Hey coders! Hope you're doing well. I'm excited to share my solutions for the LeetCode-75 series, which covers 75 essential problems to help you prepare for coding interviews.

In each post, I'll present my solution along with a detailed explanation of my approach. Feel free to leave any questions or suggestions for improvement in the comments. I'm looking forward to collaborating and discussing with you! Happy Coding!

I've added here the link for the problem: Merge Strings Alternately

Problem description

You are given two strings word1 and word2. Merge the strings by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters onto the end of the merged string.

Return the merged string.

*Example 1: *

Input: word1 = "abc", word2 = "pqr"
Output: "apbqcr"
Explanation: The merged string will be merged as so:
word1: a b c
word2: p q r
merged: a p b q c r

Example 2:

Input: word1 = "ab", word2 = "pqrs"
Output: "apbqrs"
Explanation: Notice that as word2 is longer, "rs" is appended to the end.
word1: a b
word2: p q r s
merged: a p b q r s

** Example 3:**

Input: word1 = "abcd", word2 = "pq"
Output: "apbqcd"
Explanation: Notice that as word1 is longer, "cd" is appended to the end.
word1: a b c d
word2: p q
merged: a p b q c d

SOLUTION

Intuition

Given two strings, we need to merge them by alternating characters from each string. The solution is straightforward if both strings have the same length, but they can have different lengths. We will iterate through both strings using pointers, adding characters to the result until both pointers reach the end.

Approach

1. Create a StringBuilder to store the alternated characters from both strings.
2. Create two pointers to keep track of the current position in each string.
3. Iterate both strings until both pointers reach the end of their respective strings.
4. Add elements to the StringBuilder if the string is not empty and increment the pointer
5. Return the StringBuilder

Complexity

• Time complexity:
  The time complexity is O(n) where n is the length of the longer string, as we iterate through the strings.

• Space complexity:
  The time complexity is 0(1) since we use a StringBuilder and a few variables.

Code

  public String mergeAlternately (String word1, String word2) {
    // ? Create a StringBuilder to build the result string efficiently
    StringBuilder completeWord = new StringBuilder();

    // ? Initialize two pointers to traverse both strings
    int p1 = 0;
    int p2 = 0;

    // ? Iterate through both strings until both pointers reach the end of their resépectives strings
    while (p1 < word1.length() || p2 < word2.length()) {
      // ? Append the current character from words if the pointer is within bounds
      if (p1 < word1.length()) completeWord.append(word1.charAt(p1));
      if (p2 < word2.length()) completeWord.append(word2.charAt(p2));
      p1++;
      p2++;
    }
    // ? Convert the StringBuilder to a string and return it
    return completeWord.toString();
  }