Question

Formatted question description: https://leetcode.ca/all/1370.html

 1370. Increasing Decreasing String

 Given a string s. You should re-order the string using the following algorithm:

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


 Example 3:

 Input: s = "leetcode"
 Output: "cdelotee"


 Example 4:

 Input: s = "ggggggg"
 Output: "ggggggg"
 Example 5:

 Input: s = "spo"
 Output: "ops"


 Constraints:

 1 <= s.length <= 500
 s contains only lower-case English letters.

Algorithm

The rule of arrangement is lexicographically from smallest to largest, and then from the letter with the largest lexicographic order to the smallest letter. The idea is to use a freq array to count the number of occurrences of 26 lowercase letters, and then re-splice the letters into strings according to the rules.

Time O(n)

Space O(n)

Code

Java

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

            for (int i = 25; i >= 0; i--) {
                if (freq[i] > 0) {
                    sb.append((char) (i + 'a'));
                    freq[i]--;
                    count++;
                }
            }
        }
        return sb.toString();
    }
}

Java

class Solution {
    public String sortString(String s) {
        int[] counts = new int[26];
        int length = s.length();
        for (int i = 0; i < length; i++) {
            char c = s.charAt(i);
            counts[c - 'a']++;
        }
        StringBuffer sb = new StringBuffer();
        int remaining = length;
        boolean increasing = true;
        while (remaining > 0) {
            if (increasing) {
                for (int i = 0; i < 26; i++) {
                    if (counts[i] > 0) {
                        char c = (char) ('a' + i);
                        sb.append(c);
                        counts[i]--;
                        remaining--;
                    }
                }
            } else {
                for (int i = 25; i >= 0; i--) {
                    if (counts[i] > 0) {
                        char c = (char) ('a' + i);
                        sb.append(c);
                        counts[i]--;
                        remaining--;
                    }
                }
            }
            increasing = !increasing;
        }
        return sb.toString();
    }
}

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