LeetCode Notes: Find the Difference

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Question

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.

Solution One

Analysis:

Traverse each letter of the string t to determine whether the letter exists in the string s, if it exists, delete the letter in s, otherwise this letter is the added letter in t.

Code:

/**
 * @param {string} s
 * @param {string} t
 * @return {character}
 */
var findTheDifference = function (s, t) {
  for (const c of t) {
    if (s.indexOf(c) !== -1) {
      s = s.replace(c, "");
    } else {
      return c;
    }
  }
};

Solution Two

Analysis:

Add up the Unicode encodings of all the letters of the strings s and t respectively, then subtract the sum of the coded digits of s from the sum of the coded digits of t, the result is the Unicode encoding of the added letters in t.

Code:

/**
 * @param {string} s
 * @param {string} t
 * @return {character}
 */
var findTheDifference = function (s, t) {
  let as = 0,
    at = 0;
  for (const cs of s) {
    as += cs.charCodeAt();
  }
  for (const ct of t) {
    at += ct.charCodeAt();
  }
  return String.fromCharCode(at - as);
};

Solution Three

Analysis:

  1. Traverse s and count the number of occurrences of each letter of s
  2. Traverse t, deduct the same letters appearing in t from the statistical results, once you find that t has more letters than s, find the letters added in t

Code:

/**
 * @param {string}s
 * @param {string} t
 * @return {character}
 */
var findTheDifference = function (s, t) {
  // Construct an array of length 26 filled with all 0s
  const charCount = new Array(26).fill(0);

  // Get the char code of the letter a
  const aCode = "a".charCodeAt();

  // Count the number of occurrences of each letter in s
  // For example "acce" statistic result is charCount = [1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
  for (const cs of s) {
    charCount[cs.charCodeAt() - aCode]++;
  }

  // Subtract the letters in t from the above statistics
  for (const ct of t) {
    charCount[ct.charCodeAt() - aCode]--;

    // Once it is found that the number of letters in t is exceeded, find the added letter in t
    if (charCount[ct.charCodeAt() - aCode] < 0) {
      return ct;
    }
  }
};

For problems of the same type of solution, please refer to

LeetCode Notes: Ransom Note

Solution Four

Analysis:

Use the bitwise XOR feature to solve

  1. The commutative law is satisfied: a ^ b ^ c <=> a ^ c ^ b
  2. The XOR of any number and 0 is any number: 0 ^ n => n
  3. The XOR of the same number is 0: n ^ n => 0

We take the bitwise XOR of the Unicode encodings of all the letters of the characters s and t, and the final result is the Unicode encoding of the added letters in t.

For example, s = "abcd", t = "abcde",

0 ^ a ^ b ^ c ^ d ^ a ^ b ^ c ^ d ^ e

Equivalent to

a ^ a ^ b ^ b ^ c ^ c ^ d ^ d ^ e

Equivalent to

0 ^ 0 ^ 0 ^ 0 ^ e

Equivalent to

e

Code:

/**
 * @param {string} s
 * @param {string} t
 * @return {character}
 */
var findTheDifference = function (s, t) {
  let ret = 0;
  for (const ch of s) {
    ret ^= ch.charCodeAt();
  }
  for (const ch of t) {
    ret ^= ch.charCodeAt();
  }
  return String.fromCharCode(ret);
};

For problems of the same type of solution, please refer to

LeetCode Notes: Single Number

Reference

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