Merge Strings Alternately

  • 28 Jul, 2025

📄 Problem Statement

✅ Problem: Merge Strings Alternately

You are given two strings word1 and word2. Merge them by alternating characters from each, starting with word1. If one string is longer, append the extra characters to the end of the merged result.

Examples:

  • Input: word1 = "abc", word2 = "pqr"
    Output: "apbqcr"

  • Input: word1 = "ab", word2 = "pqrs"
    Output: "apbqrs"

  • Input: word1 = "abcd", word2 = "pq"
    Output: "apbqcd"

Constraints:

  • 1 <= word1.length, word2.length <= 100
  • Both strings consist of lowercase English letters

🧠 Problem Explanation

We are to merge two strings in alternating fashion:

  • Take one character from word1, then one from word2, and repeat.
  • Once one string runs out, continue with the remaining characters of the other string.

💡 Approach

  • Use two pointers i and j for word1 and word2, respectively.
  • Loop while both pointers are in range:
    • Append word1[i] and word2[j] alternately to a result list.
  • After the main loop, append the remainder of the longer string.
  • Join the list into a final string using "".join().

🧾 Pseudocode

function: mergeAlternately
params:
  - word1: str
  - word2: str
logic:
  - initialize i = 0, j = 0, result = []
  - while i < len(word1) and j < len(word2):
      - append word1[i] to result
      - append word2[j] to result
      - increment i and j
  - append remaining part of word1[i:] and word2[j:] to result
  - return ''.join(result)

✅ Code

class Solution(object):
    def mergeAlternately(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: str
        """
        A = len(word1)
        B = len(word2)
        a, b = 0, 0
        word = 1
        s = []

        while a < A and b < B:
            if word == 1:
                s.append(word1[a])
                a += 1
                word = 2
            if word == 2:
                s.append(word2[b])
                b += 1
                word = 1

        while a < A:
            s.append(word1[a])
            a += 1

        while b < B:
            s.append(word2[b])
            b += 1

        return ''.join(s)

📊 Complexities

  • Time Complexity: O(n + m) where n = len(word1) and m = len(word2)
  • Space Complexity: O(n + m) for storing the merged result in a list

📝 Notes

  • Pattern Used: Two Pointers, Alternating Merge
  • Insights:
    • Using a toggle flag to switch between the two input strings helps keep the merge logic clean.
    • The use of ''.join() avoids performance issues due to string immutability in Python.
    • This approach naturally handles uneven string lengths.