2
    var len2 = s2.length;
 
    // Pointer to beginning
    // of string1 i.e., s1
    var pntr1 = 0;
 
    // Pointer to beginning
    // of string2 i.e., s2
    var pntr2 = 0;
 
    // Stores the final string
    var ans = "";
 
    // Traverse the strings
    while (pntr1 < len1 && pntr2 < len2) {
 
        // Append the smaller of the
        // two current characters
        if (s1[pntr1] < s2[pntr2]) {
            ans += s1[pntr1];
            pntr1++;
        }
 
        else {
            ans += s2[pntr2];
            pntr2++;
        }
    }
 
    // Append the remaining characters
    // of any of the two strings
    if (pntr1 < len1) {
        ans += s1.substr(pntr1, len1);
    }
    if (pntr2 < len2) {
        ans += s2.substr(pntr2, len2);
    }
 
    // Print the final string
    document.write( ans);
}
 
// Driver Code
var S1 = "abdcdtx";
var S2 = "achilp";
// Function Call
mergeStrings(S1, S2);
 
</script>

Output:

aabcdcdhilptx

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

Lexicographically smallest String by pair swapping

sam_2200

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Lexicographically smallest string possible by merging two sorted strings

C++14

Java

Python3

C#

Javascript

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