Write a function to check whether two given strings are Permutation of each other or not. A Permutation of a string is another string that contains same characters, only the order of characters can be different. For example, “abcd” and “dabc” are Permutation of each other.
We strongly recommend that you click here and practice it, before moving on to the solution.
Method 1 (Use Sorting)
1) Sort both strings
2) Compare the sorted strings
C++
// C++ program to check whether two strings are // Permutations of each other#include <bits/stdc++.h>using namespace std;/* function to check whether two strings are Permutation of each other */bool arePermutation(string str1, string str2){ // Get lengths of both strings int n1 = str1.length(); int n2 = str2.length(); // If length of both strings is not same, // then they cannot be Permutation if (n1 != n2) return false; // Sort both strings sort(str1.begin(), str1.end()); sort(str2.begin(), str2.end()); // Compare sorted strings for (int i = 0; i < n1; i++) if (str1[i] != str2[i]) return false; return true;}/* Driver program to test to print printDups*/int main(){ string str1 = "test"; string str2 = "ttew"; if (arePermutation(str1, str2)) printf("Yes"); else printf("No"); return 0;} |
Java
// Java program to check whether two strings are // Permutations of each other import java.util.*;class GfG {/* function to check whether two strings are Permutation of each other */static boolean arePermutation(String str1, String str2) { // Get lengths of both strings int n1 = str1.length(); int n2 = str2.length(); // If length of both strings is not same, // then they cannot be Permutation if (n1 != n2) return false; char ch1[] = str1.toCharArray(); char ch2[] = str2.toCharArray(); // Sort both strings Arrays.sort(ch1); Arrays.sort(ch2); // Compare sorted strings for (int i = 0; i < n1; i++) if (ch1[i] != ch2[i]) return false; return true; } /* Driver program to test to print printDups*/public static void main(String[] args) { String str1 = "test"; String str2 = "ttew"; if (arePermutation(str1, str2)) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python3 program to check whether two # strings are Permutations of each other# function to check whether two strings # are Permutation of each other */def arePermutation(str1, str2): # Get lengths of both strings n1 = len(str1) n2 = len(str2) # If length of both strings is not same, # then they cannot be Permutation if (n1 != n2): return False # Sort both strings a = sorted(str1) str1 = " ".join(a) b = sorted(str2) str2 = " ".join(b) # Compare sorted strings for i in range(0, n1, 1): if (str1[i] != str2[i]): return False return True# Driver Codeif __name__ == '__main__': str1 = "test" str2 = "ttew" if (arePermutation(str1, str2)): print("Yes") else: print("No")# This code is contributed by# Sahil_Shelangia |
C#
// C# program to check whether two strings are // Permutations of each otherusing System;class GfG {/* function to check whether two strings are Permutation of each other */static bool arePermutation(String str1, String str2) { // Get lengths of both strings int n1 = str1.Length; int n2 = str2.Length; // If length of both strings is not same, // then they cannot be Permutation if (n1 != n2) return false; char []ch1 = str1.ToCharArray(); char []ch2 = str2.ToCharArray(); // Sort both strings Array.Sort(ch1); Array.Sort(ch2); // Compare sorted strings for (int i = 0; i < n1; i++) if (ch1[i] != ch2[i]) return false; return true; } /* Driver code*/public static void Main(String[] args) { String str1 = "test"; String str2 = "ttew"; if (arePermutation(str1, str2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code contributed by Rajput-Ji |
Javascript
<script>// Javascript program to check whether two// strings are Permutations of each other// Function to check whether two strings are// Permutation of each otherfunction arePermutation(str1, str2){ // Get lengths of both strings let n1 = str1.length; let n2 = str2.length; // If length of both strings is not same, // then they cannot be Permutation if (n1 != n2) return false; let ch1 = str1.split(' '); let ch2 = str2.split(' '); // Sort both strings ch1.sort(); ch2.sort(); // Compare sorted strings for(let i = 0; i < n1; i++) if (ch1[i] != ch2[i]) return false; return true;}// Driver Codelet str1 = "test";let str2 = "ttew";if (arePermutation(str1, str2)) document.write("Yes");else document.write("No");// This code is contributed by suresh07</script> |
Output:
No
Time Complexity: Time complexity of this method depends upon the sorting technique used. In the above implementation, quickSort is used which may be O(n^2) in worst case. If we use a O(nLogn) sorting algorithm like merge sort, then the complexity becomes O(nLogn)
Auxiliary space: O(1).
Method 2 (Count characters)
This method assumes that the set of possible characters in both strings is small. In the following implementation, it is assumed that the characters are stored using 8 bit and there can be 256 possible characters.
1) Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
2) Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
3) Compare count arrays. If both count arrays are same, then return true.
C++
// C++ program to check whether two strings are // Permutations of each other#include <bits/stdc++.h>using namespace std;# define NO_OF_CHARS 256/* function to check whether two strings are Permutation of each other */bool arePermutation(string str1, string str2){ // Create 2 count arrays and initialize // all values as 0 int count1[NO_OF_CHARS] = {0}; int count2[NO_OF_CHARS] = {0}; int i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; str1[i] && str2[i]; i++) { count1[str1[i]]++; count2[str2[i]]++; } // If both strings are of different length. // Removing this condition will make the // program fail for strings like "aaca" // and "aca" if (str1[i] || str2[i]) return false; // Compare count arrays for (i = 0; i < NO_OF_CHARS; i++) if (count1[i] != count2[i]) return false; return true;}/* Driver program to test to print printDups*/int main(){ string str1 = "neveropen"; string str2 = "forneveropenneveropen"; if ( arePermutation(str1, str2) ) printf("Yes"); else printf("No"); return 0;} |
Java
// JAVA program to check if two strings// are Permutations of each otherimport java.io.*;import java.util.*;class GFG{ static int NO_OF_CHARS = 256; /* function to check whether two strings are Permutation of each other */ static boolean arePermutation(char str1[], char str2[]) { // Create 2 count arrays and initialize // all values as 0 int count1[] = new int [NO_OF_CHARS]; Arrays.fill(count1, 0); int count2[] = new int [NO_OF_CHARS]; Arrays.fill(count2, 0); int i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; i <str1.length && i < str2.length ; i++) { count1[str1[i]]++; count2[str2[i]]++; } // If both strings are of different length. // Removing this condition will make the program // fail for strings like "aaca" and "aca" if (str1.length != str2.length) return false; // Compare count arrays for (i = 0; i < NO_OF_CHARS; i++) if (count1[i] != count2[i]) return false; return true; } /* Driver program to test to print printDups*/ public static void main(String args[]) { char str1[] = ("neveropen").toCharArray(); char str2[] = ("forneveropenneveropen").toCharArray(); if ( arePermutation(str1, str2) ) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Nikita Tiwari. |
Python
# Python program to check if two strings are # Permutations of each otherNO_OF_CHARS = 256# Function to check whether two strings are# Permutation of each otherdef arePermutation(str1, str2): # Create two count arrays and initialize # all values as 0 count1 = [0] * NO_OF_CHARS count2 = [0] * NO_OF_CHARS # For each character in input strings, # increment count in the corresponding # count array for i in str1: count1[ord(i)]+=1 for i in str2: count2[ord(i)]+=1 # If both strings are of different length. # Removing this condition will make the # program fail for strings like # "aaca" and "aca" if len(str1) != len(str2): return 0 # Compare count arrays for i in xrange(NO_OF_CHARS): if count1[i] != count2[i]: return 0 return 1# Driver program to test the above functionsstr1 = "neveropen"str2 = "forneveropenneveropen"if arePermutation(str1, str2): print "Yes"else: print "No"# This code is contributed by Bhavya Jain |
C#
// C# program to check if two strings// are Permutations of each otherusing System;class GFG{ static int NO_OF_CHARS = 256; /* function to check whether two strings are Permutation of each other */ static bool arePermutation(char []str1, char []str2) { // Create 2 count arrays and initialize // all values as 0 int []count1 = new int [NO_OF_CHARS]; int []count2 = new int [NO_OF_CHARS]; int i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; i <str1.Length && i < str2.Length ; i++) { count1[str1[i]]++; count2[str2[i]]++; } // If both strings are of different length. // Removing this condition will make the program // fail for strings like "aaca" and "aca" if (str1.Length != str2.Length) return false; // Compare count arrays for (i = 0; i < NO_OF_CHARS; i++) if (count1[i] != count2[i]) return false; return true; } /* Driver code*/ public static void Main(String []args) { char []str1 = ("neveropen").ToCharArray(); char []str2 = ("forneveropenneveropen").ToCharArray(); if ( arePermutation(str1, str2) ) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code has been contributed by 29AjayKumar |
Javascript
<script>// Javascript program to check if two strings// are Permutations of each otherlet NO_OF_CHARS = 256; /* Function to check whether two stringsare Permutation of each other */function arePermutation(str1, str2){ // Create 2 count arrays and initialize // all values as 0 let count1 = Array(NO_OF_CHARS); let count2 = Array(NO_OF_CHARS); count1.fill(0); count2.fill(0); let i; // For each character in input strings, // increment count in the corresponding // count array for(i = 0; i < str1.length && i < str2.length; i++) { count1[str1[i]]++; count2[str2[i]]++; } // If both strings are of different length. // Removing this condition will make the program // fail for strings like "aaca" and "aca" if (str1.length != str2.length) return false; // Compare count arrays for(i = 0; i < NO_OF_CHARS; i++) if (count1[i] != count2[i]) return false; return true;}// Driver code let str1 = ("neveropen").split('');let str2 = ("forneveropenneveropen").split('');if (arePermutation(str1, str2) ) document.write("Yes");else document.write("No"); // This code is contributed by rameshtravel07</script> |
Output:
Yes
Time Complexity: O(n)
Auxiliary space: O(n).
The above implementation can be further to use only one count array instead of two. We can increment the value in count array for characters in str1 and decrement for characters in str2. Finally, if all count values are 0, then the two strings are Permutation of each other. Thanks to Ace for suggesting this optimization.
C++
// C++ function to check whether two strings are // Permutations of each otherbool arePermutation(string str1, string str2){ // Create a count array and initialize all // values as 0 int count[NO_OF_CHARS] = {0}; int i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; str1[i] && str2[i]; i++) { count[str1[i]]++; count[str2[i]]--; } // If both strings are of different length. // Removing this condition will make the // program fail for strings like "aaca" and // "aca" if (str1[i] || str2[i]) return false; // See if there is any non-zero value in // count array for (i = 0; i < NO_OF_CHARS; i++) if (count[i]) return false; return true;} |
Java
// Java function to check whether two strings are// Permutations of each otherstatic boolean arePermutation(char str1[], char str2[]){ // If both strings are of different length. // Removing this condition will make the // program fail for strings like "aaca" and // "aca" if (str1.length != str2.length) return false; // Create a count array and initialize all // values as 0 int count[] = new int[NO_OF_CHARS]; int i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; i < str1.length && i < str2.length; i++) { count[str1[i]]++; count[str2[i]]--; } // See if there is any non-zero value in // count array for (i = 0; i < NO_OF_CHARS; i++) if (count[i] != 0) return false; return true;}// This code is contributed by divyesh072019. |
Python3
# Python3 function to check whether two strings are # Permutations of each otherdef arePermutation(str1, str2): # Create a count array and initialize all # values as 0 count = [0 for i in range(NO_OF_CHARS)] i = 0 # For each character in input strings, # increment count in the corresponding # count array while(str1[i] and str2[i]): count[str1[i]] += 1 count[str2[i]] -= 1 # If both strings are of different length. # Removing this condition will make the # program fail for strings like "aaca" and # "aca" if (str1[i] or str2[i]): return False; # See if there is any non-zero value in # count array for i in range(NO_OF_CHARS): if (count[i]): return False; return True;# This code is contributed by pratham76. |
C#
// C# function to check whether two strings are // Permutations of each other static bool arePermutation(char[] str1, char[] str2) { // Create a count array and initialize all // values as 0 int[] count = new int[NO_OF_CHARS]; int i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; str1[i] && str2[i]; i++) { count[str1[i]]++; count[str2[i]]--; } // If both strings are of different length. // Removing this condition will make the // program fail for strings like "aaca" and // "aca" if (str1[i] || str2[i]) return false; // See if there is any non-zero value in // count array for (i = 0; i < NO_OF_CHARS; i++) if (count[i] != 0) return false; return true; } // This code is contributed by divyeshrabadiya07. |
Javascript
<script>// Javascript function to check whether two strings are // Permutations of each otherfunction arePermutation(str1,str2){ // Create a count array and initialize all // values as 0 let count = new Array(NO_OF_CHARS); let i; // For each character in input strings, // increment count in the corresponding // count array for (i = 0; str1[i] && str2[i]; i++) { count[str1[i]]++; count[str2[i]]--; } // If both strings are of different length. // Removing this condition will make the // program fail for strings like "aaca" and // "aca" if (str1[i] || str2[i]) return false; // See if there is any non-zero value in // count array for (i = 0; i < NO_OF_CHARS; i++) if (count[i] != 0) return false; return true;}// This code is contributed by patel2127</script> |
If the possible set of characters contains only English alphabets, then we can reduce the size of arrays to 58 and use str[i] – ‘A’ as an index for count arrays because ASCII value of ‘A’ is 65 , ‘B’ is 66, ….. , Z is 90 and ‘a’ is 97 , ‘b’ is 98 , …… , ‘z’ is 122. This will further optimize this method.
Time Complexity: O(n)
Auxiliary space: O(n).
Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
