Given two strings, str1 and str2, the task is to find and print the uncommon characters of the two given strings in sorted order without using extra space. Here, an uncommon character means that either the character is present in one string or it is present in the other string but not in both. The strings contain only lowercase characters and can contain duplicates.
Examples:
Input: str1 = “characters”, str2 = “alphabets”
Output: b c l p rInput: str1 = “neveropen”, str2 = “neveropenquiz”
Output: f i o q r u z
Approach: An approach that uses hashing has been discussed here. This problem can also be solved using bit operations.
The approach uses 2 variables that store the bit-wise OR of the left shift of 1 with each character’s ASCII code – 97 i.e. 0 for ‘a’, 1 for ‘b’, and so on. For both the strings, we get an integer after performing these bit-wise operations. Now the XOR of these two integers will give the binary bit as 1 at only those positions that denote uncommon characters. Print the character values for those positions.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print the uncommon// characters in the given string// in sorted ordervoid printUncommon(string str1, string str2){ int a1 = 0, a2 = 0; for (int i = 0; i < str1.length(); i++) { // Converting character to ASCII code int ch = int(str1[i]) - 'a'; // Bit operation a1 = a1 | (1 << ch); } for (int i = 0; i < str2.length(); i++) { // Converting character to ASCII code int ch = int(str2[i]) - 'a'; // Bit operation a2 = a2 | (1 << ch); } // XOR operation leaves only uncommon // characters in the ans variable int ans = a1 ^ a2; int i = 0; while (i < 26) { if (ans % 2 == 1) { cout << char('a' + i); } ans = ans / 2; i++; }}// Driver codeint main(){ string str1 = "neveropen"; string str2 = "neveropenquiz"; printUncommon(str1, str2); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to print the uncommon // characters in the given string // in sorted order static void printUncommon(String str1, String str2) { int a1 = 0, a2 = 0; for (int i = 0; i < str1.length(); i++) { // Converting character to ASCII code int ch = (str1.charAt(i)) - 'a'; // Bit operation a1 = a1 | (1 << ch); } for (int i = 0; i < str2.length(); i++) { // Converting character to ASCII code int ch = (str2.charAt(i)) - 'a'; // Bit operation a2 = a2 | (1 << ch); } // XOR operation leaves only uncommon // characters in the ans variable int ans = a1 ^ a2; int i = 0; while (i < 26) { if (ans % 2 == 1) { System.out.print((char) ('a' + i)); } ans = ans / 2; i++; } } // Driver code public static void main(String[] args) { String str1 = "neveropen"; String str2 = "neveropenquiz"; printUncommon(str1, str2); }}// This code contributed by Rajput-Ji |
C#
// C# implementation of the approachusing System;class GFG{ // Function to print the uncommon// characters in the given string// in sorted orderstatic void printUncommon(string str1, string str2){ int a1 = 0, a2 = 0; for (int i = 0; i < str1.Length; i++) { // Converting character to ASCII code int ch = (str1[i] - 'a'); // Bit operation a1 = a1 | (1 << ch); } for (int i = 0; i < str2.Length; i++) { // Converting character to ASCII code int ch = (str2[i] - 'a'); // Bit operation a2 = a2 | (1 << ch); } // XOR operation leaves only uncommon // characters in the ans variable int ans = a1 ^ a2; int j = 0; while (j < 26) { if (ans % 2 == 1) { Console.Write((char)('a' + j)); } ans = ans / 2; j++; }}// Driver codepublic static void Main(){ string str1 = "neveropen"; string str2 = "neveropenquiz"; printUncommon(str1, str2);}}// This code is contributed by SoM15242 |
Python3
# Python3 implementation of the approach # Function to print the uncommon # characters in the given string # in sorted order def printUncommon(str1, str2) : a1 = 0; a2 = 0; for i in range(len(str1)) : # Converting character to ASCII code ch = ord(str1[i]) - ord('a'); # Bit operation a1 = a1 | (1 << ch); for i in range(len(str2)) : # Converting character to ASCII code ch = ord(str2[i]) - ord('a'); # Bit operation a2 = a2 | (1 << ch); # XOR operation leaves only uncommon # characters in the ans variable ans = a1 ^ a2; i = 0; while (i < 26) : if (ans % 2 == 1) : print(chr(ord('a') + i),end=""); ans = ans // 2; i += 1; # Driver code if __name__ == "__main__" : str1 = "neveropen"; str2 = "neveropenquiz"; printUncommon(str1, str2); # This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approach// Function to print the uncommon// characters in the given string// in sorted orderfunction printUncommon(str1, str2){ var a1 = 0, a2 = 0; for (var i = 0; i < str1.length; i++) { // Converting character to ASCII code var ch = (str1[i].charCodeAt(0)) - 'a'.charCodeAt(0); // Bit operation a1 = a1 | (1 << ch); } for (var i = 0; i < str2.length; i++) { // Converting character to ASCII code var ch = (str2[i].charCodeAt(0)) - 'a'.charCodeAt(0); // Bit operation a2 = a2 | (1 << ch); } // XOR operation leaves only uncommon // characters in the ans variable var ans = a1 ^ a2; var i = 0; while (i < 26) { if (ans % 2 == 1) { document.write( String.fromCharCode('a'.charCodeAt(0) + i)); } ans = parseInt(ans / 2); i++; }}// Driver codevar str1 = "neveropen";var str2 = "neveropenquiz";printUncommon(str1, str2);</script> |
Output:
fioqruz
Time Complexity: O(|str1| + |str2| + 26)Auxiliary Space: O(1)
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