Lexicographically smaller string by swapping at most one character pair

07 Jul, 2021
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.
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++
Java
Python3
C#
Javascript
// C++ program check whether is// it possible to make string A// lexicographically smaller than string B #include <bits/stdc++.h>using namespace std; // Swap functionvoid 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 Bbool 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 Codeint main(){    string A = "AGAIN";    string B = "ACTION";     if (IsLexicographicallySmaller(A, B)) {        cout << "Yes"             << "\n";    }    else {        cout << "No"             << "\n";    }     return 0;}
// Java program check whether is// it possible to make String A// lexicographically smaller than String Bimport 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 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 Bdef 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 CodeA = "AGAIN"B = "ACTION" if(IsLexicographicallySmaller(A, B)):    print("Yes")else:    print("No") # This code is contributed by avanitrachhadiya2155
// C# program check whether is// it possible to make String A// lexicographically smaller// than String Busing System;class GFG{      // Swap functionstatic 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 Bstatic 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 Codepublic 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
<script>// Javascript program check whether is// it possible to make String A// lexicographically smaller than String B  // Swap functionfunction 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 Bfunction 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 Codelet A = "AGAIN";let B = "ACTION"; if (IsLexicographicallySmaller(A, B)) {    document.write("Yes");}else {    document.write("No");} // This code is contributed by ab2127</script>
Yes
 
abhaysingh290895
Akanksha_Rai
rutvik_56
avanitrachhadiya2155
ab2127
lexicographic-ordering
Strings
Strings
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