Merge Strings Alternately - Practice Coding Problems

Merge Strings Alternately - 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.

Input & Output

Example 1 — Basic Equal Length

$ Input: word1 = "abc", word2 = "pqr"

› Output: "apbqcr"

💡 Note: Merge alternately: a from word1, p from word2, b from word1, q from word2, c from word1, r from word2

Example 2 — Different Lengths

$ Input: word1 = "ab", word2 = "pqrs"

› Output: "apbqrs"

💡 Note: Merge alternately: a, p, b, q, then append remaining characters from word2: r, s

Example 3 — First String Longer

$ Input: word1 = "abcd", word2 = "pq"

› Output: "apbqcd"

💡 Note: Merge alternately: a, p, b, q, then append remaining characters from word1: c, d

Constraints

1 ≤ word1.length, word2.length ≤ 100
word1 and word2 consist of lowercase English letters

Visualization

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Understanding the Visualization

Input

Two strings word1 and word2 of potentially different lengths

Merge

Alternate taking characters starting with word1

Append

Add remaining characters from longer string to end

Key Takeaway

🎯 Key Insight: Use two pointers to efficiently merge strings while handling different lengths

Asked in

A Amazon 15 M Microsoft 12 G Google 8

Brief HTML summary: The key insight is to use two pointers to traverse both strings simultaneously, alternating character selection. Best approach is two pointers for cleaner logic. Time: O(n + m), Space: O(n + m)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Character by Character	O(n + m)	O(n + m)	Build result by checking both strings character by character in a loop
Two Pointers Optimized	O(n + m)	O(n + m)	Use separate pointers for each string to avoid redundant bounds checking

Brute Force Character by Character — Algorithm Steps

Step 1: Initialize result string and index counter
Step 2: Loop while either string has remaining characters
Step 3: Add character from word1 if available, then from word2 if available

Visualization

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Step-by-Step Walkthrough

Initialize

Start with empty result and index 0

Alternate

Add from word1 if available, then word2 if available

Continue

Repeat until both strings exhausted

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(char* word1, char* word2) {
    int len1 = strlen(word1);
    int len2 = strlen(word2);
    char* result = malloc((len1 + len2 + 1) * sizeof(char));
    int i = 0, pos = 0;
    
    while (i < len1 || i < len2) {
        if (i < len1) {
            result[pos++] = word1[i];
        }
        if (i < len2) {
            result[pos++] = word2[i];
        }
        i++;
    }
    
    result[pos] = '\0';
    return result;
}

int main() {
    char word1[1001], word2[1001];
    
    fgets(word1, sizeof(word1), stdin);
    word1[strcspn(word1, "\n")] = 0;
    
    fgets(word2, sizeof(word2), stdin);
    word2[strcspn(word2, "\n")] = 0;
    
    char* result = solution(word1, word2);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

Single pass through both strings where n and m are lengths of word1 and word2

✓ Linear Growth

Space Complexity

O(n + m)

Result string stores all characters from both input strings

⚡ Linearithmic Space

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Medium Frequency

~8 min Avg. Time

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Ln 1, Col 1

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Character by Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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