Leetcode 1768. Merge Strings Alternately

You are given two strings `word1` and `word2`. Merge the strings by adding letters in alternating order, starting with `word1`. If a string is longer than the other, append the additional letters onto the end of the merged string. Return the merged string.

Programming Skills I Easy

Description

You are given two strings word1 and word2. Merge the strings by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters onto the end of the merged string.

Return the merged string.

Example 1:

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

• The merged string will be merged as so:
  word1: a b c
  word2: p q r
  merged: a p b q c r

Example 2:

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

• Notice that as word2 is longer, “rs” is appended to the end.
  word1: a b
  word2: p q r s
  merged: a p b q r s

Example 3:

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

• Notice that as word1 is longer, “cd” is appended to the end.
  word1: a b c d
  word2: p q
  merged: a p b q c d

Constraints:

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

Solution

class Solution:
    # Iteration
    # Time Complexity: BigO(N)
    # Space Complexity: BigO(1)
    def mergeAlternately(self, word1: str, word2: str) -> str:
        l1, l2 = len(word1), len(word2)
        max = l1 if l1 > l2 else l2
        res = ""
        for i in range(max):
            if l1 > i:
                res += word1[i]
            if l2 > i:
                res += word2[i]

        return res

/**
 * Do not return anything, modify nums in-place instead.
 * Iteration
 * Time Complexity: BigO(N)
 * Space Complexity: BigO(1)
 */
function mergeAlternately(word1: string, word2: string): string {
  let maxLength = word1.length > word2.length ? word1.length : word2.length;
  let res = "";
  for (let i = 0; i < maxLength; i++) {
    if (word1[i]) res += word1[i];
    if (word2[i]) res += word2[i];
  }
  return res;
}