easy Merge Strings Alternately

Problem Statement

Given two strings, word1 and word2, return the merged string after merging each character of both strings in alternate order, starting with the word1. If one string is longer than the other, append the remaining characters of the longer string to the end of the merged string.

Examples

Example 1:
- Input: word1: "dog", word2: "cat"
- Expected Output: "dcoagt"
- Justification: The characters from both strings are alternated, starting with the first character of the first string. Since both strings are of equal length, no extra characters remain at the end.
Example 2:
- Input: word1: "hello", word2: "hi"
- Expected Output: "hheillo"
- Justification: After alternating characters from both strings, the remaining characters from the longer string ("hello") are appended to the end.
Example 3:
- Input: word1: "abc", word2: "xyz123"
- Expected Output: "axbycz123"
- Justification: Characters are alternated, and since the second string is longer, the remaining characters ("123") are appended to the end.

Try it yourself

Try solving this question here:

🎯 STRICT STANDOUT — Merge Strings Alternately (easy)

1. Why / judgment

Zip characters word1[i], word2[i] for i in 0..min-1; append remainder of longer. Pure two-pointer / index simulation — no stack, no DP. When-NOT: merge by sorted order -> merge-sort merge step; merge linked lists by value -> compare nodes.

2. Big-O derivation (K11)

O(n+m) time to build result; O(n+m) space output (unavoidable).
dog+cat->dcoagt; hello+hi->hheillo.

3. Pattern + when-NOT (K12)

Name: ALTERNATING TWO-POINTER MERGE

Recognition: interleave two strings starting with first.

When-NOT: Equal weight by runs -> different. Merge k strings round-robin -> generalize index array.

4. Edge hand-run (K13)

one empty -> other
length differ by many -> long tail

5. Interviewer follow-ups (model answers)

Q1. StringBuilder vs + in loop?
A: Use builder / list join — quadratic string + in some languages.

Q2. In-place possible?
A: Not really on immutable strings.

Q3. Related LC?
A: Do not confuse with gcd-merge strings.

6. Short drills

Drill: ab+PQRS->aPbQRS
Drill: both length 1
Drill: write index i,j version

✅ Solution Merge Strings Alternately

Problem Statement

Examples

Example 1:
- Input: word1: "dog", word2: "cat"
- Expected Output: "dcoagt"
- Justification: The characters from both strings are alternated, starting with the first character of the first string. Since both strings are of equal length, no extra characters remain at the end.
Example 2:
- Input: word1: "hello", word2: "hi"
- Expected Output: "hheillo"
- Justification: After alternating characters from both strings, the remaining characters from the longer string ("hello") are appended to the end.
Example 3:
- Input: word1: "abc", word2: "xyz123"
- Expected Output: "axbycz123"
- Justification: Characters are alternated, and since the second string is longer, the remaining characters ("123") are appended to the end.

Solution

To solve this problem, we'll iterate over both strings simultaneously, appending one character from each string to the result until we reach the end of one of the strings. This approach ensures that characters are merged in an alternating fashion, preserving the order from both input strings. After completing the alternating merge, if one string is longer, we append the remaining characters of the longer string to the merged result. This method is effective because it directly addresses the problem's requirements with a straightforward iteration, making it easy to understand and implement. By iterating over the strings simultaneously, we minimize the time complexity, ensuring that each character is visited only once, which makes the algorithm efficient for strings of varying lengths.

Step-by-Step Algorithm

Initialize an empty string result to store the merged output.
Determine the length of the shorter string to ensure the alternating merge doesn't exceed the shorter string's length.
Use a loop to iterate through both strings up to the length of the shorter string. During each iteration:
- Append the current character from the first string to result.
- Append the current character from the second string to result.
After completing the alternating merge, check if any string has remaining characters. Append the remaining characters from the longer string to result.
Return the result string.

Algorithm Walkthrough

Consider the input: word1: "abc", word2: "xyz123"

Initialize Variables:
- result is an empty string where the merged output will be stored.
- Determine the length of the shorter string. In this case, word1 has a length of 3, and word2 has a length of 6. So, the length of the shorter string is 3.
Start Looping Through Both Strings:
- For i = 0 (first iteration):
  - Append the first character of word1 ("a") to result. Now, result = "a".
  - Append the first character of word2 ("x") to result. Now, result = "ax".
- For i = 1 (second iteration):
  - Append the second character of word1 ("b") to result. Now, result = "axb".
  - Append the second character of word2 ("y") to result. Now, result = "axby".
- For i = 2 (third iteration):
  - Append the third character of word1 ("c") to result. Now, result = "axbyc".
  - Append the third character of word2 ("z") to result. Now, result = "axbycz".
Check for Remaining Characters:
- At this point, word1 is fully iterated through, but word2 still has characters left ("123").
- Since word1 is shorter and fully merged, we append the remaining characters of word2 to result. Now, result = "axbycz123".
Return the Result:
- The final merged string is "axbycz123", which is stored in result.
- Return result.

Code

java

public class Solution {

  // Method to merge two strings alternately
  public String mergeAlternately(String word1, String word2) {
    StringBuilder result = new StringBuilder(); // StringBuilder to store the result
    int minLength = Math.min(word1.length(), word2.length()); // Determine the shorter string length
    // Loop through both strings up to the length of the shorter string
    for (int i = 0; i < minLength; i++) {
      result.append(word1.charAt(i)); // Append character from word1
      result.append(word2.charAt(i)); // Append character from word2
    }
    // Append remaining characters from the longer string, if any
    if (word1.length() > minLength) {
      result.append(word1.substring(minLength));
    } else if (word2.length() > minLength) {
      result.append(word2.substring(minLength));
    }
    return result.toString(); // Convert StringBuilder to String and return
  }

  // Main method to test the algorithm with example inputs
  public static void main(String[] args) {
    Solution solution = new Solution();
    System.out.println(solution.mergeAlternately("dog", "cat")); // Example 1
    System.out.println(solution.mergeAlternately("hello", "hi")); // Example 2
    System.out.println(solution.mergeAlternately("abc", "xyz123")); // Example 3
  }
}

Complexity Analysis

Time Complexity: , where N is the length of the first string and M is the length of the second string. This is because the algorithm iterates through both strings once.
Space Complexity:| , due to the storage requirements for the resulting string which combines characters from both input strings.

🎯 STRICT STANDOUT — Solution Merge Strings Alternately

1. Why / judgment

While both remain append w1[i++], w2[j++]; append rest of w1 or w2. Hand-run hello/hi -> h,h,e,i,l,l,o. Complexity linear output size.

2. Big-O derivation (K11)

O(n+m)/O(n+m)
One empty works
Builder avoids quadratic

3. Pattern + when-NOT (K12)

Name: INTERLEAVE TWO STRINGS

Recognition: alternate merge.

When-NOT: Value-ordered merge is different problem.

4. Edge hand-run (K13)

unequal lengths
identical strings

5. Interviewer follow-ups (model answers)

Q1. Index vs iterators?
A: Either; bounds checks matter.

Q2. Memory of result?
A: Must Omega(n+m) to write answer.

Q3. Unicode surrogates?
A: Treat as language char units per problem.

6. Short drills

Drill: ab+pqrs
Drill: empty+x
Drill: same length 100

🧩 Pattern · Arrays

Recognize it: Scan once tracking what you need (running max/sum), or precompute a prefix-sum / hash → turn O(n²) into O(n).

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