Description#
You are given two strings s
and t
.
String t
is generated by random shuffling string s
and then add one more letter at a random position.
Return the letter that was added to t
.
Example 1:
Input: s = "abcd", t = "abcde"
Output: "e"
Explanation: 'e' is the letter that was added.
Example 2:
Input: s = "", t = "y"
Output: "y"
Constraints:
0 <= s.length <= 1000
t.length == s.length + 1
s
and t
consist of lowercase English letters.
Solutions#
Solution 1: Counting#
We can use a hash table or array $cnt$ to count the occurrence of each character in string $s$, then traverse string $t$. For each character, we subtract its occurrence in $cnt$. If the corresponding count is negative, it means that the occurrence of this character in $t$ is greater than in $s$, so this character is the added character.
The time complexity is $O(n)$, and the space complexity is $O(|\Sigma|)$, where $n$ is the length of the string, and $\Sigma$ represents the character set. Here the character set is all lowercase letters, so $|\Sigma|=26$.
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| class Solution:
def findTheDifference(self, s: str, t: str) -> str:
cnt = Counter(s)
for c in t:
cnt[c] -= 1
if cnt[c] < 0:
return c
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| class Solution {
public char findTheDifference(String s, String t) {
int[] cnt = new int[26];
for (int i = 0; i < s.length(); ++i) {
++cnt[s.charAt(i) - 'a'];
}
for (int i = 0;; ++i) {
if (--cnt[t.charAt(i) - 'a'] < 0) {
return t.charAt(i);
}
}
}
}
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| class Solution {
public:
char findTheDifference(string s, string t) {
int cnt[26]{};
for (char& c : s) {
++cnt[c - 'a'];
}
for (char& c : t) {
if (--cnt[c - 'a'] < 0) {
return c;
}
}
return ' ';
}
};
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| func findTheDifference(s, t string) byte {
cnt := [26]int{}
for _, ch := range s {
cnt[ch-'a']++
}
for i := 0; ; i++ {
ch := t[i]
cnt[ch-'a']--
if cnt[ch-'a'] < 0 {
return ch
}
}
}
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| function findTheDifference(s: string, t: string): string {
const cnt: number[] = Array(26).fill(0);
for (const c of s) {
++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
}
for (const c of t) {
--cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
}
for (let i = 0; ; ++i) {
if (cnt[i] < 0) {
return String.fromCharCode(i + 'a'.charCodeAt(0));
}
}
}
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| impl Solution {
pub fn find_the_difference(s: String, t: String) -> char {
let s = s.as_bytes();
let t = t.as_bytes();
let n = s.len();
let mut count = [0; 26];
for i in 0..n {
count[(s[i] - b'a') as usize] += 1;
count[(t[i] - b'a') as usize] -= 1;
}
count[(t[n] - b'a') as usize] -= 1;
char::from(
b'a' +
(
count
.iter()
.position(|&v| v != 0)
.unwrap() as u8
)
)
}
}
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| char findTheDifference(char* s, char* t) {
int n = strlen(s);
int cnt[26] = {0};
for (int i = 0; i < n; i++) {
cnt[s[i] - 'a']++;
cnt[t[i] - 'a']--;
}
cnt[t[n] - 'a']--;
for (int i = 0;; i++) {
if (cnt[i]) {
return 'a' + i;
}
}
}
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Solution 2: Summation#
We can sum the ASCII values of each character in string $t$, then subtract the sum of the ASCII values of each character in string $s$. The final result is the ASCII value of the added character.
The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.
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| class Solution:
def findTheDifference(self, s: str, t: str) -> str:
a = sum(ord(c) for c in s)
b = sum(ord(c) for c in t)
return chr(b - a)
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| class Solution {
public char findTheDifference(String s, String t) {
int ss = 0;
for (int i = 0; i < t.length(); ++i) {
ss += t.charAt(i);
}
for (int i = 0; i < s.length(); ++i) {
ss -= s.charAt(i);
}
return (char) ss;
}
}
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| class Solution {
public:
char findTheDifference(string s, string t) {
int a = 0, b = 0;
for (char& c : s) {
a += c;
}
for (char& c : t) {
b += c;
}
return b - a;
}
};
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| func findTheDifference(s string, t string) byte {
ss := 0
for _, c := range s {
ss -= int(c)
}
for _, c := range t {
ss += int(c)
}
return byte(ss)
}
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| function findTheDifference(s: string, t: string): string {
return String.fromCharCode(
[...t].reduce((r, v) => r + v.charCodeAt(0), 0) -
[...s].reduce((r, v) => r + v.charCodeAt(0), 0),
);
}
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| impl Solution {
pub fn find_the_difference(s: String, t: String) -> char {
let mut ans = 0;
for c in s.as_bytes() {
ans ^= c;
}
for c in t.as_bytes() {
ans ^= c;
}
char::from(ans)
}
}
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| char findTheDifference(char* s, char* t) {
int n = strlen(s);
char ans = 0;
for (int i = 0; i < n; i++) {
ans ^= s[i];
ans ^= t[i];
}
ans ^= t[n];
return ans;
}
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