1370. Increasing Decreasing String

EasyHash TableStringCounting

Problem Description

The given problem requires us to reorder a given string s following a specific algorithm. The algorithm involves repeatedly picking the smallest and largest characters from the string according to given rules and appending them to form a new result string. The rules for picking characters are as follows:

Select the smallest character from s and append it to the result.
Choose the next-smallest character that is greater than the last appended one and append it.
Continue with step 2 until no more characters can be picked.
Select the largest character from s and append it to the result.
Choose the next-largest character that is smaller than the last appended one and append it.
Continue with step 5 until no more characters can be picked.
Repeat steps 1 to 6 until all characters from s have been picked and appended.

The algorithm allows for any occurrence of the smallest or largest character to be chosen if there are multiple.

Intuition

The solution to this problem uses a frequency counter to keep track of how many times each character appears in the string s. We use an array counter of length 26 to represent the frequency of each lowercase English letter.

Here's how the solution is constructed:

Count the frequency of each character in the string s and store it in counter. This allows us to know how many times we need to pick each character during the reordering process.
Initialize an empty list ans to build up the result string.
Use a while loop that continues until the length of ans matches the length of the input string s.
Inside the loop, iterate over the counter from start to end to append the smallest character (if available) to ans and decrease its count.
Then iterate over the counter from end to start to append the largest character (if available) to ans and decrease its count.
The process repeats, alternating between picking the smallest and largest character until all characters are used.
Convert the list ans to a string and return it as the final reordered string.

By following these steps, the algorithm efficiently fulfills the provided reordering rules, and characters are picked in the required order to form the result string.

Solution Approach

The solution implementation utilizes a straightforward approach that involves counting, iterating, and string building.

Here's a detailed walk-through of the implementation using Python:

Character Frequency Counting: We start by creating a list of zeroes called counter to maintain a count of each letter in the string. The length of the list is 26, one for each letter of the English alphabet. We then iterate over each character of the string s, and for each character, we find its corresponding index (0 for 'a', 1 for 'b', etc.) by subtracting the ASCII value of 'a' from the ASCII value of the character. We increment the count at this index in the counter list.
```
counter = [0] * 26
for c in s:
    counter[ord(c) - ord('a')] += 1
```
Result String Assembly: We define a list ans to accumulate the characters in the order we choose them based on the algorithm rules—the resulting string after the reordering will be formed by concatenating the characters in this list.
Main Loop - Building the Result: We use a while loop to repeat the process of picking characters from the string s according to the described algorithm. The loop will continue until the length of ans becomes equal to the length of the original string s, signaling that all characters have been chosen and appended.
```
while len(ans) < len(s):
    # ...
```

Picking the Smallest Character: Inside the loop, we iterate over the counter from the start (0) to the end (25) which corresponds to characters 'a' to 'z'. If the current character's counter is not zero, indicating that it is available to be picked, we append the corresponding character to ans and decrement its count in counter.

for i in range(26):
    if counter[i]:
        ans.append(chr(i + ord('a')))
        counter[i] -= 1

Picking the Largest Character: We do the same for picking the largest character, but in reverse order, iterating over the counter from the end (25) to the start (0). Again, if the current character's counter is not zero, we append the corresponding character to ans and decrement its count.

for i in range(25, -1, -1):
    if counter[i]:
        ans.append(chr(i + ord('a')))
        counter[i] -= 1

Returning the Result: Finally, after the while loop concludes, we join the list of characters in ans using ''.join(ans) to return the final string, which is the original string s reordered according to the algorithm described.

The above steps translate the problem's algorithm into Python code in a way that is both efficient (since the actions within the loop are simple and fast) and clean, leading to a solution that straightforwardly follows the rules laid out in the problem statement.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose our input string s is "bacab".

Character Frequency Counting: We count the frequency of every character in the string.

For the string "bacab", the frequency counter counter would look like this after counting: [2, 2, 1], where 2 at the first position represents 'a', 2 at the second position is for 'b', and 1 at the third position is for 'c'.
Result String Assembly: We initialize an empty list ans to store the characters as we pick them following the algorithm.
Main Loop - Building the Result: We start the while loop since len(ans) < len(s), which is 5 in this case.
Picking the Smallest Character: On the first iteration, we look for the smallest character, which is 'a'. We add 'a' to ans, and the counter becomes [1, 2, 1].
Picking the Largest Character: We now pick the largest character, which is 'c'. After appending 'c' to ans, the counter list updates to [1, 2, 0].
Picking the Smallest Character: We pick 'a' again as it is the next available smallest character. The ans list becomes ['a', 'c', 'a'] and counter is [0, 2, 0].
Picking the Largest Character: We need to pick the largest character now, which is 'b'. After doing so, the ans list is ['a', 'c', 'a', 'b'] and counter is [0, 1, 0].
Picking the Largest Character: As per our steps, we continue to pick the largest character left, which is still 'b', updating ans to ['a', 'c', 'a', 'b', 'b'] and counter to [0, 0, 0].
Returning the Result: The while loop exits since len(ans) is now equal to len(s). We join the elements of ans to form the final string. Therefore, the final reordered string is "acabb".

By following the steps laid out in the solution approach, we successfully applied the algorithm to the example input and achieved the expected outcome. The method of counting characters, appending the smallest and largest in order, and decrementing their count ensures that the rules of the problem statement are adhered to at every step of the process.

Solution Implementation

1class Solution:
2    def sort_string(self, s: str) -> str:
3        # Initialize a list to keep track of the count of each character in the string
4        char_count = [0] * 26
5      
6        # Count the occurrences of each character in the string
7        for char in s:
8            char_count[ord(char) - ord('a')] += 1
9      
10        # Initialize a list to build the sorted string
11        sorted_chars = []
12      
13        # Continue until the sorted string's length equals the input string's length
14        while len(sorted_chars) < len(s):
15            # Traverse the `char_count` list from start to end and add each character once if it's present
16            for i in range(26):
17                if char_count[i] > 0:
18                    sorted_chars.append(chr(i + ord('a')))
19                    char_count[i] -= 1  # Decrement the count of the added character
20          
21            # Traverse the `char_count` list from end to start and add each character once if it's present
22            for i in range(25, -1, -1):
23                if char_count[i] > 0:
24                    sorted_chars.append(chr(i + ord('a')))
25                    char_count[i] -= 1  # Decrement the count of the added character
26      
27        # Join the list of characters into a string and return it
28        return ''.join(sorted_chars)
29

1class Solution {
2
3    // Method to sort the string in a custom order
4    public String sortString(String s) {
5        // Counter array to hold frequency of each character 'a'-'z'
6        int[] frequency = new int[26];
7        // Fill the frequency array with count of each character
8        for (char ch : s.toCharArray()) {
9            frequency[ch - 'a']++;
10        }
11
12        // StringBuilder to hold the result
13        StringBuilder sortedString = new StringBuilder();
14
15        // Loop until the sortedString's length is less than the original string length
16        while (sortedString.length() < s.length()) {
17            // Loop from 'a' to 'z'
18            for (int i = 0; i < 26; ++i) {
19                // Check if the character is present
20                if (frequency[i] > 0) {
21                    // Append the character to sortedString
22                    sortedString.append((char) ('a' + i));
23                    // Decrement the frequency of appended character
24                    frequency[i]--;
25                }
26            }
27            // Loop from 'z' to 'a'
28            for (int i = 25; i >= 0; --i) {
29                // Check if the character is present
30                if (frequency[i] > 0) {
31                    // Append the character to sortedString
32                    sortedString.append((char) ('a' + i));
33                    // Decrement the frequency of appended character
34                    frequency[i]--;
35                }
36            }
37        }
38        // Return the resultant sorted string
39        return sortedString.toString();
40    }
41}
42

1class Solution {
2public:
3    // Method to sort the string in a specific pattern
4    string sortString(string s) {
5        // Create a frequency counter for each letter in the alphabet
6        vector<int> frequency(26, 0);
7        for (char c : s) {
8            ++frequency[c - 'a'];  // Increment the count of the current letter
9        }
10      
11        // Initialize the answer string
12        string result = "";
13      
14        // Keep building the result until its size matches the original string size
15        while (result.size() < s.size()) {
16            // Append characters from 'a' to 'z' to the result string if they are present
17            for (int i = 0; i < 26; ++i) {
18                if (frequency[i] > 0) {  // Check if the character is present
19                    result += (i + 'a'); // Convert index to char and append
20                    --frequency[i];      // Decrement the frequency of the used character
21                }
22            }
23
24            // Append characters from 'z' to 'a' to the result string if they are present
25            for (int i = 25; i >= 0; --i) { // Start from 'z'
26                if (frequency[i] > 0) {  // Check if the character is present
27                    result += (i + 'a'); // Convert index to char and append
28                    --frequency[i];      // Decrement the frequency of the used character
29                }
30            }
31        }
32      
33        // Return the sorted string
34        return result;
35    }
36};
37

1/**
2 * Sort the string based on the custom order: ascending characters followed by descending characters
3 * @param {string} str - The original string to be sorted.
4 * @return {string} - The sorted string.
5 */
6function sortString(str: string): string {
7    let resultString: string = '';
8    const charMap: Map<string, number> = new Map();
9
10    // Count the occurrences of each character in the string
11    for (let char of str) {
12        charMap.set(char, (charMap.get(char) || 0) + 1);
13    }
14
15    const keys: string[] = Array.from(charMap.keys());
16    keys.sort(); // Sort the keys (characters) in ascending order once
17
18    // Keep constructing the string until the resultString's length equals the input string's length
19    while (resultString.length < str.length) {
20        // Append characters in ascending order to the result string
21        for (let key of keys) {
22            if (charMap.get(key)! > 0) { // Ensure the character count is not zero
23                resultString += key;
24                charMap.set(key, charMap.get(key)! - 1); // Decrement the count in the map
25            }
26        }
27
28        // Append characters in descending order to the result string
29        for (let i = keys.length - 1; i >= 0; i--) {
30            if (charMap.get(keys[i])! > 0) {
31                resultString += keys[i];
32                charMap.set(keys[i], charMap.get(keys[i])! - 1); // Decrement the count in the map
33            }
34        }
35    }
36
37    return resultString;
38}
39

Time and Space Complexity

Time Complexity

The provided Python function sortString starts with an initial counting pass over the input string s, incrementing values in counter which takes O(n) time, where n is the length of s.

After that, it enters a loop that continues until the length of ans matches that of s. Within this loop, there are two for-loops: the first iterates in ascending order, the second in descending order. Each of these for-loops iterates over the 26 possible characters (from 'a' to 'z').

For each character, if that character count is non-zero, it is appended to ans and the count decremented. Since each character in s is processed exactly once (each is appended and then decremented), and there are two passes for each character (one in ascending and one in descending order), the total count of operations inside the while-loop is 2n, leading to an additional O(n) time complexity.

Thus, the time complexity of the entire function is O(n).

Space Complexity

The space complexity includes:

The counter array which is always 26 elements long, thus it is a constant space O(1).
The ans list that will eventually grow to be the same size as s to accommodate all characters, which is O(n).

Therefore, the total space complexity is O(n), where n is the length of the input string s.

Learn more about how to find time and space complexity quickly using problem constraints.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Question 1 out of 10

What's the relationship between a tree and a graph?

A graph is a special tree

A tree is a special graph

No relationship

They are the same thing

I don’t know