Stone Game (Optimal Strategy)

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Steps

setup

Initialize dp array with zeros

Create a 4x4 dp table filled with zeros to store score differences for all intervals.

💡 Initializing the dp table is essential to store intermediate results for all subproblems.

Line:

dp = [[0]*n for _ in range(n)]

💡 We prepare a structure to hold results for every interval [i, j].

setup

Set base cases: single piles

Fill dp[i][i] with piles[i] because when interval length is 1, the player picks the only pile available.

💡 Base cases anchor the dp table with known values for intervals of length 1.

Line:

for i in range(n):
    dp[i][i] = piles[i]

💡 The score difference for a single pile interval is just the pile's value.

fill_row

Start filling intervals of length 2

Set length = 2 to compute dp for intervals of two piles.

💡 We begin solving larger intervals by increasing length from 1 to 2.

Line:

for length in range(2, n+1):

💡 We will fill dp for intervals [i, j] where j = i + length - 1.

fill_cells

Compute dp[0][1] for interval [0,1]

Calculate dp[0][1] as max(piles[0] - dp[1][1], piles[1] - dp[0][0]) = max(5-3, 3-5) = 2.

💡 We decide which end to pick to maximize the score difference for interval [0,1].

Line:

dp[0][1] = max(piles[0] - dp[1][1], piles[1] - dp[0][0])

💡 Picking the left pile yields a better score difference for the current player.

fill_cells

Compute dp[1][2] for interval [1,2]

Calculate dp[1][2] as max(piles[1] - dp[2][2], piles[2] - dp[1][1]) = max(3-4, 4-3) = 1.

💡 Choosing the best end pile for interval [1,2] to maximize score difference.

Line:

dp[1][2] = max(piles[1] - dp[2][2], piles[2] - dp[1][1])

💡 Picking the right pile is better for the current player in this interval.

fill_cells

Compute dp[2][3] for interval [2,3]

Calculate dp[2][3] as max(piles[2] - dp[3][3], piles[3] - dp[2][2]) = max(4-5, 5-4) = 1.

💡 Evaluate best pick for interval [2,3] to maximize score difference.

Line:

dp[2][3] = max(piles[2] - dp[3][3], piles[3] - dp[2][2])

💡 Picking the right pile is optimal here for the current player.

fill_row

Start filling intervals of length 3

Set length = 3 to compute dp for intervals of three piles.

💡 We now solve bigger intervals using previously computed smaller intervals.

Line:

for length in range(2, n+1):
    # length = 3 iteration

💡 We will fill dp[i][j] for intervals where j = i + 2.

fill_cells

Compute dp[0][2] for interval [0,2]

Calculate dp[0][2] as max(piles[0] - dp[1][2], piles[2] - dp[0][1]) = max(5-1, 4-2) = 4.

💡 Choose the best end pile for interval [0,2] to maximize score difference.

Line:

dp[0][2] = max(piles[0] - dp[1][2], piles[2] - dp[0][1])

💡 Picking the left pile yields a better outcome for the current player.

fill_cells

Compute dp[1][3] for interval [1,3]

Calculate dp[1][3] as max(piles[1] - dp[2][3], piles[3] - dp[1][2]) = max(3-1, 5-1) = 4.

💡 Decide the best pick for interval [1,3] to maximize score difference.

Line:

dp[1][3] = max(piles[1] - dp[2][3], piles[3] - dp[1][2])

💡 Picking the right pile is optimal for the current player here.

fill_row

Start filling intervals of length 4 (full range)

Set length = 4 to compute dp for the entire interval [0,3].

💡 We now solve the full problem using all smaller intervals computed.

Line:

for length in range(2, n+1):
    # length = 4 iteration

💡 We will compute dp[0][3] for the entire piles array.

fill_cells

Compute dp[0][3] for interval [0,3]

Calculate dp[0][3] as max(piles[0] - dp[1][3], piles[3] - dp[0][2]) = max(5-4, 5-4) = 1.

💡 Choose the best end pile for the full interval to maximize score difference.

Line:

dp[0][3] = max(piles[0] - dp[1][3], piles[3] - dp[0][2])

💡 The first player can secure a positive score difference, meaning a win.

reconstruct

Read final answer from dp[0][3]

Check if dp[0][3] > 0 to determine if the first player wins.

💡 A positive dp[0][3] means the first player can guarantee a win.

Line:

return dp[0][n-1] > 0

💡 The first player (Alex) can win with optimal play on this input.

def stoneGame(piles):
    n = len(piles)  # STEP 1
    dp = [[0]*n for _ in range(n)]  # STEP 1
    for i in range(n):  # STEP 2
        dp[i][i] = piles[i]  # STEP 2
    for length in range(2, n+1):  # STEP 3
        for i in range(n - length + 1):  # STEP 4-11
            j = i + length - 1
            dp[i][j] = max(piles[i] - dp[i+1][j], piles[j] - dp[i][j-1])  # STEP 4-11
    return dp[0][n-1] > 0  # STEP 12

if __name__ == '__main__':
    print(stoneGame([5,3,4,5]))  # True

📊

Stone Game (Optimal Strategy) - Watch the Algorithm Execute, Step by Step

Watching the dp table fill step-by-step reveals how the algorithm builds solutions for bigger intervals from smaller ones, making the optimal strategy clear.

Initializing dp array

i\w	0	1	2	3
i=0	0	?	?	?
i=1	?	0	?	?
i=2	?	?	0	?
i=3	?	?	?	0

unfilled

Initializing dp array

i\w	0	1	2	3
i=0	5	?	?	?
i=1	?	3	?	?
i=2	?	?	4	?
i=3	?	?	?	5

dp[0][0] = 5

Initializing dp array

i\w	0	1	2	3
i=0	5	?	?	?
i=1	?	3	?	?
i=2	?	?	4	?
i=3	?	?	?	5

waiting

Item 0 - wt:5 val:2

i\w	0	1	2	3
i=0	5	2	?	?
i=1	?	3	?	?
i=2	?	?	4	?
i=3	?	?	?	5

dp[0][1] = max(5-3,3-5) = 2

Item 1 - wt:3 val:1

i\w	0	1	2	3
i=0	5	2	?	?
i=1	?	3	1	?
i=2	?	?	4	?
i=3	?	?	?	5

dp[1][2] = max(3-4,4-3) = 1

Item 2 - wt:4 val:1

i\w	0	1	2	3
i=0	5	2	?	?
i=1	?	3	1	?
i=2	?	?	4	1
i=3	?	?	?	5

dp[2][3] = max(4-5,5-4) = 1

Initializing dp array

i\w	0	1	2	3
i=0	5	2	?	?
i=1	?	3	1	?
i=2	?	?	4	1
i=3	?	?	?	5

waiting

Item 0 - wt:5 val:4

i\w	0	1	2	3
i=0	5	2	4	?
i=1	?	3	1	?
i=2	?	?	4	1
i=3	?	?	?	5

dp[0][2] = max(5-1,4-2) = 4

Item 1 - wt:3 val:4

i\w	0	1	2	3
i=0	5	2	4	?
i=1	?	3	1	4
i=2	?	?	4	1
i=3	?	?	?	5

dp[1][3] = max(3-1,5-1) = 4

Initializing dp array

i\w	0	1	2	3
i=0	5	2	4	?
i=1	?	3	1	4
i=2	?	?	4	1
i=3	?	?	?	5

waiting

Item 0 - wt:5 val:1

i\w	0	1	2	3
i=0	5	2	4	1
i=1	?	3	1	4
i=2	?	?	4	1
i=3	?	?	?	5

dp[0][3] = max(5-4,5-4) = 1

Item 0 - wt:5 val:1

i\w	0	1	2	3
i=0	5	2	4	1
i=1	?	3	1	4
i=2	?	?	4	1
i=3	?	?	?	5

Answer cell dp[0][3] = 1 > 0

Key Takeaways

✓ The dp table stores the maximum score difference the current player can achieve over the opponent for every interval.

This insight is hard to see from code alone because the dp values represent differences, not absolute scores.

✓ The table is filled from smaller intervals to larger intervals, ensuring dependencies are resolved before use.

Seeing the fill order visually clarifies why the algorithm uses bottom-up tabulation.

✓ At each interval, the choice to pick the left or right pile depends on maximizing the difference after the opponent's optimal response.

Watching the max decision at each cell shows how the algorithm simulates optimal play.

Practice

1. You are given an n x n integer matrix. You want to find a path from the top row to the bottom row such that you move one step down each time, and at each step you can move to the same column, the column to the left, or the column to the right. The goal is to minimize the sum of the values along this path. Which algorithmic approach guarantees finding the minimum sum efficiently?

easy

A. Greedy algorithm that picks the minimum adjacent value at each step

B. Sorting each row and picking the smallest values independently

C. Depth-first search exploring all paths without memoization

D. Dynamic programming that builds solutions row by row using previous row results

Solution

1. Step 1: Understand the problem constraints

   The problem requires considering all possible paths moving down and diagonally, which suggests overlapping subproblems and optimal substructure.

2. Step 2: Identify the suitable algorithm

   Dynamic programming fits because it efficiently computes minimum sums for each cell based on the previous row's results, avoiding redundant calculations.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Greedy and sorting fail to consider future steps; DFS without memoization is exponential [OK]

Hint: DP uses previous row results to build solutions [OK]

Common Mistakes:

• Thinking greedy or sorting per row suffices
• Ignoring overlapping subproblems

2. You are given a printer that can print a continuous substring of identical characters in one turn. Given a string, you want to find the minimum number of turns needed to print the entire string. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic Programming over intervals with string compression and merging states

B. Greedy approach printing each character separately from left to right

C. Simple recursion without memoization or compression

D. Sliding window technique to find longest repeated substrings

Solution

1. Step 1: Understand problem constraints

   The problem requires minimizing turns to print a string where each turn prints a continuous block of identical characters.

2. Step 2: Identify suitable algorithmic pattern

   Greedy or sliding window approaches fail to merge overlapping prints optimally. Pure recursion is inefficient. Interval DP with string compression and merging states captures overlapping subproblems and optimizes turns.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Interval DP with compression is known optimal [OK]

Hint: Interval DP with compression merges overlapping prints [OK]

Common Mistakes:

• Assuming greedy printing each char separately is optimal

3. Consider the following buggy code for finding the cheapest flight within K stops. Which line contains the subtle bug that causes incorrect results on some inputs?

def findCheapestPrice(n, flights, src, dst, K):
    INF = float('inf')
    prev = [INF] * n
    prev[src] = 0
    for _ in range(K + 1):
        for u, v, w in flights:
            if prev[u] != INF:
                prev[v] = min(prev[v], prev[u] + w)
    return prev[dst] if prev[dst] != INF else -1

medium

A. Line 4: Initializing prev[src] = 0

B. Line 7: Checking if prev[u] != INF before relaxing edges

C. Line 6: Using prev array directly inside the loop instead of a separate curr array

D. Line 9: Returning -1 if prev[dst] is INF

Solution

1. Step 1: Understand the DP update mechanism

   In bottom-up DP, updates for the current iteration must not affect other relaxations in the same iteration, so a separate curr array is needed.

2. Step 2: Identify the bug

   Using prev directly causes updated costs to be used immediately in the same iteration, leading to incorrect minimum costs.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Separate curr array prevents premature updates [OK]

Hint: Must use separate curr array to avoid premature updates [OK]

Common Mistakes:

• Updating prev in-place during iteration
• Forgetting to copy arrays
• Ignoring stop constraints

4. What is the time and space complexity of the optimal space-optimized bottom-up DP solution for the Unique Paths II problem on an m x n grid with obstacles?

medium

A. Time: O(m*n), Space: O(m*n)

B. Time: O(m+n), Space: O(m+n)

C. Time: O(m*n), Space: O(n)

D. Time: O(2^(m+n)), Space: O(m*n)

Solution

1. Step 1: Analyze time complexity

   The algorithm iterates over each cell once in nested loops: O(m*n).

2. Step 2: Analyze space complexity

   Uses a single 1D dp array of length n, so space is O(n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Nested loops over m and n, dp array size n [OK]

Hint: Nested loops over m and n, dp array size n [OK]

Common Mistakes:

• Confusing recursion stack space with dp array
• Assuming full 2D dp array used
• Mistaking linear time for exponential

5. Suppose the problem changes so that you can move down to the same column or adjacent columns multiple times, but you want to find the minimum falling path sum where you can reuse rows multiple times (i.e., you can revisit rows). Which modification to the bottom-up DP approach correctly handles this variant?

hard

A. Use the original DP but memoize results to avoid recomputation

B. Use the same bottom-up DP but allow cycles by iterating until no dp changes occur

C. Switch to a shortest path algorithm like Bellman-Ford on a graph representing allowed moves

D. Apply greedy approach picking minimum adjacent values repeatedly

Solution

1. Step 1: Understand the variant with row reuse

   Allowing revisiting rows means cycles in the path graph, which breaks the acyclic assumption of DP.

2. Step 2: Identify suitable algorithm

   Model the problem as a graph with edges representing allowed moves and use Bellman-Ford to find shortest paths with possible cycles.

3. Step 3: Why other options fail

   Bottom-up DP assumes acyclic progression; greedy ignores global optimality; memoization doesn't handle cycles properly.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Bellman-Ford handles negative cycles and repeated nodes [OK]

Hint: Cycles require graph shortest path algorithms, not DP [OK]

Common Mistakes:

• Trying to reuse DP without cycle handling
• Assuming memoization solves cycles