Strange Printer

📋

Strange Printer

Edge cases:

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IDE

def strangePrinter(s: str) -> int:public int strangePrinter(String s)int strangePrinter(string s)function strangePrinter(s)

def strangePrinter(s: str) -> int:
    # Write your solution here
    pass

class Solution {
    public int strangePrinter(String s) {
        // Write your solution here
        return 0;
    }
}

#include <string>
using namespace std;

int strangePrinter(string s) {
    // Write your solution here
    return 0;
}

function strangePrinter(s) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: length of stringGreedy approach printing each character separately without merging intervals.✅ Implement interval DP with recurrence: dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k][j]) when s[k] == s[i].

Wrong: non-zero for empty stringMissing base case for empty input returning 0.✅ Add base case: if i > j return 0 in DP recursion.

Wrong: More than 1 for all identical charactersFailing to merge intervals of identical characters.✅ If s[i] == s[j], dp[i][j] = dp[i][j-1] instead of dp[i][j-1] + 1.

Wrong: Off-by-one errors causing incorrect intervalsIncorrect DP indexing or interval boundaries.✅ Ensure dp[i][j] covers s[i..j] inclusive and recursion indices are correct.

Wrong: TLE on large inputsPure recursion without memoization or compression.✅ Use memoization and compress repeated characters to reduce complexity.

✓

Test Cases

Focus on handling base cases like empty and single character strings.

Watch out for greedy approaches and off-by-one errors in intervals.

Optimize your DP with memoization and consider string compression.

t1_01basic

Input{"s":"aaabbb"}

Expected2

⏱ Performance - must finish in 2000ms

Print 'aaa' in one turn and 'bbb' in another turn.

💡 Consider how to minimize print turns by grouping same characters.

💡 Use interval DP to find minimum turns for substrings.

💡 dp[i][j] = min turns to print s[i..j], try splitting at positions where s[k] == s[i].

Why it failed: Incorrect output means the DP recurrence is wrong; ensure to consider merging intervals when s[k] == s[i]. This is the canonical example test.

✓ Correctly handles the canonical example with two print turns.

t1_02basic

Input{"s":"aba"}

Expected2

⏱ Performance - must finish in 2000ms

Print 'a' at positions 0 and 2 in two turns, 'b' in between.

💡 Check how overlapping characters can reduce print turns.

💡 Use memoization to avoid recomputing intervals.

💡 Remember to merge intervals when characters match at different positions.

Why it failed: Fails to merge intervals with same characters leading to overcounting turns; fix by checking s[k] == s[i] and merging intervals accordingly.

✓ Correctly computes minimum turns for strings with repeated characters separated by others.

t2_01edge

Input{"s":""}

Expected0

⏱ Performance - must finish in 2000ms

Empty string requires zero print turns.

💡 Check base case when string length is zero.

💡 Return 0 immediately if input is empty.

💡 Add a condition: if i > j return 0 in DP recursion.

Why it failed: Returns non-zero for empty string; fix by adding base case for empty substring returning 0.

✓ Correctly handles empty string with zero turns.

t2_02edge

Input{"s":"a"}

Expected1

⏱ Performance - must finish in 2000ms

Single character string requires one print turn.

💡 Check base case for single character substring.

💡 DP base case: if i == j return 1.

💡 Ensure recursion returns 1 for single character intervals.

Why it failed: Returns 0 or more than 1 for single character; fix base case to return 1 when i == j.

✓ Correctly returns 1 for single character strings.

t2_03edge

Input{"s":"aaaaa"}

Expected1

⏱ Performance - must finish in 2000ms

All identical characters can be printed in one turn.

💡 Check if DP merges intervals of identical characters.

💡 If s[i] == s[j], dp[i][j] can be dp[i][j-1] instead of dp[i][j-1] + 1.

💡 Optimize by compressing repeated characters or merging intervals.

Why it failed: Fails to merge identical characters leading to overcounting; fix by checking s[i] == s[j] and reducing turns accordingly.

✓ Correctly merges identical characters into one print turn.

t2_04edge

Input{"s":"ab"}

Expected2

⏱ Performance - must finish in 2000ms

Two different characters require two print turns.

💡 Check minimal turns for very short strings with different chars.

💡 DP should return 1 if same char, else 2 for length 2.

💡 Ensure base cases handle length 2 substrings correctly.

Why it failed: Returns 1 for two different chars; fix by returning 2 when s[i] != s[j] for length 2.

✓ Correctly returns 2 for two different characters.

t3_01corner

Input{"s":"ababab"}

Expected4

⏱ Performance - must finish in 2000ms

Alternating characters require multiple turns; greedy approach fails here.

💡 Greedy printing one character at a time is not optimal.

💡 Use interval DP to consider all splits and merges.

💡 Check all k where s[k] == s[i] to minimize turns.

Why it failed: Greedy approach prints each character separately causing overcount; fix by using interval DP with merging on matching characters.

✓ Correctly handles alternating characters with minimal turns.

t3_02corner

Input{"s":"aaabaa"}

Expected2

⏱ Performance - must finish in 2000ms

Repeated characters separated by different characters reduce turns by merging intervals.

💡 Check merging intervals when same characters appear separated by others.

💡 DP must consider splitting and merging intervals carefully.

💡 Use memoization to avoid recomputation of overlapping intervals.

Why it failed: Fails to merge separated identical characters causing extra turns; fix by checking s[k] == s[i] and merging intervals.

✓ Correctly merges separated identical characters to minimize turns.

t3_03corner

Input{"s":"abcabc"}

Expected5

⏱ Performance - must finish in 2000ms

Tests off-by-one errors in interval boundaries and merging logic.

💡 Verify DP indices and boundaries carefully.

💡 Ensure recursion covers full intervals including both ends.

💡 Check merging logic for characters at interval ends.

Why it failed: Off-by-one errors cause incorrect interval coverage; fix by ensuring dp[i][j] covers s[i..j] inclusive.

✓ Correctly handles interval boundaries and merging.

t4_01performance

Input{"s":"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, O(n^3) DP must complete in 2s.

💡 Optimize with memoization and pruning.

💡 Consider string compression to reduce effective length.

💡 Bottom-up DP with careful indexing improves performance.

Why it failed: TLE due to O(n^3) complexity without memoization or compression; fix by implementing memoized DP and compressing repeated characters.

✓ Efficient DP implementation runs within time limits.

Practice

1. Consider the following Python function implementing the bottom-up DP solution for the Burst Balloons problem. What is the value of dp[1][4] after the completion of the outer loops when the input is [3,1,5,8]?

def maxCoins(nums):
    n = len(nums)
    nums = [1] + nums + [1]
    dp = [[0] * (n + 2) for _ in range(n + 2)]
    for length in range(2, n + 2):
        for i in range(0, n + 2 - length):
            j = i + length
            for k in range(i + 1, j):
                coins = dp[i][k] + nums[i] * nums[k] * nums[j] + dp[k][j]
                if coins > dp[i][j]:
                    dp[i][j] = coins
    return dp[0][n + 1]

easy

A. 15

B. 45

C. 40

D. 167

Solution

1. Step 1: Identify dp indices

   dp[1][4] corresponds to bursting balloons in nums indices 1 to 3 (original array indices 0 to 2), i.e. balloons [3,1,5].

2. Step 2: Calculate dp[1][4]

   By bursting last balloon k in (1,4), check k=2 and k=3: - k=2: dp[1][2] + nums[1]*nums[2]*nums[4] + dp[2][4] = 0 + 3*1*8 + 40 = 64 - k=3: dp[1][3] + nums[1]*nums[3]*nums[4] + dp[3][4] = 15 + 3*5*8 + 0 = 135 Max is 135, so dp[1][4] = 135 after the loops.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   dp[1][4] = 135 matches manual calculation [OK]

Hint: dp[i][j] stores max coins for bursting balloons between i and j [OK]

Common Mistakes:

• Confusing dp indices with nums indices
• Off-by-one errors in loops
• Miscomputing coins for k choices

2. You are given a triangular array of numbers where each element can only move to one of the two adjacent numbers in the row below. The goal is to find the minimum sum path from the top to the bottom. Which algorithmic approach guarantees an optimal solution efficiently?

easy

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

B. Divide and conquer by splitting the triangle into two halves and solving independently

C. Pure brute force recursion exploring all paths without memoization

D. Dynamic Programming using bottom-up tabulation to build minimum path sums

Solution

1. Step 1: Understand problem constraints and properties

   The problem requires finding a minimum path sum with overlapping subproblems and optimal substructure, which suits dynamic programming.

2. Step 2: Identify suitable algorithm

   Bottom-up DP tabulation efficiently computes minimum sums from the bottom row up, avoiding recomputation and ensuring optimality.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP tabulation solves overlapping subproblems efficiently [OK]

Hint: DP bottom-up tabulation ensures optimal substructure [OK]

Common Mistakes:

• Assuming greedy choice is always optimal
• Using brute force recursion without memoization
• Trying to split triangle independently ignoring dependencies

3. You are given a string representing a sequence of colored balls on a board and a multiset of balls in hand. Your goal is to clear the board by inserting balls from your hand such that any group of three or more consecutive balls of the same color is removed, possibly triggering chain reactions. Which algorithmic approach guarantees finding the minimum number of insertions needed to clear the board optimally?

easy

A. Interval dynamic programming with memoization exploiting optimal substructure

B. Topological sorting of ball groups followed by greedy removals

C. Pure brute force recursion without memoization exploring all insertions

D. Greedy insertion of balls at the earliest possible position to remove groups

Solution

1. Step 1: Understand problem structure

   The problem requires minimizing insertions to clear the board, which involves exploring overlapping subproblems and optimal substructure.

2. Step 2: Identify suitable algorithm

   Interval DP with memoization efficiently explores all intervals and hand states, guaranteeing minimal moves by considering all splits and removals.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Interval DP is known for solving interval removal problems optimally [OK]

Hint: Interval DP handles overlapping subproblems optimally [OK]

Common Mistakes:

• Assuming greedy insertion always yields minimal moves
• Using brute force without memoization leads to exponential time
• Misapplying topological sort which is unrelated here

4. Identify the bug in the following code snippet for minimum score triangulation of a polygon:

medium

A. Line missing dp[i][j] = float('inf') before minimization

B. Line initializing dp array with zeros

C. Loop boundaries for k from i+1 to j-1

D. Return statement returning dp[0][n-1]

Solution

1. Step 1: Check dp initialization inside loops

   dp[i][j] must be set to infinity before checking for minimal cost; otherwise, dp[i][j] starts at 0 and may never update correctly.

2. Step 2: Confirm other lines are correct

   dp array initialization, loop boundaries, and return statement are correct and standard.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Without dp[i][j] = float('inf'), minimal cost calculation is incorrect [OK]

Hint: Always initialize dp[i][j] before minimization [OK]

Common Mistakes:

• Forgetting dp initialization
• Off-by-one in loops
• Mixing indices i,j,k

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