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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 <= 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 $|\Sigma|=26$.

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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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)
      }
  }
  
  

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