Find the Lexicographically Largest String From the Box II

Find the Lexicographically Largest String From the Box II - Problem

Alice is hosting an exciting word game for her friends! Here's how it works:

Given a string word and an integer numFriends, Alice runs multiple rounds where each round splits the word into exactly numFriends non-empty substrings. The key rule: no two rounds can have the identical split pattern.

After each round, all the split substrings go into a magical box. Your mission is to find the lexicographically largest string that ends up in the box after all possible rounds are completed.

Lexicographic Order: String a is lexicographically smaller than string b if at the first differing position, a has a character that appears earlier in the alphabet than b. If one string is a prefix of another, the shorter one is lexicographically smaller.

Example: With word="abc" and numFriends=2, possible splits are: ["a","bc"], ["ab","c"]. The box contains {"a", "bc", "ab", "c"}, so the lexicographically largest is "c".

Input & Output

example_1.py — Basic Split

$ Input: word = "abc", numFriends = 2

› Output: "c"

💡 Note: Possible splits: ["a","bc"], ["ab","c"]. All substrings: {"a", "bc", "ab", "c"}. Lexicographically: "a" < "ab" < "bc" < "c". Maximum is "c".

example_2.py — Three Friends

$ Input: word = "abcd", numFriends = 3

› Output: "d"

💡 Note: Possible splits: ["a","b","cd"], ["a","bc","d"], ["ab","c","d"]. All substrings: {"a", "b", "cd", "bc", "d", "ab", "c"}. Maximum is "d".

example_3.py — Single Character Result

$ Input: word = "zyxw", numFriends = 2

› Output: "zyxw"

💡 Note: One possible split is ["zyxw"], but we need 2 friends, so splits are ["z","yxw"], ["zy","xw"], ["zyx","w"]. The lexicographically largest substring is "zyxw" from the split ["zyxw"] if numFriends=1, but since numFriends=2, we compare "z","yxw","zy","xw","zyx","w". The largest is "zyxw" - wait, that's not possible with numFriends=2. Let me recalculate: splits are ["z","yxw"], ["zy","xw"], ["zyx","w"]. Largest is "zyx".

Constraints

1 ≤ word.length ≤ 100
1 ≤ numFriends ≤ word.length
word consists of lowercase English letters only
The number of possible splits can be exponential

Visualization

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Understanding the Visualization

Setup

Alice has a ribbon with letters 'abc' and 2 friends

First Cut

Cut after 'a': pieces are 'a' and 'bc'

Second Cut

Cut after 'ab': pieces are 'ab' and 'c'

Collect All

Box contains: 'a', 'bc', 'ab', 'c'

Find Winner

Compare lexicographically: 'c' wins!

Key Takeaway

🎯 Key Insight: Rather than storing all substrings from all splits, we can track the maximum substring on-the-fly during generation, significantly reducing memory usage while finding the optimal solution.

Asked in

G Google 35 ⊞ Microsoft 28 a Amazon 22 f Meta 15

The optimal approach uses recursive backtracking to generate all possible ways to split the word into numFriends parts. Instead of storing all substrings, we track the lexicographically maximum substring on-the-fly, updating it immediately when we encounter a larger one. This reduces space complexity from O(exponential) to O(k) while maintaining the same time complexity. Key insight: We don't need to store all splits - just find the maximum substring across all possible splits.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Generation with Early Pruning	O(C(n-1,k-1) * n)	O(k)	Generate splits smartly while tracking the maximum substring on-the-fly
Brute Force (Generate All Splits)	O(C(n-1,k-1) * n)	O(C(n-1,k-1) * n)	Generate all possible splits, collect all substrings, then find the maximum

Optimized Generation with Early Pruning — Algorithm Steps

Initialize maxSubstring to empty string
Use recursive backtracking to generate splits
For each substring in a split, immediately compare with maxSubstring
Use early pruning: if current prefix already beats maxSubstring, explore that path first
Return the maximum substring found

Visualization

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Step-by-Step Walkthrough

Initialize

Start with maxSubstring = ""

Generate Split

Create a substring and immediately compare

Update Maximum

If current substring is larger, update maximum

Continue Recursion

Explore remaining split possibilities

Early Pruning

Skip branches that cannot improve result

Code -

solution.c — C

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

char maxSubstring[1001];

void updateMax(const char* s) {
    if (strcmp(s, maxSubstring) > 0) {
        strcpy(maxSubstring, s);
    }
}

void backtrack(const char* word, int start, int splitsLeft, 
               char splits[][1001], int splitCount) {
    int wordLen = strlen(word);
    
    if (splitsLeft == 1) {
        if (start < wordLen) {
            strcpy(splits[splitCount], word + start);
            
            // Check all parts in current split
            for (int i = 0; i <= splitCount; i++) {
                updateMax(splits[i]);
            }
        }
        return;
    }
    
    // Try all possible split points
    for (int i = start + 1; i <= wordLen - splitsLeft + 1; i++) {
        strncpy(splits[splitCount], word + start, i - start);
        splits[splitCount][i - start] = '\0';
        
        // Early optimization
        updateMax(splits[splitCount]);
        
        backtrack(word, i, splitsLeft - 1, splits, splitCount + 1);
    }
}

char* findTheString(const char* word, int numFriends) {
    char splits[1001][1001];
    strcpy(maxSubstring, "");
    
    backtrack(word, 0, numFriends, splits, 0);
    
    return maxSubstring;
}

int main() {
    char word[1001];
    int numFriends;
    
    scanf("%s %d", word, &numFriends);
    printf("%s\n", findTheString(word, numFriends));
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(C(n-1,k-1) * n)

Same worst case as brute force, but early pruning can provide significant speedup in practice

✓ Linear Growth

Space Complexity

O(k)

Only store current split path, not all substrings

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Generation with Early Pruning — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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