Problem Description

You are given two strings word1 and word2. Your task is to construct a new string called merge by repeatedly choosing characters from either string following these rules:

The Merging Process:

• Start with an empty merge string
• While at least one of word1 or word2 still has characters:
  • You can take the first character from word1, append it to merge, and remove it from word1
  • OR you can take the first character from word2, append it to merge, and remove it from word2
• Continue until both strings are empty

The Goal: Return the lexicographically largest possible merge string you can create.

What is Lexicographically Larger? A string is lexicographically larger than another if at the first position where they differ, it has a character that comes later in the alphabet. For example, "abcd" is lexicographically larger than "abcc" because at position 3, 'd' > 'c'.

Example Walkthrough: If word1 = "abc" and word2 = "def":

• You could take from word2 first: merge = "d", word2 = "ef"
• Then from word2 again: merge = "de", word2 = "f"
• Continue optimally to get the largest possible result

The key insight is that at each step, you should choose the option that leads to the lexicographically largest final string. This means not just comparing single characters, but considering the entire remaining portions of each string to make the optimal choice.

Intuition

The key insight is that when deciding which character to take next, we shouldn't just compare the immediate characters at the current positions. Instead, we need to consider the entire remaining portions of both strings.

Why? Consider this example: word1 = "ba" and word2 = "bb". If we only compare the first characters ('b' vs 'b'), they're equal. But if we look at the entire remaining strings ("ba" vs "bb"), we see that "bb" is lexicographically larger than "ba". Taking from word2 first gives us a better result.

This leads to the greedy strategy: at each step, compare the remaining suffixes word1[i:] and word2[j:]. Choose to take from whichever string has the lexicographically larger suffix. This ensures that we're always making the choice that leads to the largest possible final string.

The beauty of this approach is that Python's string comparison naturally handles the suffix comparison for us. When we write word1[i:] > word2[j:], it compares the strings character by character from left to right, which is exactly what we need.

After the main loop, when one string is exhausted, we simply append whatever remains of the other string to our result, since there's no choice left to make.

This greedy approach works because at each decision point, choosing the lexicographically larger suffix guarantees that we're maximizing the result at that position without compromising future positions - the optimal substructure property that makes greedy algorithms work.

Learn more about Greedy and Two Pointers patterns.

Solution Approach

The implementation uses a two-pointer greedy approach with suffix comparison:

Algorithm Steps:

1. Initialize Two Pointers: Set i = 0 for word1 and j = 0 for word2 to track our current position in each string. Create an empty list ans to build our result.

2. Main Merging Loop: While both strings have remaining characters (i < len(word1) and j < len(word2)):

   • Compare the suffixes word1[i:] with word2[j:]
   • If word1[i:] > word2[j:], take the character from word1:
     • Append word1[i] to ans
     • Increment i by 1
   • Otherwise, take the character from word2:
     • Append word2[j] to ans
     • Increment j by 1

3. Handle Remaining Characters: After the loop, one string might still have characters:

   • Append word1[i:] to ans (empty string if i reached the end)
   • Append word2[j:] to ans (empty string if j reached the end)

4. Build Final Result: Join all characters in ans using "".join(ans) to create the final merged string.

Why This Works:

The suffix comparison word1[i:] > word2[j:] is the critical operation. Python compares these strings lexicographically from left to right. By always choosing the string with the larger suffix, we ensure that:

• We get the largest possible character at the current position
• We maintain the best possible options for future positions

Time Complexity: O(n * (m + n)) where n and m are the lengths of word1 and word2. In the worst case, we perform n + m iterations, and each suffix comparison takes O(m + n) time.

Space Complexity: O(m + n) for storing the result string.

Example Walkthrough

Let's walk through the solution with word1 = "cab" and word2 = "ba":

Initial state:

• i = 0, j = 0
• ans = []
• Remaining: word1[0:] = "cab", word2[0:] = "ba"

Step 1: Compare suffixes "cab" vs "ba"

• "cab" > "ba" lexicographically (first character 'c' > 'b')
• Take from word1: append 'c' to ans
• Move pointer: i = 1
• Result so far: ans = ['c']

Step 2: Compare suffixes "ab" vs "ba"

• "ab" < "ba" lexicographically (second character 'b' < 'a' after both start with same char)
• Take from word2: append 'b' to ans
• Move pointer: j = 1
• Result so far: ans = ['c', 'b']

Step 3: Compare suffixes "ab" vs "a"

• "ab" > "a" lexicographically (has more characters after matching prefix)
• Take from word1: append 'a' to ans
• Move pointer: i = 2
• Result so far: ans = ['c', 'b', 'a']

Step 4: Compare suffixes "b" vs "a"

• "b" > "a" lexicographically ('b' > 'a')
• Take from word1: append 'b' to ans
• Move pointer: i = 3
• Result so far: ans = ['c', 'b', 'a', 'b']

Step 5: word1 is exhausted (i = 3), append remaining from word2

• Append word2[1:] = "a" to ans
• Final result: ans = ['c', 'b', 'a', 'b', 'a']

Final merged string: "cbaba"

This example demonstrates why suffix comparison is crucial. In Step 2, even though both strings start with the same character at their current positions ('a' and 'b'), comparing the full suffixes "ab" vs "ba" correctly identifies that taking from word2 leads to a better result.

Solution Implementation

Time and Space Complexity

Time Complexity: O((n + m) * min(n, m)) where n = len(word1) and m = len(word2).

The algorithm uses a two-pointer approach that iterates through both strings. In the worst case, we traverse all characters from both strings, giving us O(n + m) iterations of the while loop. However, within each iteration, we perform string slicing comparisons word1[i:] and word2[j:].

String comparison in Python takes O(min(len(word1[i:]), len(word2[j:]))) time in the worst case. In the beginning, these slices can be as long as O(n) and O(m) respectively. On average, the comparison cost is O(min(n - i, m - j)).

Since we perform up to n + m comparisons and each comparison can take up to O(min(n, m)) time in the worst case, the overall time complexity is O((n + m) * min(n, m)).

Space Complexity: O(n + m)

The space complexity consists of:

• The ans list which stores all characters from both input strings, requiring O(n + m) space
• The string slices word1[i:] and word2[j:] created during comparisons, which in Python create new string objects requiring O(n + m) space in total
• The final string created by "".join(ans) which requires O(n + m) space

Therefore, the total space complexity is O(n + m).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Greedy Character-by-Character Comparison

The Mistake: A common initial approach is to simply compare the current characters at positions i and j:

# INCORRECT approach
if word1[i] > word2[j]:
    result.append(word1[i])
    i += 1
elif word2[j] > word1[i]:
    result.append(word2[j])
    j += 1
else:
    # When characters are equal, just pick one arbitrarily
    result.append(word1[i])
    i += 1

Why It Fails: This fails when the current characters are equal. Consider word1 = "ca" and word2 = "cb":

• Both start with 'c'
• The incorrect approach might arbitrarily choose from word1, giving us "cca" + "b" = "ccab"
• But the optimal choice is to take from word2 first, giving us "cbc" + "a" = "cbca"
• "cbca" > "ccab" lexicographically

The Solution: Always compare the entire remaining suffixes word1[i:] vs word2[j:], not just single characters. This ensures we consider the full context when making decisions.

Pitfall 2: Inefficient String Concatenation

The Mistake: Building the result string using repeated string concatenation:

# INEFFICIENT approach
result = ""
while i < len(word1) and j < len(word2):
    if word1[i:] > word2[j:]:
        result += word1[i]  # String concatenation creates new string each time
        i += 1
    else:
        result += word2[j]
        j += 1

Why It's Problematic: In Python, strings are immutable. Each += operation creates a new string object, copying all previous characters. This leads to O((m+n)²) time complexity for string building alone.

The Solution: Use a list to collect characters and join them once at the end:

result = []
# ... append characters to list
return "".join(result)  # Single O(m+n) operation

Pitfall 3: Incorrect Handling of Remaining Characters

The Mistake: Using a loop to append remaining characters one by one:

# After main loop
while i < len(word1):
    result.append(word1[i])
    i += 1
while j < len(word2):
    result.append(word2[j])
    j += 1

Why It's Suboptimal: While this works correctly, it's unnecessarily verbose and less efficient than using string slicing.

The Solution: Use string slicing to append all remaining characters at once:

result.append(word1[i:])  # Appends remaining portion or empty string
result.append(word2[j:])  # Appends remaining portion or empty string

This is cleaner and more efficient since slicing operations are optimized in Python.