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Formatted question description: https://leetcode.ca/all/1768.html

# 1768. Merge Strings Alternately

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"
Explanation: 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"
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   s


Example 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   d


Constraints:

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

## Solution

Merge the two strings as long as both strings have remaining letters. When a string does not have remaining letters, append the remaining letters of the other string to the merged string.

• class Solution {
public String mergeAlternately(String word1, String word2) {
StringBuffer sb = new StringBuffer();
int length1 = word1.length(), length2 = word2.length();
int index1 = 0, index2 = 0;
while (index1 < length1 && index2 < length2) {
sb.append(word1.charAt(index1++));
sb.append(word2.charAt(index2++));
}
while (index1 < length1)
sb.append(word1.charAt(index1++));
while (index2 < length2)
sb.append(word2.charAt(index2++));
return sb.toString();
}
}

############

class Solution {
public String mergeAlternately(String word1, String word2) {
int m = word1.length(), n = word2.length();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < m || i < n; ++i) {
if (i < m) {
ans.append(word1.charAt(i));
}
if (i < n) {
ans.append(word2.charAt(i));
}
}
return ans.toString();
}
}

• // OJ: https://leetcode.com/problems/merge-strings-alternately/
// Time: O(M + N)
// Space: O(1)
class Solution {
public:
string mergeAlternately(string a, string b) {
int i = 0, j = 0, M = a.size(), N = b.size();
string ans;
while (i < M || j < N) {
if (i < M) ans += a[i++];
if (j < N) ans += b[j++];
}
return ans;
}
};

• class Solution:
def mergeAlternately(self, word1: str, word2: str) -> str:
return ''.join(a + b for a, b in zip_longest(word1, word2, fillvalue=''))

############

class Solution:
def mergeAlternately(self, word1: str, word2: str) -> str:
cur1 = 0
cur2 = 0
res = ""
while cur1 < len(word1) and cur2 < len(word2):
res += word1[cur1] + word2[cur2]
cur1 += 1
cur2 += 1
if cur1 != len(word1):
res += word1[cur1:]
if cur2 != len(word2):
res += word2[cur2:]
return res

• func mergeAlternately(word1 string, word2 string) string {
m, n := len(word1), len(word2)
ans := make([]byte, 0, m+n)
for i := 0; i < m || i < n; i++ {
if i < m {
ans = append(ans, word1[i])
}
if i < n {
ans = append(ans, word2[i])
}
}
return string(ans)
}

• function mergeAlternately(word1: string, word2: string): string {
const res = [];
const n = Math.max(word1.length, word2.length);
for (let i = 0; i < n; i++) {
word1[i] && res.push(word1[i]);
word2[i] && res.push(word2[i]);
}
return res.join('');
}


• impl Solution {
pub fn merge_alternately(word1: String, word2: String) -> String {
let s1 = word1.as_bytes();
let s2 = word2.as_bytes();
let n = s1.len().max(s2.len());
let mut res = vec![];
for i in 0..n {
if s1.get(i).is_some() {
res.push(s1[i]);
}
if s2.get(i).is_some() {
res.push(s2[i]);
}
}
String::from_utf8(res).unwrap()
}
}