Strange Printer

📋

Strange Printer

Edge cases:

</>

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. You have a grid where two players start at the top-left corner and move to the bottom-right corner simultaneously, collecting cherries on their paths. Both players can only move right or down, and some cells contain thorns that block passage. The goal is to maximize the total cherries collected by both players without double-counting cherries if they land on the same cell. Which algorithmic approach guarantees finding the optimal solution for this problem?

easy

A. Greedy algorithm that always picks the next cell with the most cherries for each player independently.

B. Divide and conquer by splitting the grid into two halves and solving each half independently.

C. Dynamic programming that tracks both players' positions simultaneously and uses a 3D state to avoid double counting and invalid paths.

D. Simple BFS traversal from start to end for each player separately, then summing their maximum cherries.

Solution

1. Step 1: Understand the problem constraints

   Two players move simultaneously on the grid, collecting cherries without double counting and avoiding thorns. Both must take the same number of steps at any point.

2. Step 2: Identify the suitable algorithmic pattern

   Dynamic programming with a 3D state (step count and both players' row positions) tracks both players' positions simultaneously, ensuring valid states and avoiding double counting.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP with simultaneous states handles constraints correctly [OK]

Hint: Two simultaneous paths require joint DP state [OK]

Common Mistakes:

• Assuming independent paths and summing results
• Using greedy to pick local max cherries
• Splitting grid and solving halves independently

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. Consider the following buggy code for Paint House (K Colors). Which line contains the subtle bug that can cause adjacent houses to have the same color, violating constraints?

medium

A. Line with 'min_prev = min(dp[x] for x in range(k))'

B. Line with 'dp = costs[0][:]' initialization

C. Line with 'new_dp = [0]*k' inside the loop

D. Line with 'return min(dp)' at the end

Solution

1. Step 1: Identify constraint violation

   The problem requires that adjacent houses cannot have the same color. The code must exclude the current color when computing min_prev.

2. Step 2: Locate bug in min_prev calculation

   The line 'min_prev = min(dp[x] for x in range(k))' includes the current color's dp value, allowing same color adjacency, violating constraints.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Excluding current color in min calculation is essential [OK]

Hint: Check if min excludes current color to avoid same-color adjacency [OK]

Common Mistakes:

• Including current color in min calculation
• Incorrect dp initialization
• Returning min(dp) without proper updates

5. Suppose the Unique Paths problem is modified so that you can move right, down, or diagonally (down-right) at each step. Which of the following changes correctly adapts the space-optimized DP solution to count all unique paths?

hard

A. Use a 2D DP array where dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1]

B. Update dp[j] as dp[j] + dp[j - 1] + dp[j - 2] in the inner loop

C. Keep the 1D dp array and update dp[j] = dp[j] + dp[j - 1] only, ignoring diagonal moves

D. Use recursion without memoization to handle the extra diagonal move

Solution

1. Step 1: Understand diagonal move impact

   Diagonal moves depend on the top-left neighbor dp[i-1][j-1], which cannot be tracked with a single 1D array.

2. Step 2: Choose correct DP structure

   A 2D DP array is needed to access dp[i-1][j-1] directly; update rule: dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1].

3. Step 3: Reject incorrect options

   Update dp[j] as dp[j] + dp[j - 1] + dp[j - 2] in the inner loop incorrectly accesses dp[j-2] in 1D array which doesn't represent diagonal; Keep the 1D dp array and update dp[j] = dp[j] + dp[j - 1] only, ignoring diagonal moves ignores diagonal; Use recursion without memoization to handle the extra diagonal move is inefficient.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Diagonal requires 2D DP for correct neighbor access [OK]

Hint: Diagonal moves require 2D DP for top-left neighbor [OK]

Common Mistakes:

• Trying to handle diagonal with 1D DP
• Ignoring diagonal moves in update