Explanation

To solve this problem, we can iterate over both input strings simultaneously and merge them by adding letters in alternating order. We can keep track of the current index in both strings and append the characters to the result string accordingly. If one string is longer than the other, we can simply append the remaining characters to the result string.

Algorithmic Idea

Initialize two pointers i and j for word1 and word2 respectively.
Iterate over both strings and merge them by adding letters in alternating order.
If one string is longer than the other, append the remaining characters to the result string.
Return the merged string.

Time Complexity

The time complexity of this approach is O(n), where n is the maximum length of word1 and word2.

Space Complexity

The space complexity is also O(n) since we are using a StringBuilder to store the merged string.

View Leetcode Problem

class Solution {
    public String mergeAlternately(String word1, String word2) {
        StringBuilder result = new StringBuilder();
        int i = 0, j = 0;
        
        while (i < word1.length() || j < word2.length()) {
            if (i < word1.length()) {
                result.append(word1.charAt(i));
                i++;
            }
            if (j < word2.length()) {
                result.append(word2.charAt(j));
                j++;
            }
        }
        
        return result.toString();
    }
}

Code Editor (Testing phase)

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