Solution 1 - Two Pointers

Alternate characters while both remain, then append leftover.

TimeComplexity: O(n+m)

SpaceComplexity: O(n+m)

""" 1768.1 - Merge Strings Alternately - Solution 1 - Two Pointers """

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        out = []
        i = 0
        while i < len(word1) or i < len(word2):
            if i < len(word1): out.append(word1[i])
            if i < len(word2): out.append(word2[i])
            i += 1
        return "".join(out)

function mergeAlternately(word1, word2){
 let out=[];
 for(let i=0;i<Math.max(word1.length, word2.length);i++){
  if(i<word1.length) out.push(word1[i]);
  if(i<word2.length) out.push(word2[i]);
 }
 return out.join('');
}

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

package main

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

func main() {}

fn merge_alternately(word1: &str, word2: &str) -> String {
	let a: Vec<char> = word1.chars().collect();
	let b: Vec<char> = word2.chars().collect();
	let mut out = String::new();
	for i in 0..a.len().max(b.len()) {
		if i < a.len() { out.push(a[i]); }
		if i < b.len() { out.push(b[i]); }
	}
	out
}
fn main() {}

1768 - Merge Strings Alternately

Solution 1 - Two Pointers