Adding Spaces to a String

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ArraysStringsTwo Pointers

You are given a 0-indexed string s and a 0-indexed integer array spaces that describes the indices in the original string where spaces will be added. Each space should be inserted before the character at the given index.

For example, given s = "EnjoyYourCoffee" and spaces = [5, 9], we place spaces before 'Y' and 'C', which are at indices 5 and 9 respectively. Thus, we obtain "Enjoy Your Coffee".

Return the modified string after the spaces have been added.

Example 1:

Input: s = "LeetcodeHelpsMeLearn", spaces = [8,13,15]
Output: "Leetcode Helps Me Learn"
Explanation: 
The indices 8, 13, and 15 correspond to the underlined characters in "LeetcodeHelpsMeLearn".
We then place spaces before those characters.

Example 2:

Input: s = "icodeinpython", spaces = [1,5,7,9]
Output: "i code in py thon"
Explanation:
The indices 1, 5, 7, and 9 correspond to the underlined characters in "icodeinpython".
We then place spaces before those characters.

Example 3:

Input: s = "spacing", spaces = [0,1,2,3,4,5,6]
Output: " s p a c i n g"
Explanation:
We are also able to place spaces before the first character of the string.

Constraints:

1 <= s.length <= 3 * 10⁵
s consists only of lowercase and uppercase English letters.
1 <= spaces.length <= 3 * 10⁵
0 <= spaces[i] <= s.length - 1
All the values of spaces are strictly increasing.

Try coding it in LeetCode

Adding Spaces to a String

Medium

Asked by:

13 views

Topics:

ArraysStringsTwo Pointers

You are given a 0-indexed string s and a 0-indexed integer array spaces that describes the indices in the original string where spaces will be added. Each space should be inserted before the character at the given index.

For example, given s = "EnjoyYourCoffee" and spaces = [5, 9], we place spaces before 'Y' and 'C', which are at indices 5 and 9 respectively. Thus, we obtain "Enjoy Your Coffee".

Return the modified string after the spaces have been added.

Example 1:

Input: s = "LeetcodeHelpsMeLearn", spaces = [8,13,15]
Output: "Leetcode Helps Me Learn"
Explanation: 
The indices 8, 13, and 15 correspond to the underlined characters in "LeetcodeHelpsMeLearn".
We then place spaces before those characters.

Example 2:

Input: s = "icodeinpython", spaces = [1,5,7,9]
Output: "i code in py thon"
Explanation:
The indices 1, 5, 7, and 9 correspond to the underlined characters in "icodeinpython".
We then place spaces before those characters.

Example 3:

Input: s = "spacing", spaces = [0,1,2,3,4,5,6]
Output: " s p a c i n g"
Explanation:
We are also able to place spaces before the first character of the string.

Constraints:

1 <= s.length <= 3 * 10⁵
s consists only of lowercase and uppercase English letters.
1 <= spaces.length <= 3 * 10⁵
0 <= spaces[i] <= s.length - 1
All the values of spaces are strictly increasing.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected size of the input string, and what are the practical limits of the `spaces` array? Are there any memory constraints to consider?
Can the `spaces` array be empty, or can any of the values in the `spaces` array be out of bounds relative to the string's length?
What should the function return if the input string is null or empty, or if the `spaces` array is null?
Is the `spaces` array guaranteed to be sorted in ascending order? Does it contain duplicate values?
Are we expected to create a new string or can we modify the original string in place (assuming strings are mutable in the given language)?

Brute Force Solution

Approach

The brute force method involves exploring every single possible way to insert spaces into the string at the specified positions. We generate all combinations of placing spaces and then choose the best one. Think of it as trying out every possibility, no matter how inefficient it might seem at first.

Here's how the algorithm would work step-by-step:

Consider each possible place where a space can be added in the string.
Start by adding no spaces, then one space in the first possible position, then one space in the second possible position, and so on, until one space has been tested in all the locations.
Now, test all combinations of adding two spaces, three spaces, and so on, at every possible pair, triplet, or more of locations.
For each combination of spaces, create a new string with the spaces inserted in those positions.
After generating all possible strings, analyze each one according to the rules of the problem. Pick the string that fulfills the desired criteria (e.g., the shortest string, the most balanced string).
If multiple strings meet the desired criteria, return any one of them.

Code Implementation

def add_spaces_brute_force(input_string, space_indices):
    possible_strings = []
    number_of_indices = len(space_indices)

    # Generate all possible combinations of spaces
    for i in range(2**number_of_indices):
        current_string = list(input_string)
        spaces_to_add = []

        # Determine which indices should have spaces based on the binary representation of i
        for index_position in range(number_of_indices):
            if (i >> index_position) & 1:
                spaces_to_add.append(space_indices[index_position])

        # Sort the indices in reverse order to avoid shifting issues when inserting
        spaces_to_add.sort(reverse=True)

        # Insert the spaces into the string
        for index in spaces_to_add:
            current_string.insert(index, ' ')

        possible_strings.append("".join(current_string))

    # This part would normally involve criteria to select the best string,
    # but since it's not specified, we return the first one
    if possible_strings:
        return possible_strings[0]

    #Handle the case when the input string is empty
    return input_string

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible combination of inserting spaces at the given positions. With 'n' possible positions to insert a space, each position can either have a space or not. This leads to 2^n possible combinations of strings to generate and analyze. Therefore, the time complexity is exponential, specifically O(2^n), as we need to consider all possible subsets of space insertion positions.

Space Complexity

O(2^N) – The brute force method generates all possible combinations of adding spaces to the string. In the worst case, every position in the string could potentially have a space, leading to 2^N possible combinations of strings, where N is the length of the input string. Each of these strings needs to be stored temporarily to analyze and select the desired one. Therefore, the space complexity is dominated by the need to store all possible string combinations, leading to O(2^N) auxiliary space.

Optimal Solution

Approach

The efficient solution focuses on building the final string piece by piece. It figures out how many words fit on each line and then spaces them out accordingly, rather than trying all possible combinations.

Here's how the algorithm would work step-by-step:

Examine the list of spaces where we need to add.
Create a new string to hold our final result.
Walk through the original string character by character.
If the current spot in the string matches one of the space positions, add a space to our result string.
Append the current character from the original string to our result string.
After going through the whole original string, the result string will have the added spaces.

Code Implementation

def add_spaces_to_string(original_string, spaces_to_add): 
    string_with_spaces = ""
    space_index = 0

    for i, character in enumerate(original_string):
        # Add space if the current index is in spaces_to_add
        if space_index < len(spaces_to_add) and i == spaces_to_add[space_index]:
            string_with_spaces += " "
            space_index += 1

        string_with_spaces += character

    return string_with_spaces

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the original string of length n character by character. Inside this loop, it checks if the current index is present in the list of space positions. This check can be done in constant time if the space positions are stored in a hash set. Appending to the result string is also typically a constant-time operation. Therefore, the dominant operation is the single pass through the original string, resulting in O(n) time complexity.

Space Complexity

O(N) – The primary contributor to space complexity is the creation of a new string to hold the final result. In the worst-case scenario, where every character in the original string needs to be followed by a space, the new string will be approximately twice the size of the original. Therefore, the algorithm uses auxiliary space proportional to the length of the original string, denoted as N, to store the resulting string. Hence, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty string or throw an IllegalArgumentException, depending on the desired behavior defined in the requirements.
Null or empty array of space insertion indices	Return the original string unchanged, since no spaces need to be added.
Empty string and empty array of indices	Return an empty string, as there is nothing to process.
Space insertion indices out of bounds (negative or greater than string length)	Filter out invalid indices or throw an exception to indicate invalid input.
Duplicate space insertion indices	The algorithm should only insert a space once at the specific location even if the index appears multiple times; use a set to deduplicate the indices.
Space insertion indices in reverse order	Sort the indices in ascending order before iterating to ensure correct insertion.
Very long input string and large number of space insertion indices leading to possible memory issues.	Use a StringBuilder to efficiently construct the modified string and avoid excessive memory allocation.
Adjacent or overlapping space insertion indices (e.g., inserting spaces at indices 1 and 2).	After inserting a space, increment the subsequent indices to account for the increased string length caused by the space insertion.