Question

1370. Increasing Decreasing String

中文文档

Description

You are given a string s. Reorder the string using the following algorithm:

1. Pick the smallest character from s and append it to the result.
2. Pick the smallest character from s which is greater than the last appended character to the result and append it.
3. Repeat step 2 until you cannot pick more characters.
4. Pick the largest character from s and append it to the result.
5. Pick the largest character from s which is smaller than the last appended character to the result and append it.
6. Repeat step 5 until you cannot pick more characters.
7. Repeat the steps from 1 to 6 until you pick all characters from s.

In each step, If the smallest or the largest character appears more than once you can choose any occurrence and append it to the result.

Return _the result string after sorting _s_ with this algorithm_.

 

Example 1:

Input: s = "aaaabbbbcccc"
Output: "abccbaabccba"
Explanation: After steps 1, 2 and 3 of the first iteration, result = "abc"
After steps 4, 5 and 6 of the first iteration, result = "abccba"
First iteration is done. Now s = "aabbcc" and we go back to step 1
After steps 1, 2 and 3 of the second iteration, result = "abccbaabc"
After steps 4, 5 and 6 of the second iteration, result = "abccbaabccba"

Example 2:

Input: s = "rat"
Output: "art"
Explanation: The word "rat" becomes "art" after re-ordering it with the mentioned algorithm.

 

Constraints:

• 1 <= s.length <= 500
• s consists of only lowercase English letters.

Solutions

Solution 1: Counting + Simulation

First, we use a hash table or an array $cnt$ of length $26$ to count the number of occurrences of each character in the string $s$.

Then, we enumerate the letters $[a,…,z]$. For the current enumerated letter $c$, if $cnt[c] > 0$, we append the letter $c$ to the end of the answer string and decrease $cnt[c]$ by one. We repeat this step until $cnt[c] = 0$. Then we enumerate the letters $[z,…,a]$ in reverse order and perform similar operations. If the length of the answer string equals the length of $s$, then we have completed all the concatenation operations.

The time complexity is $O(n \times |\Sigma|)$, and the space complexity is $O(|\Sigma|)$. Where $n$ is the length of the string $s$, and $\Sigma$ is the character set. In this problem, the character set is all lowercase letters, so $|\Sigma| = 26$.

class Solution:
  def sortString(self, s: str) -> str:
    cnt = Counter(s)
    cs = ascii_lowercase + ascii_lowercase[::-1]
    ans = []
    while len(ans) < len(s):
      for c in cs:
        if cnt[c]:
          ans.append(c)
          cnt[c] -= 1
    return "".join(ans)

class Solution {
  public String sortString(String s) {
    int[] cnt = new int[26];
    int n = s.length();
    for (int i = 0; i < n; ++i) {
      cnt[s.charAt(i) - 'a']++;
    }
    StringBuilder sb = new StringBuilder();
    while (sb.length() < n) {
      for (int i = 0; i < 26; ++i) {
        if (cnt[i] > 0) {
          sb.append((char) ('a' + i));
          --cnt[i];
        }
      }
      for (int i = 25; i >= 0; --i) {
        if (cnt[i] > 0) {
          sb.append((char) ('a' + i));
          --cnt[i];
        }
      }
    }
    return sb.toString();
  }
}

class Solution {
public:
  string sortString(string s) {
    int cnt[26]{};
    for (char& c : s) {
      ++cnt[c - 'a'];
    }
    string ans;
    while (ans.size() < s.size()) {
      for (int i = 0; i < 26; ++i) {
        if (cnt[i]) {
          ans += i + 'a';
          --cnt[i];
        }
      }
      for (int i = 25; i >= 0; --i) {
        if (cnt[i]) {
          ans += i + 'a';
          --cnt[i];
        }
      }
    }
    return ans;
  }
};

func sortString(s string) string {
  cnt := [26]int{}
  for _, c := range s {
    cnt[c-'a']++
  }
  n := len(s)
  ans := make([]byte, 0, n)
  for len(ans) < n {
    for i := 0; i < 26; i++ {
      if cnt[i] > 0 {
        ans = append(ans, byte(i)+'a')
        cnt[i]--
      }
    }
    for i := 25; i >= 0; i-- {
      if cnt[i] > 0 {
        ans = append(ans, byte(i)+'a')
        cnt[i]--
      }
    }
  }
  return string(ans)
}

function sortString(s: string): string {
  const cnt: number[] = Array(26).fill(0);
  for (const c of s) {
    ++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
  }
  const ans: string[] = [];
  while (ans.length < s.length) {
    for (let i = 0; i < 26; ++i) {
      if (cnt[i]) {
        ans.push(String.fromCharCode(i + 'a'.charCodeAt(0)));
        --cnt[i];
      }
    }
    for (let i = 25; i >= 0; --i) {
      if (cnt[i]) {
        ans.push(String.fromCharCode(i + 'a'.charCodeAt(0)));
        --cnt[i];
      }
    }
  }
  return ans.join('');
}

/**
 * @param {string} s
 * @return {string}
 */
var sortString = function (s) {
  const cnt = Array(26).fill(0);
  for (const c of s) {
    ++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
  }
  const ans = [];
  while (ans.length < s.length) {
    for (let i = 0; i < 26; ++i) {
      if (cnt[i]) {
        ans.push(String.fromCharCode(i + 'a'.charCodeAt(0)));
        --cnt[i];
      }
    }
    for (let i = 25; i >= 0; --i) {
      if (cnt[i]) {
        ans.push(String.fromCharCode(i + 'a'.charCodeAt(0)));
        --cnt[i];
      }
    }
  }
  return ans.join('');
};