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2109. Adding Spaces to a String

Description

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.

Solutions

Solution 1: Two Pointers

We can use two pointers \(i\) and \(j\) to point to the beginning of the string \(s\) and the array \(\textit{spaces}\), respectively. Then, we iterate through the string \(s\) from the beginning to the end. When \(i\) equals \(\textit{spaces}[j]\), we add a space to the result string, and then increment \(j\) by \(1\). Next, we add \(s[i]\) to the result string, and then increment \(i\) by \(1\). We continue this process until we have iterated through the entire string \(s\).

The time complexity is \(O(n + m)\), and the space complexity is \(O(n + m)\), where \(n\) and \(m\) are the lengths of the string \(s\) and the array \(spaces\), respectively.

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class Solution:
    def addSpaces(self, s: str, spaces: List[int]) -> str:
        ans = []
        j = 0
        for i, c in enumerate(s):
            if j < len(spaces) and i == spaces[j]:
                ans.append(' ')
                j += 1
            ans.append(c)
        return ''.join(ans)

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class Solution {
    public String addSpaces(String s, int[] spaces) {
        StringBuilder ans = new StringBuilder();
        for (int i = 0, j = 0; i < s.length(); ++i) {
            if (j < spaces.length && i == spaces[j]) {
                ans.append(' ');
                ++j;
            }
            ans.append(s.charAt(i));
        }
        return ans.toString();
    }
}

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class Solution {
public:
    string addSpaces(string s, vector<int>& spaces) {
        string ans = "";
        for (int i = 0, j = 0; i < s.size(); ++i) {
            if (j < spaces.size() && i == spaces[j]) {
                ans += ' ';
                ++j;
            }
            ans += s[i];
        }
        return ans;
    }
};

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func addSpaces(s string, spaces []int) string {
    var ans []byte
    for i, j := 0, 0; i < len(s); i++ {
        if j < len(spaces) && i == spaces[j] {
            ans = append(ans, ' ')
            j++
        }
        ans = append(ans, s[i])
    }
    return string(ans)
}

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function addSpaces(s: string, spaces: number[]): string {
    const ans: string[] = [];
    for (let i = 0, j = 0; i < s.length; i++) {
        if (i === spaces[j]) {
            ans.push(' ');
            j++;
        }
        ans.push(s[i]);
    }
    return ans.join('');
}

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