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

1768. Merge Strings Alternately

Level

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()
        }
    }
    
    

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