To my satisfaction, this approach passed all test cases. The two-pointer strategy effectively handled the merging process and the subsequent appending of remaining characters.
While the initial solution worked, I identified a potential optimization. Instead of maintaining two separate pointers, I could iterate based on the maximum length of the two strings. By checking if the current index is within the bounds of each string, I can directly append characters without unnecessary checks. This streamlined approach improves efficiency.
Time complexity: O(m n), where m and n are the lengths of word1 and word2, respectively. This is because we iterate through each character in both strings once.
Space complexity: O(m n) as well, since we create a new string to store the merged result.
Given two strings, word1 and word2, the task is to merge them by alternating characters. The process begins with word1 and continues until one string is exhausted. Any remaining characters from the longer string are appended to the end of the merged string.
Given the problem's simplicity, I immediately recognized a two-pointer approach as the most suitable solution. My initial pseudocode outlined the following steps:
1.Initialize two pointers, one for each string.
2.Iterate through both strings, alternatingly adding characters to a new string until one string is empty.
3.Append the remaining characters from the non-empty string to the new string.
To my satisfaction, this approach passed all test cases. The two-pointer strategy effectively handled the merging process and the subsequent appending of remaining characters.
While the initial solution worked, I identified a potential optimization. Instead of maintaining two separate pointers, I could iterate based on the maximum length of the two strings. By checking if the current index is within the bounds of each string, I can directly append characters without unnecessary checks. This streamlined approach improves efficiency.
Time complexity: O(m n), where m and n are the lengths of word1 and word2, respectively. This is because we iterate through each character in both strings once.
Space complexity: O(m n) as well, since we create a new string to store the merged result.
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