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2109. Adding Spaces to a String - LeetCode Daily | Medium | String | Efficient | Beg Easy Solution

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This is a solution to the Daily Leetcode problem "Adding Spaces to a String" using Brute Force Approach in Java in a very easy to understand beginner level explanation.
The goal of making these Leetcode problems solutions is to help me & others with deeper understanding & being able to see new & existing patterns in a particular problem using multiple approaches from least efficient to most efficient approach so that we can better explain our approach to the interviews & properly handle edge cases in coding interviews and not just copy paste every problem's solution without properly understanding the problem first like i used to do for a long time.
Kindly stay tuned for more content like this in the future😃
Problem Link(Easy) :
EDIT😅 :
Forgot to explain Time Complexity & Space Complexity in the video through using this Brute Force approach!
Complexity : Its a Efficient Solution
1. Time Complexity: We initialize a StringBuilder with the original string s, which takes O(n) space.
2. Space Complexity :
- Initialization of StringBuilder takes O(n) space.
- The final string that we return will have n + m characters (since we're adding m spaces). So, The space required to store this string is O(n + m).
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The goal of making these Leetcode problems solutions is to help me & others with deeper understanding & being able to see new & existing patterns in a particular problem using multiple approaches from least efficient to most efficient approach so that we can better explain our approach to the interviews & properly handle edge cases in coding interviews and not just copy paste every problem's solution without properly understanding the problem first like i used to do for a long time.
Kindly stay tuned for more content like this in the future😃
Problem Link(Easy) :
EDIT😅 :
Forgot to explain Time Complexity & Space Complexity in the video through using this Brute Force approach!
Complexity : Its a Efficient Solution
1. Time Complexity: We initialize a StringBuilder with the original string s, which takes O(n) space.
2. Space Complexity :
- Initialization of StringBuilder takes O(n) space.
- The final string that we return will have n + m characters (since we're adding m spaces). So, The space required to store this string is O(n + m).
My LinkedIn Profile :
My LeetCode Profile :
My Leetcode Solutions Repo :
My Leetcode Solutions Notion Tracker :
My Github Profile :