3403. Find the Lexicographically Largest String From the Box I

MediumTwo PointersStringEnumeration

Problem Description

You are given a string word, and an integer numFriends.

Alice is organizing a game for her numFriends friends. There are multiple rounds in the game, where in each round:

word is split into numFriends non-empty strings, such that no previous round has had the exact same split.
All the split words are put into a box.

Find the lexicographically largest string from the box after all the rounds are finished.

Intuition

To solve this problem, the key observation is to find the largest possible part of the string word that can be created through the specified splitting. Each part must be non-empty, and no previous round should repeat the same split configuration. The challenge is to achieve the lexicographically largest substring after considering all possible partitions.

Given numFriends splits, if numFriends equals 1, the whole string word itself is the largest as there are no sub-divisions. However, when numFriends is greater than 1, we aim to find the largest slice of the string that can be achieved when dividing the string into numFriends parts.

The approach is to iterate through the string while calculating possible parts that can be formed from the current index. We consider the maximum length these parts can be by leveraging the total permissible splits minus the ones already made (n - numFriends + 1). We compare each possible slice's lexicographical order to select the biggest one as the answer.

Learn more about Two Pointers patterns.

Solution Approach

To implement the solution for finding the lexicographically largest substring after splitting the string word into numFriends parts, the following approach is taken:

Edge Case Check:
- If numFriends is 1, the answer is simply the entire string word, since no splitting is required, and no other combination will surpass the word itself.
Iterative Comparison:
- Start by determining the length n of the string word.
- Initialize an empty string ans to hold the current largest substring.
Sliding Window Approach:
- Loop through each position i of word (for i in range(n)), treating each as a potential starting point of a substring.
- Calculate k, the maximum valid length of the substring starting at position i, given as min(n - i, n - numFriends + 1). This ensures the substring remains within bounds and considers the maximum permissible length.
- Extract the substring word[i : i + k] and compare it with ans to possibly update ans to the new maximal substring.
Output:
- After processing, ans holds the lexicographically largest substring after all possible configurations defined by numFriends have been considered.

This algorithm efficiently processes the string in linear time, maximizing the substring through dynamic sliding window techniques and leveraging lexicographical comparisons.

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Example Walkthrough

Let's go through a small example to illustrate the solution approach using the problem description and solution strategy provided:

Suppose we have the string word = "abcde" and numFriends = 2.

Edge Case Check:
- Since numFriends is not 1, we proceed to the iterative comparison step.
Iterative Comparison:
- Determine the length of word, n = 5.
- Initialize an empty string ans to hold the largest substring found.
Sliding Window Approach:
- We loop through each position in word and consider it as a potential starting point for slicing:
- Index 0 (starting at 'a'):
  - Calculate k = min(5 - 0, 5 - 2 + 1) = 4.
  - The substring from index 0 with length 4 is "abcd".
  - This becomes our initial ans = "abcd" since it's the first substring we've verified.
- Index 1 (starting at 'b'):
  - Calculate k = min(5 - 1, 5 - 2 + 1) = 4.
  - The substring from index 1 with length 4 is "bcde".
  - Compare "bcde" with ans = "abcd". Since "bcde" is lexicographically larger, update ans = "bcde".
- Index 2 (starting at 'c'):
  - Calculate k = min(5 - 2, 5 - 2 + 1) = 3.
  - The substring from index 2 with length 3 is "cde".
  - Compare "cde" with ans = "bcde". Since "cde" is not larger, keep ans = "bcde".
- Index 3 (starting at 'd'):
  - Calculate k = min(5 - 3, 5 - 2 + 1) = 2.
  - The substring from index 3 with length 2 is "de".
  - Compare "de" with ans = "bcde". As "de" is not larger, keep ans = "bcde".
- Index 4 (starting at 'e'):
  - Calculate k = min(5 - 4, 5 - 2 + 1) = 1.
  - The substring from index 4 with length 1 is "e".
  - Compare "e" with ans = "bcde". Since "e" is not larger, maintain ans = "bcde".
Output:
- After processing all splits, ans = "bcde" is the lexicographically largest substring after considering all possible splits for numFriends = 2.

Solution Implementation

1class Solution:
2    def answerString(self, word: str, numFriends: int) -> str:
3        # If there is only one friend, return the original word unchanged
4        if numFriends == 1:
5            return word
6      
7        n = len(word)  # Determine the length of the word
8        ans = ""  # Initialize the answer as an empty string
9      
10        # Loop through each character in `word`
11        for i in range(n):
12            # Calculate the maximum length of the substring to be considered
13            k = min(n - i, n - numFriends + 1)
14            # Update `ans` with the lexicographically largest substring found
15            ans = max(ans, word[i: i + k])
16      
17        return ans  # Return the lexicographically largest substring found
18

1class Solution {
2    public String answerString(String word, int numFriends) {
3        // If there is only one friend, return the whole word as is.
4        if (numFriends == 1) {
5            return word;
6        }
7      
8        int wordLength = word.length();
9        String answer = "";
10      
11        // Iterate over each character of the word.
12        for (int i = 0; i < wordLength; ++i) {
13            // Determine the maximum substring length that can be taken starting from the current position.
14            int maxSubstringLength = Math.min(wordLength - i, wordLength - numFriends + 1);
15          
16            // Extract the substring starting at position i with the determined length.
17            String currentSubstring = word.substring(i, i + maxSubstringLength);
18          
19            // Compare the current substring with the answer found so far and update if it is greater.
20            if (answer.compareTo(currentSubstring) < 0) {
21                answer = currentSubstring;
22            }
23        }
24      
25        // Return the largest lexicographical substring found.
26        return answer;
27    }
28}
29

1#include <string>
2#include <algorithm>
3
4class Solution {
5public:
6    // Method to determine the maximum lexicographical substring
7    // that can be formed given the number of friends (numFriends).
8    string answerString(string word, int numFriends) {
9        // If there is only one friend, return the original word
10        if (numFriends == 1) {
11            return word;
12        }
13
14        int n = word.size(); // Get the length of the input word
15        string answer; // Initialize an empty string to store the result
16
17        // Iterate over each character in the word
18        for (int i = 0; i < n; ++i) {
19            // Determine the maximum length of substring we can extract
20            int k = std::min(n - i, n - numFriends + 1);
21
22            // Extract the substring from the current position with length k
23            string currentSubstring = word.substr(i, k);
24
25            // Update the answer with the maximum lexicographical substring
26            answer = std::max(answer, currentSubstring);
27        }
28
29        return answer; // Return the result
30    }
31};
32

1// Function to compute the answer string based on the input word and number of friends.
2function answerString(word: string, numFriends: number): string {
3    // If there is only one friend, return the original word, as no comparison is required.
4    if (numFriends === 1) {
5        return word;
6    }
7
8    // Initialize a variable to store the best possible substring answer.
9    let ans: string = '';
10
11    // Get the length of the input word.
12    const wordLength = word.length;
13
14    // Iterate over each character in the word.
15    for (let index = 0; index < wordLength; ++index) {
16        // Calculate the maximum length of the substring starting from the current index.
17        const maxLength = Math.min(wordLength - index, wordLength - numFriends + 1);
18
19        // Slice out the substring from the current index, with calculated maximum length.
20        const substring = word.slice(index, index + maxLength);
21
22        // Update the answer if the current substring is lexicographically greater.
23        if (ans < substring) {
24            ans = substring;
25        }
26    }
27
28    // Return the best found substring.
29    return ans;
30}
31

Time and Space Complexity

The provided code performs operations that can be analyzed as follows:

The outer loop runs n times, where n is the length of the input word.
Inside the loop, k is calculated as min(n - i, n - numFriends + 1), and a substring of length k is created and potentially assigned to ans using max().

This means the substring operation and comparison occur in constant time if Python's string slicing is assumed to be O(k) where k is the slice length. However, since k can range up to n, in the worst case when numFriends is 1, each iteration may involve operations involving the length of the substring.

Thus, the overall time complexity is O(n^2), due to the nested operations of slicing and comparing potentially every substring combination as the outer loop iterates.

The space complexity is primarily O(1), since the space required for variables like ans is minimal and does not scale with input size, excluding the input word itself, which is predetermined space.

Therefore:

Time Complexity: O(n^2)
Space Complexity: O(1)

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