LeetCode 2109. Adding Spaces to a String Solution in Java, C++, Python & More | Explanation + Code

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 spaces, respectively. Then, we iterate through the string s from the beginning to the end. When i equals 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.

PythonJavaC++GoTypeScript

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)(code-box)

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(); } }(code-box)

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; } };(code-box)

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) }(code-box)

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(''); }(code-box)