1700 - 1799
1768 - Merge Strings Alternately (Easy)
Problem Link
https://leetcode.com/problems/merge-strings-alternately
Problem Statement
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"
Explanation: The merged string will be merged as so:
word1: a b c
word2: p q r
merged: a p b q c rExample 2:
Input: word1 = "ab", word2 = "pqrs"
Output: "apbqrs"
Explanation: 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 sExample 3:
Input: word1 = "abcd", word2 = "pq"
Output: "apbqcd"
Explanation: 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 dConstraints:
1 <= word1.length, word2.length <= 100word1andword2consist of lowercase English letters.
Approach 1: Two Pointers
We define two pointers, pointer_for_word1and pointer_for_word2 . We then iterate the letters from both words until we have exhausted one of them. After that, we exhaust the remaining word (if both words have different lengths).
def mergeAlternately(self, word1: str, word2: str) -> str:
#initialize the 2 pointers for word1 and word2
pointer_for_word1 = 0
pointer_for_word2 = 0
#initialize the 2 lengths of word1 and word2
length_of_word1 = len(word1)
length_of_word2 = len(word2)
#initialize result list
result = []
#we iterate each letter in word1 and word2 until we have exhausted one of them
while(pointer_for_word1 < length_of_word1 and pointer_for_word2 < length_of_word2):
#add a letter from word1, then a letter from word2
result.append(word1[pointer_for_word1])
result.append(word2[pointer_for_word2])
#move the two pointers
pointer_for_word1 += 1
pointer_for_word2 += 1
#if word1 is not exhausted, add its remaining letters to result
while(pointer_for_word1 < length_of_word1):
result.append(word1[pointer_for_word1])
pointer_for_word1 += 1
#if word2 is not exhausted, add its remaining letters to result
while(pointer_for_word2 < length_of_word2):
result.append(word2[pointer_for_word2])
pointer_for_word2 += 1
#return the string of the result list
return "".join(result)