Solution

Our initial strategy revolves around the concept of interleaving: taking two sequences and combining them into one by alternating elements from each. The primary concern is ensuring that we handle strings of unequal length gracefully, appending any remaining characters from the longer string after the interleaving process.

Test Cases

// MARK: - Test Cases

let solution = Solution()

// Test Case 1
let str1_1 = "ABCABC"
let str2_1 = "ABC"
print("Test Case 1: \\(solution.mergeAlternately(str1_1, str2_1))") 
// Expected Output: "AABBCCABC"

// Test Case 2
let str1_2 = "ABABAB"
let str2_2 = "ABAB"
print("Test Case 2: \\(solution.mergeAlternately(str1_2, str2_2))") 
// Expected Output: "AABBAABBAB"

// Test Case 3
let str1_3 = "LEET"
let str2_3 = "CODE"
print("Test Case 3: \\(solution.mergeAlternately(str1_3, str2_3))") 
// Expected Output: "LCEOEDTE"

// Test Case 4
let str1_4 = "Hello"
let str2_4 = "World"
print("Test Case 4: \\(solution.mergeAlternately(str1_4, str2_4))") 
// Expected Output: "HWeolrllod"

// Test Case 5
let str1_5 = "12345"
let str2_5 = "abc"
print("Test Case 5: \\(solution.mergeAlternately(str1_5, str2_5))") 
// Expected Output: "1a2b3c45"

Approaches

Approach 1: Iterative Character Appending

Description

• Step 1: Initialize an empty string to store the result.
• Step 2: Determine the minimum length of the two input strings.
• Step 3: Loop through both strings up to the minimum length, appending one character from each to the result.
• Step 4: Append the remaining characters from the longer string to the result.

Complexity Analysis

• Time Complexity: O(n + m), where n and m are the lengths of word1 and word2, respectively.
• Space Complexity: O(n + m), primarily for storing the result string.

class Solution {
    func mergeAlternately(_ word1: String, _ word2: String) -> String {
        var mergedString = ""
        let minLength = min(word1.count, word2.count)

        let char1 = Array(word1)
        let char2 = Array(word2)

        for i in 0..<minLength {
            mergedString.append(char1[i])
            mergedString.append(char2[i])
        }

        if word1.count > minLength {
            mergedString.append(String(char1[minLength...]))
        } else if word2.count > minLength {
            mergedString.append(String(char2[minLength...]))
        }

        return mergedString
    }
}

Approach 2: Zip and FlatMap

Description

• Utilize Swift's zip function to combine the characters of word1 and word2 into tuples, then flatten this structure into a single string using flatMap. Finally, append any remaining characters from the longer string.