Lexicographically smaller string by swapping at most one character pair

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Given two strings A and B of all uppercase letters, the task is to find whether is it possible to make string A strictly lexicographically smaller than string B by swapping at most one pair of characters in A.
Examples:

Input: A = “AGAIN”, B = “ACTION”
Output: Yes
Explanation:
We can make string A strictly lexicographically smaller than string B by swapping G and A (AAGIN)
AAGIN is lexicographically smaller than ACTION
Input: A = “APPLE” B = “AAAAAPPPLLE”
Output: No

Approach:

Sort string A.
We can find the first position where A and sorted(A) doesn’t match.
We then find the letter that should be in that position and swap it with the letter in the sorted(A).
If there are multiple choices, it is better to take the one that occurs last, since it makes the resulting string smallest.
Now, compare string A and string B.

Below is the implementation of the above approach.

C++

// C++ program check whether is
// it possible to make string A
// lexicographically smaller than string B
 
#include <bits/stdc++.h>
using namespace std;
 
// Swap function
void swap(char& x, char& y)
{
    char temp = x;
    x = y;
    y = temp;
}
 
// Function that finds whether is
// it possible to make string A
// lexicographically smaller than string B
bool IsLexicographicallySmaller(
    string A, string B)
{
    // Condition if string A
    // is already smaller than B
    if (A < B) {
        return true;
    }
    string temp = A;
 
    // Sorting temp string
    sort(temp.begin(), temp.end());
 
    int index = -1;
 
    for (int i = 0; i < A.length(); i++) {
        // Condition for first changed
        // character of string A and temp
        if (A[i] != temp[i]) {
            index = i;
            break;
        }
    }
 
    // Condition if string A
    // is already  sorted
    if (index == -1) {
        return false;
    }
 
    int j;
 
    // Finding first changed character
    // from  last of string A
    for (int i = 0; i < A.length(); i++) {
        if (A[i] == temp[index])
            j = i;
    }
 
    // Swap the two characters
    swap(A[index], A[j]);
 
    // Condition if string A
    // is smaller   than B
    if (A < B) {
        return true;
    }
 
    else {
        return false;
    }
}
 
// Driver Code
int main()
{
    string A = "AGAIN";
    string B = "ACTION";
 
    if (IsLexicographicallySmaller(A, B)) {
        cout << "Yes"
             << "\n";
    }
    else {
        cout << "No"
             << "\n";
    }
 
    return 0;
}

Java

// Java program check whether is
// it possible to make String A
// lexicographically smaller than String B
import java.util.*;
 
class GFG{
     
     
    // Swap function
    static String swap(String str, int i, int j)
    {
        char[] tempArr = str.toCharArray();
        char temp = tempArr[i];
        tempArr[i] = tempArr[j];
        tempArr[j] = temp;
        return String.valueOf(tempArr);
    }
     
    // Function that finds whether is
    // it possible to make String A
    // lexicographically smaller than String B
    static boolean IsLexicographicallySmaller(String A, String B)
    {
        // Condition if String A
        // is already smaller than B
        if (A.compareTo(B) < 0) {
            return true;
        }
        String temp = A;
        char p[] = temp.toCharArray();
 
        // Sorting temp String
        Arrays.sort(p);
        temp=String.valueOf(p);
        int index = -1;
     
        for (int i = 0; i < A.length(); i++) {
 
            // Condition for first changed
            // character of String A and temp
            if (A.charAt(i) != temp.charAt(i)) {
                index = i;
                break;
            }
        }
     
        // Condition if String A
        // is already sorted
        if (index == -1) {
            return false;
        }
     
        int j = 0;
     
        // Finding first changed character
        // from last of String A
        for (int i = 0; i < A.length(); i++) {
            if (A.charAt(i) == temp.charAt(index))
                j = i;
        }
     
        // Swap the two characters
        A = swap(A, index, j);
     
        // Condition if String A
        // is smaller than B
        if (A.compareTo(B) < 0) {
            return true;
        }
     
        else {
            return false;
        }
    }
     
    // Driver Code
    public static void main(String args[])
    {
        String A = "AGAIN";
        String B = "ACTION";
     
        if (IsLexicographicallySmaller(A, B)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
     
    }
}
 
// This code is contributed by AbhiThakur

Python3

# Python3 program check 
# it possible to make string
# A lexicographically smaller 
# than string B
 
# Function that finds whether is
# it possible to make string A
# lexicographically smaller than 
# string B
def IsLexicographicallySmaller(A, B):
 
    # Condition if string A
    # is already smaller 
    # than B
    if(A < B):
        return True
    temp = A
 
    # Sorting temp string
    temp = ''.join(sorted(temp))
    index =- 1
 
    for i in range(len(A)):
 
        # Condition for first 
        # changed character of 
        # string A and temp
        if(A[i] != temp[i]):
            index = i
            break
 
    # Condition if string A
    # is already  sorted
    if(index == -1):
        return False
       
    j = 0
 
    # Finding first changed
    # character from last 
    # of string A
    for i in range(len(A)):
        if(A[i] == temp[index]):
            j = i
             
    A = list(A)
 
    # Swap the two characters
    A[index], A[j] = A[j], A[index]
    A = ''.join(A)
 
    # Condition if string A
    # is smaller than B
    if(A < B):
        return True
    else:
        return False
 
# Driver Code
A = "AGAIN"
B = "ACTION"
 
if(IsLexicographicallySmaller(A, B)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program check whether is
// it possible to make String A
// lexicographically smaller 
// than String B
using System;
class GFG{
      
// Swap function
static string swap(string str, 
                   int i, int j)
{
  char[] tempArr = str.ToCharArray();
  char temp = tempArr[i];
  tempArr[i] = tempArr[j];
  tempArr[j] = temp;
  return new string(tempArr);
}
      
// Function that finds whether is
// it possible to make String A
// lexicographically smaller than String B
static bool IsLexicographicallySmaller(string A, 
                                       string B)
{
  // Condition if String A
  // is already smaller than B
  if (A.CompareTo(B) < 0) 
  {
    return true;
  }
  string temp = A;
  char []p = temp.ToCharArray();
 
  // Sorting temp String
  Array.Sort(p);
  temp=new string(p);
  int index = -1;
 
  for (int i = 0; i < A.Length; i++) 
  {
    // Condition for first changed
    // character of String A and temp
    if (A[i] != temp[i]) 
    {
      index = i;
      break;
    }
  }
 
  // Condition if String A
  // is already sorted
  if (index == -1) 
  {
    return false;
  }
 
  int j = 0;
 
  // Finding first changed character
  // from last of String A
  for (int i = 0; i < A.Length; i++)
  {
    if (A[i] == temp[index])
      j = i;
  }
 
  // Swap the two characters
  A = swap(A, index, j);
 
  // Condition if String A
  // is smaller than B
  if (A.CompareTo(B) < 0) 
  {
    return true;
  }
  else
  {
    return false;
  }
}
      
// Driver Code
public static void Main(string []args)
{
  string A = "AGAIN";
  string B = "ACTION";
  if (IsLexicographicallySmaller(A, B)) 
  {
    Console.Write("Yes");
  }
  else
  {
    Console.Write("No");
  }
}
}
 
// This code is contributed by Rutvik_56

Javascript

<script>
// Javascript program check whether is
// it possible to make String A
// lexicographically smaller than String B
 
 // Swap function
function swap(str,i,j)
{
    let tempArr = str.split("");
        let temp = tempArr[i];
        tempArr[i] = tempArr[j];
        tempArr[j] = temp;
        return (tempArr).join("");
}
 
// Function that finds whether is
    // it possible to make String A
    // lexicographically smaller than String B
function IsLexicographicallySmaller(A, B)
{
 
    // Condition if String A
        // is already smaller than B
        if (A < (B) ) {
            return true;
        }
        let temp = A;
        let p = temp.split("");
  
        // Sorting temp String
        p.sort();
        temp=(p).join("");
        let index = -1;
      
        for (let i = 0; i < A.length; i++) {
  
            // Condition for first changed
            // character of String A and temp
            if (A[i] != temp[i]) {
                index = i;
                break;
            }
        }
      
        // Condition if String A
        // is already sorted
        if (index == -1) {
            return false;
        }
      
        let j = 0;
      
        // Finding first changed character
        // from last of String A
        for (let i = 0; i < A.length; i++) {
            if (A[i] == temp[index])
                j = i;
        }
      
        // Swap the two characters
        A = swap(A, index, j);
      
        // Condition if String A
        // is smaller than B
        if (A < (B) ) {
            return true;
        }
      
        else {
            return false;
        }
}
 
// Driver Code
let A = "AGAIN";
let B = "ACTION";
 
if (IsLexicographicallySmaller(A, B)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by ab2127
</script>

Output:

Yes

Time Complexity: O(N log N), for sorting the given string.
Auxiliary Space: O(N), for storing the given string in extra variable temp.

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