Given two strings str1 and str2, the task is to check if any permutation of the given strings str1 and str2 is possible such that the character at each index of one string is greater than or equal to the other string.
Examples:
Input: A = “abc”, B = “xya”
Output: Yes
Explanation:
“ayx” is a permutation of B = “xya” which can break to string “abc” which is a permutation of A = “abc”.Input: A = “abe”, B = “acd”
Output: “No”
Naive Approach: The idea is to generate all the permutation of one string and check if each character of any permutation is greater than the other string then print “YES” else print “NO”.
Time Complexity: O(N^2)
Auxiliary Space: O(1)
Efficient Approach: Since we have to check if each character of permutation of one string is greater than or equals to the permutation of another string or not. The idea is to sort both the strings in alphabetical order. Now iterate a loop over all the character of the string if all the string of string str1 is less than str2 or all the character of string str2 is less than str1 then print Yes else print No.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if permutation of // one string can break permutation of // another string or not bool CanBreak(string A, string B) { // Sort both the strings // in alphabetical order sort(A.begin(), A.end()); sort(B.begin(), B.end()); bool ans1 = true, ans2 = true; // Iterate over the strings for (int i = 0; i < A.length(); i++) { // If any of the string can break // other then mark ans as false; if (A[i] < B[i]) ans1 = false; if (B[i] < A[i]) ans2 = false; } // If any of the string can break return ans1 || ans2; } // Driver Code int main() { // Given string A and B string A = "abc", B = "xya"; // Function Call if (CanBreak(A, B)) cout << "Yes"; else cout << "No"; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to check if permutation of // one String can break permutation of // another String or not static boolean CanBreak(String A, String B) { // Sort both the Strings // in alphabetical order A = sortString(A); B = sortString(B); boolean ans1 = true, ans2 = true; // Iterate over the Strings for(int i = 0; i < A.length(); i++) { // If any of the String can break // other then mark ans as false; if (A.charAt(i) < B.charAt(i)) ans1 = false; if (B.charAt(i) < A.charAt(i)) ans2 = false; } // If any of the String can break return ans1 || ans2; } static String sortString(String inputString) { // Convert input string to char array char tempArray[] = inputString.toCharArray(); // Sort tempArray Arrays.sort(tempArray); // Return new sorted string return new String(tempArray); } // Driver Code public static void main(String[] args) { // Given String A and B String A = "abc", B = "xya"; // Function call if (CanBreak(A, B)) System.out.print("Yes"); else System.out.print("No"); } } // This code is contributed by gauravrajput1 |
Python3
# Python3 program for # the above approach # Function to check if # permutation of one string # can break permutation of # another string or not def CanBreak(A, B): # Sort both the strings # in alphabetical order A = "".join(sorted(A)) B = "".join(sorted(B)) ans1 = True ans2 = True # Iterate over the strings for i in range(len(A)): # If any of the string # can break other then # mark ans as false; if(A[i] < B[i]): ans1 = False if(B[i] < A[i]): ans2 = False # If any of the string # can break return (ans1 or ans2) # Driver Code # Given string A and B A = "abc"B = "xya" # Function Call if(CanBreak(A, B)): print("Yes") else: print("No") # This code is contributed by avanitrachhadiya2155 |
C#
// C# program for the above approach using System; class GFG{ // Function to check if permutation of // one String can break permutation of // another String or not static bool CanBreak(String A, String B) { // Sort both the Strings // in alphabetical order A = sortString(A); B = sortString(B); bool ans1 = true, ans2 = true; // Iterate over the Strings for(int i = 0; i < A.Length; i++) { // If any of the String can break // other then mark ans as false; if (A[i] < B[i]) ans1 = false; if (B[i] < A[i]) ans2 = false; } // If any of the String can break return ans1 || ans2; } static String sortString(String inputString) { // Convert input string to char array char []tempArray = inputString.ToCharArray(); // Sort tempArray Array.Sort(tempArray); // Return new sorted string return new String(tempArray); } // Driver Code public static void Main(String[] args) { // Given String A and B String A = "abc", B = "xya"; // Function call if (CanBreak(A, B)) Console.Write("Yes"); else Console.Write("No"); } } // This code is contributed by gauravrajput1 |
Javascript
<script> // JavaScript program for the above approach // Function to check if permutation of // one String can break permutation of // another String or not function CanBreak(A, B) { // Sort both the Strings // in alphabetical order A = sortString(A); B = sortString(B); let ans1 = true, ans2 = true; // Iterate over the Strings for(let i = 0; i < A.length; i++) { // If any of the String can break // other then mark ans as false; if (A[i] < B[i]) ans1 = false; if (B[i] < A[i]) ans2 = false; } // If any of the String can break return ans1 || ans2; } function sortString(inputString) { // Convert input string to char array let tempArray = inputString.split(''); // Sort tempArray tempArray.sort(); // Return new sorted string return new String(tempArray); } // Driver Code // Given String A and B let A = "abc", B = "xya"; // Function call if (CanBreak(A, B)) document.write("Yes"); else document.write("No"); </script> |
Output:
Yes
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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