Question

389. Find the Difference

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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 $|\Sigma|=26$.

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

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 ' ';
  }
};

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 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
        )
    )
  }
}

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;
    }
  }
}

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 $O(1)$.

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)

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;
  }
};

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)
  }
}

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;
}