Find the Lexicographically Largest String From the Box I

Problem Description

Given a string word and an integer numFriends, imagine that in many rounds you split word into exactly numFriends non‐empty parts (each round using a split that has never been used before). All parts from every round are collected into a “box.” Your task is to return the lexicographically largest string present in the box. Note that each round’s split is produced by choosing numFriends–1 split points (between characters) in word. For example, for word = "dbca" with numFriends = 2, the possible splits are: ["d","bca"], ["db","ca"], and ["dbc","a"] – and the answer is "dbc".

Key Insights

• Each round splits the word into numFriends contiguous non‐empty substrings.
• The parts that appear are exactly the pieces determined by a choice of numFriends – 1 cut points in word.
• Not every contiguous substring of word appears; a substring can be a segment only when its start and end positions coincide with partition boundaries that allow the remaining parts to be non‐empty.
• We can show that a substring s[i:j] can be part of a valid partition if and only if there exists an index m (its “segment index” in the partition, 0‑indexed) such that:
  • If it is the first segment (m = 0): i must be 0 and there must be at least numFriends – 1 characters remaining (i.e. j ≤ n – (numFriends – 1)).
  • If it is the last segment (m = numFriends – 1): j must equal n and there are at least numFriends – 1 characters before (i ≥ numFriends – 1).
  • For a middle segment (0 < m < numFriends – 1): there must be at least m characters before and numFriends – 1 – m characters after s[i:j].
• In summary, s[i:j] can be a segment if there exists an m such that: max(0, numFriends – 1 – (n – j)) ≤ min(i, numFriends – 1).
• Thus, rather than enumerating all partitions, we can “simulate” over all possible segments s[i:j] that can occur with the proper lower‐bound on i (which is L = max(0, numFriends – 1 – (n – j)) for a fixed end index j).
• Finally, we iterate over all allowed segments and update the maximum lexicographical string.

Space and Time Complexity

Time Complexity: O(n^2 * S) in the worst case where n = word.length and S is the cost to compare two substrings. (n is at most 5000 so with efficient comparison—via built‐in methods or by hashing—the solution passes.) Space Complexity: O(n) extra space to store the best result and, if optimizing comparisons, O(n) space for prefix hashes.

Solution

We solve the problem by “simulating” all possible ways a contiguous substring could be one segment in a valid partition. For every ending index j from 1 to n (n = word.length), we determine the smallest starting index i (call it L) for which s[i:j] can appear. The rule for L is:   L = max(0, numFriends – 1 – (n – j)) Then for every starting index i from L to j – 1 we consider the substring word[i:j]. We compare it to the current best candidate (initialized as the empty string) and update if it is lexicographically larger. Key points in the approach include:  • Using a double loop to enumerate all segments that could be produced in some valid split.  • The lower bound L on the starting index i for a fixed j guarantees that the chosen substring can “fit” as a segment (either first, middle, or last) in a valid partition.  • Standard string comparison is used to update the best candidate. The algorithm uses O(n^2) iterations and, if needed, optimizations such as hashing or precomputed LCP (Longest Common Prefix) arrays could speed up substring comparisons.

Code Solutions

Below are code implementations in Python, JavaScript, C++, and Java with explanatory comments.