Leetcode 1754: Largest Merge Of Two Strings
You are given two strings,
word1 and word2. You need to construct a string merge by choosing characters from both strings. At each step, you can either append the first character of word1 to merge (if word1 is non-empty) or append the first character of word2 to merge (if word2 is non-empty). The goal is to form the lexicographically largest string possible from these choices.Problem
Approach
Steps
Complexity
Input: You are given two strings, `word1` and `word2`, both consisting of lowercase English letters.
Example: Input: word1 = 'abcde', word2 = 'acd'
Constraints:
• 1 <= word1.length, word2.length <= 3000
• word1 and word2 consist only of lowercase English letters.
Output: Return the lexicographically largest merge string you can construct.
Example: Output: 'abcdeacd'
Constraints:
• The resulting string is guaranteed to be lexicographically the largest possible merge.
Goal: To find the lexicographically largest merge string, we need to make the optimal choice at each step of picking from `word1` or `word2`.
Steps:
• 1. Compare the first characters of `word1` and `word2`.
• 2. If the character in `word1` is larger, append it to `merge`. Otherwise, append the character from `word2`.
• 3. If one string becomes empty, append the remaining characters from the other string to `merge`.
Goal: Ensure that the solution works for large inputs and performs within time limits.
Steps:
• The solution should be optimized to handle inputs with lengths up to 3000.
Assumptions:
• Both `word1` and `word2` are non-empty strings consisting of lowercase English letters.
• Input: Input: word1 = 'abcd', word2 = 'acef'
• Explanation: At each step, we compare the first characters of `word1` and `word2`. The largest lexicographical choice is appended to `merge`. The result is 'abcdeaf'.
• Input: Input: word1 = 'abc', word2 = 'aaa'
• Explanation: Here, we always take from `word1` as its characters are larger than those in `word2`, resulting in 'abcaaa'.
Approach: The approach involves comparing the first characters of both strings and greedily choosing the lexicographically larger one, ensuring the largest possible merge string.
Observations:
• We can use a greedy approach to always select the larger character between the two strings at each step.
• The solution is based on the principle of comparing characters in a lexicographical manner.
Steps:
• 1. Initialize an empty string `merge`.
• 2. While both `word1` and `word2` are non-empty, compare the first characters of both strings.
• 3. Append the larger character to `merge` and remove it from the respective string.
• 4. When one string becomes empty, append the remaining characters of the other string to `merge`.
Empty Inputs:
• If either `word1` or `word2` is empty, simply return the other string.
Large Inputs:
• The solution should handle cases where `word1` and `word2` are close to 3000 characters in length.
Special Values:
• If both strings are identical, the result is the string concatenated to itself.
Constraints:
• The solution should perform efficiently with the maximum input size.
string largestMerge(string w1, string w2) {
if (w1 == "" || w2 == "")
return w1 + w2;
else if (w1 < w2)
return w2[0] + largestMerge(w1, w2.substr(1));
else return w1[0] + largestMerge(w1.substr(1), w2);
}
1 : Function Declaration
string largestMerge(string w1, string w2) {
Declares the recursive function to create the largest lexicographical merge of two strings.
2 : Base Case
if (w1 == "" || w2 == "")
Checks if either string is empty, which serves as the base case for recursion.
3 : Return Statement
return w1 + w2;
If one string is empty, appends the other string and ends the recursion.
4 : Conditional Check
else if (w1 < w2)
Checks if the first string is lexicographically smaller than the second string.
5 : Recursive Call
return w2[0] + largestMerge(w1, w2.substr(1));
Appends the first character of `w2` and makes a recursive call with the remaining part of `w2`.
6 : Recursive Call
else return w1[0] + largestMerge(w1.substr(1), w2);
Appends the first character of `w1` and makes a recursive call with the remaining part of `w1`.
Best Case: O(n + m), where n is the length of `word1` and m is the length of `word2`.
Average Case: O(n + m), as each character from both strings is processed once.
Worst Case: O(n + m), where n and m are the lengths of the strings.
Description: The time complexity is linear in the sum of the lengths of the two strings, as each character is processed once.
Best Case: O(n + m), as the space used is proportional to the length of the result string.
Worst Case: O(n + m), where n and m are the lengths of `word1` and `word2`.
Description: The space complexity is linear, as the merge string needs to store all the characters from both input strings.
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