Optimize password transform and discount scheduling
Company: Citadel
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: Medium
Interview Round: Take-home Project
Part 1 — Transform to palindromic k-periodic string:
Input: a lowercase string currentPassword and integer k.
Constraints: 1 <= k < len(currentPassword) <= 2*10^5, and len(currentPassword) is divisible by k.
Operation: you may change any character to any lowercase letter; the cost is the number of positions changed.
Goal: convert currentPassword to newPassword such that
(a) newPassword is a palindrome, and
(b) newPassword[i] = newPassword[i + k] for all valid i (0-based). Return the minimum number of character changes required and describe the algorithm with time and space complexity.
Example: currentPassword = "abzzbz", k = 3 → one optimal newPassword is "zbzzbz"; answer = 1.
Part 2 — Minimize processing cost with daily discount:
Input: integer n; arrays filterCost[1..n], startDay[1..n], endDay[1..n]; and integer discountPrice. For each day d, the set of active images is { i | startDay[i] <= d <= endDay[i] }. Without a discount, day d costs sum(filterCost[i] over active images i). You may apply the discount at most once per day; if applied on day d, the total cost that day is exactly discountPrice.
Goal: process all n images over their required day ranges to minimize the total cost across all days; return the minimum total cost modulo 1,000,000,007. Provide the algorithm and its time and space complexity.
Example: n = 3; filterCost = [2,3,4]; startDay = [1,1,2]; endDay = [2,3,4]; discountPrice = 6. Per-day sums: day1=5, day2=9, day3=7, day4=4. Using the discount on days 2 and 3 gives total 5 + 6 + 6 + 4 = 21.
Quick Answer: Optimize password transform and discount scheduling evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.
Solution
# Solution Alignment
The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.
## Problem Restatement
Part 1 — Transform to palindromic k-periodic string: Input: a lowercase string currentPassword and integer k. Constraints: 1 <= k < len(currentPassword) <= 2*10^5, and len(currentPassword) is divisible by k. Operation: you may change any character to any lowercase letter; the cost is the number of positions changed. Goal: convert currentPassword to newPassword such that (a) newPassword is a palindrome, and (b) newPassword[i] = newPassword[i + k] for all valid i (0-based). Return the minimum number of character changes required and describe the algorithm with time and space complexity. Example: currentPassword = "abzzbz", k = 3 → one optimal newPassword is "zbzzbz"; answer = 1. Part 2 — Minim...
## Recommended Approach
Define a state that captures exactly the remaining decision information. Fill base cases first, then transition from smaller subproblems to larger ones. For games, use score difference or minimax DP; for counting, sum the valid predecessor states.
## Correctness
The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.
## Complexity
Typical DP time is number_of_states times transition_cost. Space can often be reduced when transitions only need the previous layer or diagonal.
## Edge Cases and Tests
Empty input, length 1, invalid symbols, negative values where allowed, ties under optimal play, and large counts requiring modulo arithmetic.