Leetcode 389. Find the Difference
Problem Explanation
Given two strings s
and t
, we want to find out the letter that was added in t
. String t
is created by shuffling string s
and adding an extra letter in a random position. Both strings contain only lowercase letters.
Walk Through an Example
For example, if s = "abcd"
and t = "abcde"
, the letter 'e'
was added in t
.
Approach Explanation
The problem can be solved using XOR ('^') operation in bit manipulation. The XOR operation can return a unique result given two inputs. If we take XOR of zero and a bit, it returns that bit
1 2 3 0 ^ 1 = 1 4 0 ^ 0 = 0
If we take XOR of two same bits, it returns 0
1 2 3 1 ^ 1 = 0 4 0 ^ 0 = 0
So if we take XOR of all the bits together, we get the unique bit, which is the letter added in 't'.
ASCII Illustration
1 2 3String s = "abcd" 4 5 a b c d 6 โ โ โ โ 7 1101 1110 1100 1101 => XOR operation on all bits of s 8 9 10String t = "abcde" 11 12 a b c d e 13 โ โ โ โ โ 14 1101 1110 1100 1101 1100 => XOR operation on all bits of t 15 16XOR operation on all bits of t and s = added letter 'e'
Solutions
Now, let's look at how to implement this solution in various programming languages.
Python Solution
1 2 python 3class Solution: 4 def findTheDifference(self, s: str, t: str) -> str: 5 ans = 0 6 # XOR operation on all characters in string s 7 for c in s: 8 ans ^= ord(c) 9 # XOR operation on all characters in string t 10 for c in t: 11 ans ^= ord(c) 12 # return the added character. 13 return chr(ans)
Java Solution
1
2 java
3class Solution {
4 public char findTheDifference(String s, String t) {
5 char ans = 0;
6 // XOR operation on all characters in string s
7 for (char c : s.toCharArray()) {
8 ans ^= c;
9 }
10 // XOR operation on all characters in string t
11 for (char c : t.toCharArray()) {
12 ans ^= c;
13 }
14 // return the added character
15 return ans;
16 }
17}
Javascript Solution
1
2javascript
3var findTheDifference = function(s, t) {
4 let ans = 0;
5 // XOR operation on all characters in string s
6 for (let i = 0; i < s.length; i++) {
7 ans ^= s.charCodeAt(i);
8 }
9 // XOR operation on all characters in string t
10 for (let i = 0; i < t.length; i++) {
11 ans ^= t.charCodeAt(i);
12 }
13 // return the added character
14 return String.fromCharCode(ans);
15};
C++ Solution
1
2 C++
3class Solution {
4public:
5 char findTheDifference(string s, string t) {
6 char ans = 0;
7 // XOR operation on all characters in string s
8 for (char c : s){
9 ans ^= c;
10 }
11 // XOR operation on all characters in string t
12 for (char c : t) {
13 ans ^= c;
14 }
15 // return the added character
16 return ans;
17 }
18};
C# Solution
1
2 C#
3public class Solution {
4 public char FindTheDifference(string s, string t) {
5 char ans = (char)0;
6 // XOR operation on all characters in string s
7 for (int i = 0; i < s.Length; i++) {
8 ans ^= s[i];
9 }
10 // XOR operation on all characters in string t
11 for (int i = 0; i < t.Length; i++) {
12 ans ^= t[i];
13 }
14 // return the added character
15 return ans;
16 }
17}
Final Thoughts
In each solution, the XOR operation helps to safely ignore all the characters that are common in both strings and leaves behind the character that was added in t
. Thus the function returns the character that was added in string t
.The mentioned approach makes use of bit manipulation, which is efficient but might not be straightforward for everyone. Alternatively, another solution could be to use a hash table or dictionary. Hashing works by assigning each unique character to a unique key in a table.
The dictionary-based solution would work as follows:
- Initialize an empty dictionary.
- Iterate through string 's', for each character 'c', add it to the dictionary. If 'c' already exists in the dictionary, increment its count.
- Iterate through string 't', for each character 'c', check if it is in the dictionary. If it is, decrease its count; if it's not, 'c' is the character that was added.
- Return the character added in 't'.
Here are the implementations of the dictionary-based solution:
Python Solution
1
2 python
3from collections import defaultdict
4
5def findTheDifference(s: str, t: str) -> str:
6 letters = defaultdict(int)
7 # count the letters in s
8 for c in s:
9 letters[c] += 1
10 # decrease the count of the letters found in t
11 for c in t:
12 if letters[c] == 0:
13 return c
14 letters[c] -= 1
15
Java Solution
1
2 java
3import java.util.*;
4
5public class Main {
6 public static char findTheDifference(String s, String t) {
7 int[] alphabet = new int[26];
8 for(char c : s.toCharArray()){
9 alphabet[c - 'a']++;
10 }
11
12 for(char c : t.toCharArray()){
13 if(--alphabet[c - 'a'] < 0){
14 return c;
15 }
16 }
17 return 0;
18 }
19}
JavaScript Solution
1 2 javascript 3var findTheDifference = function(s, t) { 4 let charCount = {}; 5 for (let i=0; i < s.length; i++){ 6 charCount[s[i]] = charCount[s[i]]+1 || 1; 7 } 8 9 for (let i=0; i < t.length; i++){ 10 if (!charCount[t[i]]) return t[i]; 11 charCount[t[i]] -= 1; 12 } 13};
These solutions also provide an efficient way to find the additional character. They are easier to understand but require more space (for storing the dictionary) compared to the XOR operation.
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