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389. Find the Difference
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 <= 1000t.length == s.length + 1sandtconsist 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 $|\Sigma|=26$.
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 { 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); } } } } -
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 ' '; } }; -
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 -
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 } } } -
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)); } } } -
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) } } -
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; } } -
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; } }; -
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; } } } -
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; } -
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) -
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) } -
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), ); } -
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) } }