I started by thinking about it as some kind of BFS or greedy sweep and got confused fast.
Model the problem as a dynamic programming problem where the state is the position of each token, but since tokens are indistinguishable and move independently, we can treat each token separately and combine results. For each token, compute the maximum coins it can collect from each starting position using DP, then assign tokens to starting positions to maximize total coins without conflicts. Alternatively, use DP over the board with state representing the positions of all tokens, but optimize by noting tokens move in fixed steps and cannot share cells.
Pro tip: Clarify whether tokens can start on any empty cell or are given initial positions; if not specified, assume you can place tokens optimally. Also, consider that tokens might block each other, so the order of moves matters.
Ask clarifying questions: Are token starting positions given? Can tokens move multiple times? Do they collect coins only on landing, not passing over? Can tokens move left? Confirm that tokens move exactly 3 cells right, cannot share cells, and collect coins only when landing on a coin cell.
For a single token, define dp[i] as the maximum coins collectable starting from position i. Transition: dp[i] = coin[i] + max(dp[j] for j > i and j-i is a multiple of 3? Actually, token moves exactly 3 cells right each move, so from i it can go to i+3, i+6, etc. So dp[i] = coin[i] + max(dp[i+3], dp[i+6], ...) but careful: if it moves to i+3, it collects coin at i+3, so dp[i] = coin[i] + max_{k>=1} dp[i+3k]? Actually, if it moves to i+3, it collects coin at i+3, so dp[i] = coin[i] + max(dp[i+3], dp[i+6], ...) but that would double count? Let's think: If token starts at i, it collects coin at i if any. Then it can move to i+3, collecting coin there, then from i+3 it can move to i+6, etc. So the total coins from starting at i is coin[i] + max over next positions j = i+3, i+6, ... of (coins collected from j). So dp[i] = coin[i] + max_{k>=1} dp[i+3k]. But this allows skipping positions, which is fine because token can choose to move multiple times. However, if it moves from i to i+3, it cannot then go back, so it's a path. So dp[i] = coin[i] + max(0, max_{k>=1} dp[i+3k]). But note that if it moves to i+3, it collects coin at i+3, so dp[i+3] already includes coin at i+3. So that works.
If tokens are indistinguishable and can be placed anywhere initially, we need to choose starting positions for each token such that their paths do not intersect (cannot share a cell at any time). Since tokens move only right by 3, their paths are sets of positions congruent modulo 3. Two tokens starting at positions with different residues mod 3 will never collide because they move by 3, so they stay in their residue class. Tokens in the same residue class could collide if they land on the same cell. To avoid collision, we can order them and ensure they maintain distance. This becomes a problem of selecting up to T starting positions (T = number of tokens) from each residue class, with the constraint that if two tokens are in the same residue class, their paths cannot overlap. Since they move right, we can think of them as moving along a line; to avoid collision, we can assign them to non-overlapping intervals. This can be solved by DP over the board, considering tokens one by one, or by noting that if we have multiple tokens, we can just take the best T paths that are disjoint. But since tokens are identical, we can use DP with state (position, tokens_used) and ensure that when we place a token, we mark its path as used. However, the board size might be small enough for brute force or DP with bitmask if tokens are few.
If the number of tokens is small (e.g., <= 3), we can use DP with state as tuple of token positions. If tokens are many, we can use the fact that tokens in different residue classes are independent, so we can solve each residue class separately and then combine by choosing the best allocation of tokens to residue classes. For each residue class, we have a 1D array of cells (every 3rd cell). Tokens move along this array by jumping any number of steps? Actually, in the original board, moving 3 cells right corresponds to moving 1 step right in the residue class array. So in the residue class array, a token can move any number of steps right? Wait: In original, token moves exactly 3 cells right, so from position i to i+3. That is one step in the residue class array. But can it move multiple times? Yes, so it can move any number of steps right in the residue class array, but each move is exactly one step? Actually, if it moves from i to i+3, that's one move. Then from i+3 to i+6, another move. So it can move any number of steps, but each step is exactly one cell in the residue class array. So in the residue class array, a token can move any number of cells to the right, but it collects coins only on landing cells. So it's like a path that can skip cells? No, if it moves from i to i+3, it lands on i+3, so it collects coin there. It cannot skip i+3 and go to i+6 in one move. So it must land on every intermediate cell? Actually, the problem says 'Tokens can only move exactly 3 cells to the right', so each move is exactly 3 cells. So from i, it can move to i+3, then from i+3 to i+6, etc. So it visits every cell in its residue class that is a multiple of 3 away from start? Not necessarily: it could stop at i+3 and not move further. So it can choose to stop at any point. So the path is a contiguous segment in the residue class array starting at the initial position and going right, but it can stop at any point. So it collects coins on all cells it lands on, which are exactly the cells from start to stop in steps of 3. So it's a contiguous subarray in the residue class array. So for a single token, the maximum coins it can collect from a starting position is the maximum sum of a contiguous subarray starting at that position? Actually, it collects coins on all cells it lands on, which are start, start+3, start+6, ... up to some end. So the total coins is the sum of coins on that arithmetic progression. So to maximize, for each starting position, we want to choose an end to maximize the sum. That's like maximum subarray sum starting at i in the residue class array. But note that the token can also choose not to move at all, so it collects only the coin at start. So dp[i] = coin[i] + max(0, dp[i+3])? Actually, if it moves to i+3, it collects coin at i+3 and then can continue. So dp[i] = coin[i] + max(0, dp[i+3])? But careful: if it moves to i+3, it collects coin at i+3, so the total from i is coin[i] + (coins from i+3 onwards). But dp[i+3] already includes coin at i+3. So dp[i] = coin[i] + max(0, dp[i+3])? That would mean it can either stop after i (collect only coin[i]) or move to i+3 and then collect dp[i+3]. But if it moves to i+3, it collects coin at i+3, so total = coin[i] + dp[i+3]. But dp[i+3] includes coin at i+3, so that's correct. However, if it moves to i+3, it cannot skip i+3. So the recurrence is dp[i] = coin[i] + max(0, dp[i+3])? But wait, if it moves to i+3, it must land on i+3, so it collects coin there. So the total is coin[i] + dp[i+3]. But dp[i+3] is the max coins starting from i+3, which includes coin at i+3. So that works. But what if it moves to i+6 directly? It can't, because it must move exactly 3 cells at a time. So it must land on i+3 first. So the recurrence is correct: dp[i] = coin[i] + max(0, dp[i+3]). But this assumes that if it moves, it must continue optimally from i+3. However, it could also move to i+3 and then stop,
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Pretty natural extension once you have the mod-3 insight.
First, clarify the original problem and the current solution for k=3, then generalize the approach by replacing the constant 3 with a variable k. Discuss how the time and space complexity change, and consider edge cases such as k=1 or k larger than the array size.
Pro tip: Demonstrate awareness that the optimal solution may depend on k (e.g., using a heap for small k vs. sorting for large k) and mention that you would discuss trade-offs with the interviewer before coding.
Restate the problem for k=3 and briefly explain the current algorithm, including its time and space complexity.
Adapt the solution by replacing the constant 3 with k, ensuring the logic works for any positive integer k.
Determine how the time and space complexity depend on k, and discuss whether the solution remains optimal for different k values.
If k is very small or very large, evaluate if a different data structure or algorithm (e.g., heap, sorting) would be more efficient.
Test edge cases like k=1, k greater than the input size, and k=0 if applicable, and ensure the solution handles them correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I said yes without really thinking it through, and then paused.
First, clarify that the question is about the coin change problem where we maximize total value, not minimize coin count. Then, explain that the greedy approach (which works for minimizing count with canonical coin systems) does not generally work for maximizing value; instead, dynamic programming is required. Finally, outline the DP solution and discuss its complexity and trade-offs.
Pro tip: Mention that the greedy approach fails for maximizing value even with canonical coin systems, and provide a counterexample to illustrate. This shows deep understanding and avoids a common pitfall.
Restate the problem: given coin denominations and a target amount, maximize the total value of coins used (each coin can be used at most once? or unlimited?). Clarify constraints: are coins unlimited (unbounded) or limited (0/1)? This affects the solution.
Explain that for minimizing coin count, greedy works for canonical systems (e.g., US coins). For maximizing value, greedy (e.g., always pick the largest coin that fits) does not work in general. Provide a counterexample: coins {1, 3, 4}, target 6. Greedy picks 4 then 1+1 = 6 (value 6), but optimal is 3+3 = 6 (value 6) — actually same value? Wait, need a better example. For maximizing value with unlimited coins, if we want to reach exactly target, the maximum value is always the target if possible. So the problem might be: maximize value without exceeding target? Or with limited coins? The question is ambiguous. Assume it's: given coin denominations and a target amount, maximize total value of coins used (each coin can be used at most once? or unlimited?) but you cannot exceed the target? Actually, typical problem: given coins with values, maximize total value with total weight <= capacity? That's knapsack. But here it's coins and target amount. Possibly it's: you have coins of different denominations, you want to select a subset to maximize total value subject to total sum <= target? Or exactly target? The question says 'maximize total value instead of coin count', so likely it's the coin change problem where you want to make change for a target amount, but instead of minimizing number of coins, you want to maximize the total value of coins used. But if you must make exactly the target, the total value is fixed (the target). So that doesn't make sense. Perhaps it's: you have a set of coins with values, and you want to pick coins to maximize total value without exceeding a target sum? That's the knapsack problem. Or maybe it's: you have unlimited coins of each denomination, and you want to make change for a target amount, but you want to maximize the number of coins? No, that's different. The question likely refers to a variant where coins have different values (not just denominations) and you want to maximize total value, perhaps with a constraint on the number of coins? Or it's about the greedy algorithm for coin change: if coins have different values, does greedy still work for maximizing value? I think the intended interpretation is: In the classic coin change problem, we minimize the number of coins to make a target amount. Now consider a variant where we want to maximize the total value of coins used to make the target amount. But if we must make exactly the target, the total value is the target. So that's trivial. So maybe it's: we have a set of coins with values, and we want to select a subset to maximize total value subject to the total number of coins used? That doesn't make sense either. Perhaps it's: we have coins of different denominations, and we want to make change for a target amount, but we want to maximize the total value of coins used, meaning we can use more coins than necessary? But the total value is still the target. So the only way to maximize value is to use coins that sum to more than the target? That would be overpaying. So maybe the question is about the knapsack problem: given items with weights and values, maximize value subject to weight capacity. But it says 'coins had different values', so coins have values and weights? Typically coins have denominations which are both weight and value? I think the question is a follow-up to a previous discussion about coin change where we minimize coin count. The interviewer asks: what if coins had different values and you wanted to maximize total value instead of coin count? Does the same approach still work? This is likely referring to the greedy algorithm for coin change. For minimizing coin count, greedy works for some coin systems. For maximizing value, greedy does not work. So the answer should explain that the greedy approach does not work for maximizing value, and dynamic programming is needed. Provide an example where greedy fails. For instance, coins with values {1, 3, 4} and we want to make change for 6, but we want to maximize the total value? That doesn't make sense. So maybe the problem is: we have coins with values, and we want to select a subset to maximize total value without exceeding a target sum. That's the 0/1 knapsack problem. Greedy by value/weight ratio works for fractional knapsack but not for 0/1. So the answer: the same greedy approach does not work; we need dynamic programming. So I'll frame it as: the problem is to maximize total value subject to a constraint (e.g., total weight <= capacity, or total number of coins <= k). The greedy approach fails; DP is required. I'll keep it general.
Outline a dynamic programming approach: define state dp[i][v] = maximum value achievable using first i coins with total value v (or total weight v). Transition: either skip coin i or take it. Complexity O(n * V) where V is the target value or capacity. Mention that this is similar to the knapsack problem.
Compare greedy vs DP: greedy is faster (O(n log n)) but not always correct; DP is slower (O(nV)) but guarantees optimality. Mention that if the problem has special structure (e.g., coin denominations are canonical), greedy might work, but for general values, DP is needed.
Summarize that the approach depends on the exact problem: for maximizing value with constraints, greedy often fails, and DP is the standard solution. Highlight that understanding the problem's properties (e.g., matroid structure) can determine if greedy works.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the original problem's token movement rules and the non-crossing argument. Then, analyze how adding a left-by-3 move affects the reachable state space and whether tokens can cross paths. Finally, discuss the implications for algorithm design and complexity.
Pro tip: Relate the change to classic problems like the 15-puzzle or token swapping on graphs, and mention that adding moves can break invariants such as parity or ordering, which are often used in non-crossing proofs.
Restate the original token movement rules and the non-crossing argument to ensure a common understanding. Identify the invariants or properties that prevent tokens from crossing.
Consider how allowing a token to move left by 3 changes the state graph. Determine if this new move enables tokens to bypass each other or alters the reachable configurations.
Assess whether the new move expands the set of reachable states. Check if previously unreachable configurations become reachable, and whether the problem becomes easier or harder.
Investigate if tokens can now cross paths. Look for counterexamples where tokens swap order, and consider how this affects any parity-based or ordering-based invariants.
Explain how the change impacts algorithm design, such as BFS/DFS state space size, potential for greedy solutions, or complexity class changes. Mention trade-offs like increased branching factor.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Explain that you would augment the state-space search (e.g., BFS) to store parent pointers, then backtrack from the goal to reconstruct the move sequence. The length is bounded by the number of states along the shortest path, which is at most the total number of reachable states.
Pro tip: Mention that storing parent pointers adds O(V) space, but you can optimize by storing only the move that led to each state, and that the sequence length is optimal because BFS explores level by level.
Use BFS for unweighted graphs to guarantee the shortest path in terms of number of moves. If moves have weights, use Dijkstra or A*.
When exploring a new state, record the previous state and the move that led to it. This can be done with a parent map or by storing the move in the queue.
Once the goal state is reached, backtrack from the goal to the start using the parent pointers, collecting the moves in reverse order.
The length is at most the number of states in the shortest path, which is bounded by the total number of reachable states (V). In the worst case, it's O(V).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.