SAMSUNG 49-Inch Odyssey G9

Because earth is not flat

Description

Given two strings, word1 and word2, the task is to merge them alternately. The function should combine the two strings by adding one character from word1 followed by one character from word2, continuing this process until one or both strings are exhausted. The remaining characters from the longer string, if any, are then appended to the end of the result.

Intuition

The intuition behind this problem is straightforward. Since the goal is to merge two strings in an alternating fashion, we can iterate through both strings simultaneously, appending one character from each string to the result string. If one string is longer, we simply append the remaining characters from the longer string after the loop ends.

Approach

1. Initialize a result string: Start with an empty string that will store the final merged result.
2. Iterate through the strings: Use a loop to iterate through the indices of the strings, adding one character from each string to the result string alternately.
   • If the current index is within the bounds of word1, append the character from word1.
   • If the current index is within the bounds of word2, append the character from word2.
3. Handle remaining characters: If one string is longer than the other, the remaining characters of the longer string will be appended after the loop.
4. Return the result: Once the loop is complete, return the result string.

Complexity

Time Complexity:

The time complexity is (O(\max(n, m))), where (n) is the length of word1 and (m) is the length of word2. This is because the function iterates through both strings, processing each character once.

Space Complexity:

The space complexity is (O(n + m)), as the function creates a new string that contains all characters from both word1 and word2.

Code

C++

class Solution {
public:
    string mergeAlternately(string word1, string word2) {
        string result = "";
        int len1 = word1.length();
        int len2 = word2.length();
        
        for (int i = 0; i < max(len1, len2); i++) {
            if (i < len1) result += word1[i];
            if (i < len2) result += word2[i];
        }
        return result;
    }
};

Python

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        result = ""
        len1, len2 = len(word1), len(word2)
        
        for i in range(max(len1, len2)):
            if i < len1:
                result += word1[i]
            if i < len2:
                result += word2[i]
                
        return result

Java

class Solution {
    public String mergeAlternately(String word1, String word2) {
        StringBuilder result = new StringBuilder();
        int len1 = word1.length();
        int len2 = word2.length();
        
        for (int i = 0; i < Math.max(len1, len2); i++) {
            if (i < len1) result.append(word1.charAt(i));
            if (i < len2) result.append(word2.charAt(i));
        }
        return result.toString();
    }
}

JavaScript

var mergeAlternately = function (word1, word2) {

    let result = '';
    for (let i = 0; i < Math.max(word1.length, word2.length); i++) {
        if (i < word1.length) result += word1[i];
        if (i < word2.length) result += word2[i];
    }
    return result;
};