Ph.D. Qual Study Assistant

1. The Problem

Given a string `S` of `n` characters with no spaces, and a function `Quality(word)` that returns a score for any given string, we need to insert spaces into `S` to partition it into a sequence of words `w_1, w_2, ..., w_k`. The goal is to find a partition that maximizes the sum of the qualities of its words: `Quality(w_1) + Quality(w_2) + ... + Quality(w_k)`.

2. The Idea: Dynamic Programming

The problem has optimal substructure. The best segmentation of a string `S` must contain the best segmentation of one of its prefixes. This allows us to build an optimal solution from optimal solutions to smaller subproblems.

Let `MaxQ[i]` be the maximum possible quality score for a segmentation of the prefix `S[1...i]` (the first `i` characters of `S`).

• Base Case: For an empty prefix, the quality is 0. So, `MaxQ[0] = 0`.
• Recurrence Relation: To compute `MaxQ[i]`, we consider every possible last word. The last word could be `S[j...i]` for any `1 <= j <= i`. If we make `S[j...i]` the last word, the total quality is the maximum quality of the preceding part of the string (`S[1...j-1]`) plus the quality of this last word. We want to choose the split point `j` that maximizes this sum.

We compute `MaxQ[i]` for `i = 1, 2, ..., n`. The final answer is the value `MaxQ[n]`.

3. The Algorithm

1. Create a 1D array, `MaxQ`, of size `n+1`. Initialize `MaxQ[0] = 0` and all other entries to negative infinity.
2. Create a 1D array, `Split`, of size `n+1` to store the optimal split position for each prefix. `Split[i] = j` means the best segmentation of `S[1...i]` has `S[j...i]` as its last word.
3. Iterate with `i` from 1 to `n` (the length of the prefix we are solving for).
   • Iterate with `j` from 1 to `i` (the potential start of the last word).
     • Calculate `current_quality = MaxQ[j-1] + Quality(S.substring(j-1, i))`.
     • If `current_quality > MaxQ[i]`, update `MaxQ[i] = current_quality` and `Split[i] = j-1`.
4. The maximum total quality is `MaxQ[n]`. The segmentation can be reconstructed by backtracking from `n` using the `Split` array.

4. Complexity Analysis

Time Complexity: Polynomial

Let `n` be the length of the string `S`. The algorithm has two nested loops. The outer loop (`i`) runs `n` times, and the inner loop (`j`) runs up to `n` times. Inside the inner loop, we extract a substring and call the `Quality` function. Let's assume creating a substring of length `k` takes O(k) time and the `Quality` function runs in `P(k)` time, where `P` is a polynomial. The total complexity is roughly the sum over `i` and `j` of `O(i-j) + P(i-j)`. This results in a total time complexity of O(n² * (Avg_Substring_Time + Avg_Quality_Time)). Since `Quality` is polynomial and substring creation is at most O(n), the entire algorithm is guaranteed to be polynomial, satisfying the problem's requirement. A common implementation results in O(n²).

Space Complexity: O(n)

We use two 1D arrays, `MaxQ` and `Split`, both of size `n+1`, which requires O(n) space.