Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
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 <= 100word1 and word2 consist of lowercase English letters.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
"abc" "pqr"
"ab" "pqrs"
"abcd" "pq"
zigzag-iterator)minimum-additions-to-make-valid-string)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1768: Merge Strings Alternately
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();
}
}
// Accepted solution for LeetCode #1768: Merge Strings Alternately
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)
}
# Accepted solution for LeetCode #1768: Merge Strings Alternately
class Solution:
def mergeAlternately(self, word1: str, word2: str) -> str:
return ''.join(a + b for a, b in zip_longest(word1, word2, fillvalue=''))
// Accepted solution for LeetCode #1768: Merge Strings Alternately
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()
}
}
// Accepted solution for LeetCode #1768: Merge Strings Alternately
function mergeAlternately(word1: string, word2: string): string {
const ans: string[] = [];
const [m, n] = [word1.length, word2.length];
for (let i = 0; i < m || i < n; ++i) {
if (i < m) {
ans.push(word1[i]);
}
if (i < n) {
ans.push(word2[i]);
}
}
return ans.join('');
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.