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 need to find the cheapest cost to travel from a source city to a destination city with at most K stops, given a list of flights with costs. Which algorithmic approach guarantees finding the optimal solution efficiently under these constraints?

easy

A. Greedy algorithm using Dijkstra's shortest path without modification

B. Topological sort followed by single pass relaxation of edges

C. Simple depth-first search exploring all paths without pruning

D. Dynamic programming using a bottom-up approach iterating over stops

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding the cheapest flight with at most K stops, which limits path length and requires considering multiple paths.

2. Step 2: Identify suitable algorithm

   Greedy Dijkstra fails because it doesn't limit stops; DFS is exponential; topological sort requires DAG which flights graph may not be. Bottom-up DP iterates over stops and relaxes edges, guaranteeing optimal cost within K stops.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Bottom-up DP with stops limit ensures optimal solution [OK]

Hint: DP with stops limit ensures optimal cost [OK]

Common Mistakes:

• Using Dijkstra without stop limit
• Trying DFS without pruning
• Assuming DAG for topological sort

2. You are given a grid where some cells are blocked and others are free. You need to find the number of unique paths from the top-left corner to the bottom-right corner, moving only down or right, but you cannot pass through blocked cells. Which algorithmic approach guarantees an efficient and correct solution for this problem?

easy

A. Dynamic Programming that builds solutions using previously computed subproblems while skipping blocked cells

B. Pure brute force recursion exploring all paths without memoization

C. Greedy algorithm that always moves right if possible, else down

D. Dijkstra's shortest path algorithm treating grid cells as graph nodes

Solution

1. Step 1: Understand problem constraints

   The problem requires counting all unique paths avoiding obstacles, which involves overlapping subproblems and optimal substructure.

2. Step 2: Identify suitable algorithm

   Dynamic Programming efficiently computes the number of paths by reusing results and handling obstacles by zeroing paths through blocked cells.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP handles obstacles and overlapping subproblems correctly [OK]

Hint: DP handles obstacles and overlapping subproblems correctly [OK]

Common Mistakes:

• Thinking greedy can find all paths
• Using brute force without pruning
• Confusing shortest path with counting paths

3. What is the time complexity of the bottom-up dynamic programming solution for the Dungeon Game problem on an m x n grid, and why might some candidates mistakenly think it is higher?

medium

A. O(2^{m+n}) because all paths are explored recursively

B. O(m + n) since only one path is considered at a time

C. O(m * n * max(|dungeon[i][j]|)) due to health value range affecting computations

D. O(m * n) because each cell is computed once using constant time operations

Solution

1. Step 1: Identify loops

   The bottom-up DP uses nested loops over m rows and n columns, each cell computed once.

2. Step 2: Analyze per-cell work

   Each dp[i][j] calculation is O(1), involving min and max operations.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP table size m*n and constant work per cell [OK]

Hint: DP fills m*n table once, no recursion overhead [OK]

Common Mistakes:

• Confusing brute force recursion with DP
• Thinking health values affect complexity

4. The following code attempts to solve the Strange Printer problem using bottom-up DP. Which line contains a subtle bug that can cause incorrect results on inputs like "aba"?

medium

A. Line 18: dp[i][j] = dp[i][j - 1] + 1 assignment

B. Line 13: dp[i][i] = 1 initialization

C. Line 4: compressed.append(c) without checking for duplicates

D. Line 21: if s[k] == s[j] condition

Solution

1. Step 1: Identify compression step

   Line 4 appends every character without checking if it equals the last compressed character, so no compression occurs.

2. Step 2: Understand impact of missing compression

   Without compression, the DP runs on the full string length, causing inefficiency and possibly incorrect merges, especially on inputs like "aba" where compression reduces complexity.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Compression step must skip duplicates to optimize DP [OK]

Hint: Compression must skip consecutive duplicates to avoid TLE and errors [OK]

Common Mistakes:

• Forgetting to compress string before DP

5. The following code attempts to compute the number of unique paths in an m x n grid using a space-optimized DP approach:

def uniquePaths(m, n):
    dp = [1] * n
    for i in range(1, m):
        for j in range(1, n):
            dp[j] = dp[j] + dp[j]
    return dp[-1]

What is the bug in this code?

medium

A. Line 3 should start loop from 0 instead of 1

B. Line 4 incorrectly doubles dp[j] instead of adding dp[j-1]

C. Line 5 should return dp[0] instead of dp[-1]

D. Line 2 initializes dp with wrong size

Solution

1. Step 1: Analyze inner loop update

   Line 4 updates dp[j] by adding dp[j] to itself, doubling the value instead of adding dp[j-1].

2. Step 2: Identify correct update

   The correct update is dp[j] += dp[j - 1] to accumulate paths from left and top cells.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Doubling dp[j] breaks path count logic -> bug at line 4 [OK]

Hint: Check dp update uses dp[j-1], not dp[j] twice [OK]

Common Mistakes:

• Using dp[j] + dp[j] instead of dp[j] + dp[j-1]
• Off-by-one errors in loops