1768. Merge Strings Alternately

Description of Problem

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

Solution

Code (Rust)

impl Solution {
    pub fn merge_alternately(word1: String, word2: String) -> String {

        // Since `word1` and `word2` contains only English letters, we can act on &[u8] and Vec<u8>
        let mut v = vec![];
        let (w1, m, w2, n) = (word1.as_bytes(), word1.len(), word2.as_bytes(), word2.len());
        let (mut b, mut i, mut j) = (false, 0, 0);

        // Merging
        while( i < m && j < n) {
            match b {
                false => {
                    v.push(w1[i]);
                    i+=1;
                    b = true;
                },
                true => {
                    v.push(w2[j]);
                    j+=1;
                    b = false;
                }
            }
        }

        // Appending remamining characters
        while( i < m ) {
            v.push(w1[i]);
            i+=1;
        }

        // Appending remamining characters
        while( j < n ) {
            v.push(w2[j]);
            j+=1;
        }

        return String::from_utf8(v).unwrap();

    }
}

Complexity

• \(m\) is length of word1
• \(n\) is length of word2

Time Complexity

• \(T(m,n) = O(m+n) \)

Auxiliary Space

• \(S(m,n) = O(1) \)
  • As output vector does not included in Auxiliary Space