Strange Printer

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Steps

setup

Compress the input string

We iterate over the input string 'aaabbb' and build a compressed string by removing consecutive duplicates, resulting in 'ab'.

💡 Compression reduces problem size by merging consecutive identical characters, simplifying DP computation.

Line:

compressed = []
for c in s:
    if not compressed or compressed[-1] != c:
        compressed.append(c)

💡 The compressed string 'ab' captures the essential character changes, reducing complexity.

setup

Initialize DP table with zeros

Create a 2x2 DP table for the compressed string 'ab', initializing all cells to 0.

💡 DP table size matches compressed string length; initialization prepares for base cases.

Line:

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

💡 DP table is ready to store minimum turns for all substrings.

setup

Set base cases dp[i][i] = 1

For each single character substring, set dp[i][i] = 1 since one turn is needed to print one character.

💡 Base cases anchor the DP; single characters require exactly one print turn.

Line:

for i in range(n):
    dp[i][i] = 1

💡 Diagonal cells represent single characters and are set to 1.

fill_row

Start filling substrings of length 2

Begin processing substrings of length 2: substring s[0..1] = 'ab'.

💡 We fill DP for increasing substring lengths to ensure dependencies are computed.

Line:

for length in range(2, n + 1):
    for i in range(n - length + 1):
        j = i + length - 1

💡 We will compute dp[0][1] next, using smaller substrings.

fill_cells

Initialize dp[0][1] with dp[0][0] + 1

Set dp[0][1] = dp[0][0] + 1 = 1 + 1 = 2 as a starting point before checking for better splits.

💡 This assumes printing s[0..0] then printing s[1] separately.

Line:

dp[i][j] = dp[i][j - 1] + 1

💡 Initial value represents printing substring by printing s[i..j-1] then s[j] separately.

compare

Check if s[k] == s[j] for k=0

Compare s[0] = 'a' with s[1] = 'b'. They differ, so no update to dp[0][1].

💡 Matching characters allow merging print turns, reducing total turns.

Line:

for k in range(i, j):
    if s[k] == s[j]:
        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j - 1])

💡 No character match means no improvement over initial dp[0][1].

fill_row

Start filling substrings of length 3 (not applicable here)

Since compressed string length n=2, no substrings of length 3 exist, so we proceed to final answer.

💡 DP table is fully computed for all substring lengths.

Line:

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

💡 All necessary dp values are computed.

reconstruct

Return the final answer dp[0][n-1]

Return dp[0][1] = 2 as the minimum number of turns to print the compressed string 'ab'.

💡 This value represents the minimum turns for the entire original string after compression.

Line:

return dp[0][n - 1]

💡 The answer 2 matches the expected output for 'aaabbb'.

reconstruct

Summary: DP table interpretation

The DP table shows minimal turns for substrings: single chars need 1 turn, 'ab' needs 2 turns.

💡 This confirms the algorithm's correctness and the role of compression.

💡 DP table cells correspond to minimal printing turns for substrings.

reconstruct

End of trace

The algorithm has completed and returned the minimal number of turns needed.

💡 This final step confirms the algorithm's output matches the problem's expected result.

💡 DP approach efficiently solves the Strange Printer problem.

def strangePrinter(s: str) -> int:
    # STEP 1: Compress string
    compressed = []
    for c in s:
        if not compressed or compressed[-1] != c:
            compressed.append(c)
    s = ''.join(compressed)
    n = len(s)

    # STEP 2: Initialize DP table
    dp = [[0] * n for _ in range(n)]

    # STEP 3: Base cases dp[i][i] = 1
    for i in range(n):
        dp[i][i] = 1

    # STEP 4: Fill DP for substrings length 2 to n
    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            # STEP 5: Initialize dp[i][j]
            dp[i][j] = dp[i][j - 1] + 1
            # STEP 6: Check for better splits
            for k in range(i, j):
                if s[k] == s[j]:
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j - 1])

    # STEP 8: Return answer
    return dp[0][n - 1]

if __name__ == '__main__':
    print(strangePrinter("aaabbb"))  # Output: 2
    print(strangePrinter("aba"))     # Output: 2

📊

Strange Printer - Watch the Algorithm Execute, Step by Step

Watching each step helps you understand how interval DP works and how overlapping subproblems are combined to build the solution.

Item 0 - wt:0 val:0

i\w

Item 0 - wt:0 val:0

i\w	0	1
i=0	0	0
i=1	0	0

Item 0 - wt:0 val:1

i\w	0	1
i=0	1	0
i=1	0	1

dp[0][0] = 1

Item 0 - wt:0 val:0

i\w	0	1
i=0	1	?
i=1	0	1

Compute dp[0][1]

Item 0 - wt:0 val:2

i\w	0	1
i=0	1	2
i=1	0	1

dp[0][1] = dp[0][0] + 1 = 2

Item 0 - wt:0 val:2

i\w	0	1
i=0	1	2
i=1	0	1

s[0] != s[1], no update

Item 0 - wt:0 val:2

i\w	0	1
i=0	1	2
i=1	0	1

Final dp[0][1] = 2

Item 0 - wt:0 val:2

i\w	0	1
i=0	1	2
i=1	0	1

Answer dp[0][1] = 2

Item 0 - wt:0 val:2

i\w	0	1
i=0	1	2
i=1	0	1

Final DP table

Item 0 - wt:0 val:2

i\w	0	1
i=0	1	2
i=1	0	1

Answer cell

Key Takeaways

✓ Compression reduces the problem size by merging consecutive identical characters.

This step is crucial but often overlooked; it simplifies the DP and improves efficiency.

✓ DP table is filled by increasing substring length, ensuring all dependencies are computed before use.

Understanding the order of filling helps grasp how interval DP builds solutions from smaller subproblems.

✓ Matching characters at positions k and j allow merging print turns, reducing the total count.

This insight explains why the DP checks s[k] == s[j] and updates dp[i][j] accordingly.

Practice

1. You are given a grid of non-negative integers representing costs. Starting from the top-left corner, you want to reach the bottom-right corner by moving only down or right, minimizing the total cost along the path. Which algorithmic approach guarantees finding the minimum total cost efficiently?

easy

A. A greedy algorithm that always moves to the adjacent cell with the smallest cost

B. Dynamic programming that builds up solutions from smaller subproblems using a grid-based state

C. Pure brute force recursion exploring all possible paths without memoization

D. Divide and conquer by splitting the grid into halves and solving independently

Solution

1. Step 1: Understand problem constraints

   The problem requires minimizing path cost with only down or right moves, which naturally forms overlapping subproblems.

2. Step 2: Identify suitable algorithmic pattern

   Dynamic programming efficiently solves overlapping subproblems by storing intermediate results, unlike greedy which can fail on some grids, brute force which is exponential, or divide and conquer which doesn't handle dependencies well.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP uses subproblem solutions to build the answer bottom-up [OK]

Hint: DP handles overlapping subproblems and optimal substructure [OK]

Common Mistakes:

• Assuming greedy always works for grid path problems
• Thinking brute force is efficient enough
• Believing divide and conquer applies without overlapping subproblems

2. Consider the following Python code for computing unique paths in a grid:

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

print(uniquePaths(3, 3))

What is the value of dp after the outer loop completes its second iteration (i=2)?

easy

A. [1, 3, 6]

B. [1, 2, 3]

C. [1, 3, 5]

D. [1, 4, 10]

Solution

1. Step 1: Initialize dp array

   Initially, dp = [1, 1, 1] for n=3.

2. Step 2: Trace iterations

   First outer iteration (i=1): - j=1: dp[1] = dp[1] + dp[0] = 1 + 1 = 2 - j=2: dp[2] = dp[2] + dp[1] = 1 + 2 = 3 After i=1, dp = [1, 2, 3] Second outer iteration (i=2): - j=1: dp[1] = dp[1] + dp[0] = 2 + 1 = 3 - j=2: dp[2] = dp[2] + dp[1] = 3 + 3 = 6 After i=2, dp = [1, 3, 6]

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   dp array accumulates path counts correctly -> [1,3,6] [OK]

Hint: Track dp updates carefully per iteration [OK]

Common Mistakes:

• Off-by-one errors in loop indices
• Mis-updating dp[j] with wrong dp[j-1] value

3. The following code attempts to implement the space-optimized DP solution for Maximal Square. Identify the line containing the subtle bug that causes incorrect results on some inputs.

def maximalSquare(matrix):
    if not matrix or not matrix[0]:
        return 0
    rows, cols = len(matrix), len(matrix[0])
    dp = [0] * (cols + 1)
    max_side = 0
    prev = 0
    for i in range(rows):
        for j in range(1, cols + 1):
            temp = dp[j]
            if matrix[i][j-1] == '1':
                dp[j] = 1 + min(dp[j], dp[j-1], prev)  # Bug here
                max_side = max(max_side, dp[j])
            else:
                dp[j] = 0
            prev = temp
    return max_side * max_side

medium

A. Line with 'dp[j] = 1 + min(dp[j], dp[j-1], dp[j-1])'

B. Line with 'temp = dp[j]'

C. Line with 'prev = temp'

D. Line with 'max_side = max(max_side, dp[j])'

Solution

1. Step 1: Identify the DP recurrence

   The recurrence requires min(dp[j], dp[j-1], prev), but code uses dp[j-1] twice, missing 'prev'.

2. Step 2: Understand impact

   Using dp[j-1] twice ignores the diagonal value 'prev', causing incorrect dp values and max square size.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Correct recurrence uses three distinct values [OK]

Hint: Check if all three DP states are used in min() [OK]

Common Mistakes:

• Forgetting to use 'prev' in min()
• Overwriting dp before using dependencies
• Incorrect dp initialization

4. The following code attempts to solve the Triangle minimum path sum problem using bottom-up DP. Identify the line that contains a subtle bug causing incorrect results on some inputs.

medium

A. for i in range(len(triangle) - 2, -1, -1):

B. dp[j] = triangle[i][j] + min(dp[j + 1], dp[j])

C. dp = triangle[-1][:]

D. return dp[0]

Solution

1. Step 1: Examine dp update line

   The original code has dp[j] = triangle[i][j] + min(dp[j], dp[j + 1]). The buggy code swaps the min arguments to min(dp[j + 1], dp[j]) which is harmless since min is commutative.

2. Step 2: Check for index out of bounds risk

   Accessing dp[j + 1] is safe because dp length is one more than current row length, so no out-of-bounds here.

3. Step 3: Identify subtle bug

   The subtle bug is that dp is updated in place from left to right, which can cause dp[j + 1] to be already updated when used in min, leading to incorrect results.

4. Step 4: Explanation of bug

   Because dp is updated from left to right, dp[j + 1] may have been updated in the current iteration, so min(dp[j], dp[j + 1]) uses a mixed state of dp values.

5. Step 5: Correct fix

   To fix, update dp from right to left so dp[j + 1] is not yet updated when used.

6. Final Answer:

   Option B -> Option B

7. Quick Check:

   Updating dp in wrong order causes subtle bugs [OK]

Hint: Bug usually in dp update line with min or indexing [OK]

Common Mistakes:

• Swapping min arguments (harmless but suspicious)
• Index out of bounds on dp[j+1]
• Updating dp in wrong order causing overwritten values

5. Suppose the problem is modified so that you can take unlimited flights (reuse edges) but still want the cheapest price within K stops. Which modification to the bottom-up DP approach correctly handles this variant?

hard

A. Use Bellman-Ford algorithm with K+1 iterations allowing edge reuse in each iteration

B. Switch to Dijkstra's algorithm since unlimited reuse removes stop constraints

C. Modify DP to allow updating curr[v] multiple times per iteration by relaxing edges repeatedly until no changes

D. Use the same bottom-up DP but increase iterations to K+1 without changes

Solution

1. Step 1: Understand the unlimited reuse variant

   Unlimited reuse means cycles are allowed, so shortest path with stop constraints resembles Bellman-Ford relaxation over K+1 iterations.

2. Step 2: Identify correct algorithm

   Bellman-Ford naturally handles edge reuse and negative cycles (if any), iterating K+1 times to relax edges, matching problem constraints.

3. Step 3: Why other options fail

   Use the same bottom-up DP but increase iterations to K+1 without changes ignores edge reuse effect; Switch to Dijkstra's algorithm since unlimited reuse removes stop constraints ignores stop constraints; Modify DP to allow updating curr[v] multiple times per iteration by relaxing edges repeatedly until no changes suggests repeated relaxation within iteration, which is inefficient and incorrect.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Bellman-Ford with K+1 iterations handles unlimited reuse correctly [OK]

Hint: Bellman-Ford handles edge reuse with K+1 relaxations [OK]

Common Mistakes:

• Using Dijkstra ignoring stops
• Not increasing iterations
• Trying repeated relaxations inside iteration