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 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
// 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
# 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#
// 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 |
Javascript
<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> |
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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