Rearrange String k Distance Apart

📋

Rearrange String k Distance Apart

Edge cases:

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IDE

def rearrangeString(s: str, k: int) -> str:public String rearrangeString(String s, int k)string rearrangeString(string s, int k)function rearrangeString(s, k)

def rearrangeString(s, k):
    # Write your solution here
    pass

class Solution {
    public String rearrangeString(String s, int k) {
        // Write your solution here
        return "";
    }
}

#include <string>
using namespace std;

string rearrangeString(string s, int k) {
    // Write your solution here
    return "";
}

function rearrangeString(s, k) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: Non-empty string when rearrangement is impossibleNot detecting when max frequency characters cannot be spaced k apart✅ Add a check: if max frequency > (n + k - 1) / k, return empty string

Wrong: Original string returned even when k > 0 and rearrangement neededIgnoring k distance constraint or treating k=0 case incorrectly✅ Return original string only if k == 0; otherwise enforce rearrangement

Wrong: Characters repeated too close togetherGreedy approach without cooldown queue allowing reuse before k distance✅ Use a cooldown queue to delay reuse of characters until k positions later

Wrong: Empty string returned for single character inputNot handling single character or empty input base cases✅ Return input string immediately if length is 0 or 1

Wrong: Unnecessary rearrangement when all characters are uniqueNot checking for uniqueness and large k values✅ Return original string if all characters are unique

✓

Test Cases

Focus on handling base cases and boundary conditions carefully.

Think about common algorithmic pitfalls like greedy traps and usage constraints.

Optimize your solution to handle large inputs efficiently using heaps and queues.

t1_01basic

Input{"s":"aabbcc","k":3}

Expected"abcabc"

⏱ Performance - must finish in 2000ms

Characters 'a', 'b', and 'c' are spaced at least 3 apart in the rearranged string.

💡 Think about using a max heap to always pick the most frequent character available.

💡 Use a cooldown queue to ensure characters are spaced at least k apart before reuse.

💡 Implement a max heap with a queue that holds characters until their cooldown expires.

Why it failed: Output does not maintain k distance between identical characters. Fix by enforcing cooldown queue logic with max heap to space characters properly.

✓ Correct rearrangement with all characters spaced at least k apart.

t1_02basic

Input{"s":"aaabc","k":3}

Expected"abaca"

⏱ Performance - must finish in 2000ms

Rearranged string 'abaca' spaces 'a' characters at least 3 apart.

💡 Check character frequencies and prioritize placing the most frequent characters first.

💡 Use a max heap and cooldown queue to manage character placement and spacing.

💡 Ensure that after placing a character, it is not reused until k positions later.

Why it failed: Fails to space 'a' characters at least k apart, likely reusing characters too soon. Fix by implementing cooldown queue correctly.

✓ Correctly spaced characters with valid rearrangement.

t2_01edge

Input{"s":"","k":2}

Expected""

⏱ Performance - must finish in 2000ms

Empty string input returns empty string output.

💡 Consider what happens when the input string is empty.

💡 Your code should handle empty input gracefully without errors.

💡 Return the input string immediately if it is empty.

Why it failed: Code crashes or returns non-empty output on empty input. Fix by adding a base case to return empty string if input is empty.

✓ Handles empty input correctly by returning empty string.

t2_02edge

Input{"s":"a","k":1}

Expected"a"

⏱ Performance - must finish in 2000ms

Single character string with k=1 returns the original string.

💡 With only one character, spacing constraints are trivially satisfied.

💡 Check if your code handles single character strings without unnecessary processing.

💡 Return the input string directly if length is 1 or k is 0.

Why it failed: Fails or returns empty string for single character input. Fix by returning input string immediately when length is 1.

✓ Correctly returns single character string as is.

t2_03edge

Input{"s":"aa","k":2}

Expected""

⏱ Performance - must finish in 2000ms

Impossible to rearrange 'aa' with k=2 since characters must be at least 2 apart.

💡 If the most frequent character count is too high, rearrangement is impossible.

💡 Check if max frequency * (k - 1) exceeds string length minus max frequency.

💡 Return empty string if rearrangement is not possible due to frequency constraints.

Why it failed: Returns non-empty string when rearrangement is impossible. Fix by detecting impossibility when max frequency is too high and returning empty string.

✓ Correctly returns empty string for impossible rearrangement.

t2_04edge

Input{"s":"abcdef","k":6}

Expected"abcdef"

⏱ Performance - must finish in 2000ms

All characters unique and k equals string length, so original string is valid.

💡 When all characters are unique, any permutation is valid.

💡 Check if your code handles large k values correctly when no duplicates exist.

💡 Return original string if no rearrangement needed.

Why it failed: Fails or rearranges unnecessarily when all characters are unique. Fix by returning original string if no duplicates.

✓ Correctly returns original string when all characters are unique.

t3_01corner

Input{"s":"aaabbbcc","k":3}

Expected"abcabcab"

⏱ Performance - must finish in 2000ms

Greedy approach that picks highest frequency first but must maintain cooldown to avoid clustering.

💡 Beware of greedy approaches that pick max frequency without cooldown enforcement.

💡 Use a cooldown queue to delay reuse of characters until k distance is met.

💡 Implement max heap with cooldown queue to avoid placing same characters too close.

Why it failed: Greedy approach clusters same characters too close, violating k distance. Fix by adding cooldown queue to delay reuse.

✓ Correctly spaces characters using max heap and cooldown queue.

t3_02corner

Input{"s":"aaabc","k":0}

Expected"aaabc"

⏱ Performance - must finish in 2000ms

k=0 means no spacing constraint, original string returned.

💡 Check if your code handles k=0 as a special case.

💡 When k=0, no rearrangement is needed; return original string.

💡 Add a condition to return s immediately if k is 0.

Why it failed: Code attempts rearrangement even when k=0, causing unnecessary complexity or wrong output. Fix by returning s immediately if k=0.

✓ Correctly returns original string when k=0.

t3_03corner

Input{"s":"aaabbc","k":3}

Expected"abacab"

⏱ Performance - must finish in 2000ms

Tests correct handling of 0/1 constraint, ensuring characters are used once per cooldown period.

💡 Distinguish between 0/1 usage per cooldown and unbounded usage.

💡 Ensure each character is only used once before cooldown expires.

💡 Use a queue to hold characters until they can be reused after k positions.

Why it failed: Incorrectly reuses characters multiple times before cooldown ends. Fix by enforcing single use per cooldown with a queue.

✓ Correctly enforces 0/1 usage per cooldown period.

t4_01performance

Input{"s":"aaaaaaaaaabbbbbbbbbbccccccccccdddddddddd","k":10}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=40 and k=10 to test O(n log m) heap-based solution performance within 2 seconds.

💡 Brute force backtracking will time out on large inputs.

💡 Use a max heap and cooldown queue for efficient O(n log m) solution.

💡 Avoid recomputing frequencies repeatedly; use efficient data structures.

Why it failed: Brute force or naive solutions time out due to factorial or exponential complexity. Fix by implementing max heap with cooldown queue for O(n log m) complexity.

✓ Efficient heap-based solution runs within time limits.

Practice

1. You need to find the median of every contiguous subarray of size k in an integer array efficiently. Which approach best guarantees an optimal solution for this problem?

easy

A. Use a dynamic programming approach to precompute medians for all subarrays

B. Sort each window independently and pick the median

C. Maintain two balanced heaps (max-heap and min-heap) with lazy removal to track the median as the window slides

D. Use a sliding window with a simple queue and recompute median by scanning the window each time

Solution

1. Step 1: Understand the problem constraints

   Finding median in each sliding window requires efficient insertion, deletion, and median retrieval.

2. Step 2: Evaluate approaches

   Sorting each window (A) is too slow (O(nk log k)). DP (C) doesn't apply well to median. Simple queue (D) can't maintain order efficiently. Balanced two heaps (B) maintain order and allow O(log k) updates.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Two heaps maintain median efficiently [OK]

Hint: Two heaps efficiently maintain median in sliding window [OK]

Common Mistakes:

• Thinking sorting each window is efficient enough
• Assuming DP can precompute medians
• Using simple queue without order maintenance

2. You are given a list of intervals, each with a start and end time. For each interval, you want to find the interval with the smallest start time that is greater than or equal to the end time of the current interval. Which approach guarantees an optimal solution in terms of time complexity?

easy

A. Use two max-heaps: one for interval starts and one for interval ends, to efficiently find the next interval.

B. Use a greedy algorithm that picks the earliest finishing interval first.

C. Sort intervals by start time and use dynamic programming to find the next interval.

D. Use a brute force nested loop to compare each interval with every other interval.

Solution

1. Step 1: Understand problem requirements

   The problem requires finding, for each interval, the next interval with the smallest start time ≥ current interval's end time.

2. Step 2: Evaluate approaches

   Brute force (nested loops) is correct but inefficient (O(n²)). Sorting + binary search is efficient but requires extra steps. The two max-heaps approach efficiently tracks max starts and ends to find next intervals in O(n log n) time, guaranteeing optimality.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two heaps approach matches the problem's need for efficient next interval lookup [OK]

Hint: Two heaps efficiently track max starts and ends [OK]

Common Mistakes:

• Assuming greedy earliest finish works
• Using DP unnecessarily
• Brute force accepted as optimal

3. Consider the following Python code snippet implementing the brute force approach for sliding window median. What is the output of result after processing nums = [1, 3, -1, -3] with k = 3?

easy

A. [1.0, -1.0]

B. [1.0, 3.0]

C. [1.0, -1.0, -3.0]

D. [1.0, 0.0]

Solution

1. Step 1: Trace first window [1, 3, -1]

   Sorted window is [-1, 1, 3], median is middle element 1.0.

2. Step 2: Trace second window [3, -1, -3]

   Sorted window is [-3, -1, 3], median is middle element -1.0.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two medians computed correctly for windows of size 3 [OK]

Hint: Sort each window and pick middle element for odd k [OK]

Common Mistakes:

• Off-by-one error in window slicing
• Incorrect median index for odd k
• Confusing median calculation for even k

4. Consider the following buggy code snippet for the IPO problem. Which line contains the subtle bug that causes incorrect results or inefficiency?

medium

A. Line pushing projects into min_capital_heap without sorting

B. Line popping from min_capital_heap in the while loop

C. Line pushing profits into max_profit_heap with negation

D. Line breaking the loop when max_profit_heap is empty

Solution

1. Step 1: Identify heap initialization

   Projects are pushed into min_capital_heap without sorting by capital, which is required for efficient affordability checks.

2. Step 2: Understand impact

   Without sorting, min_capital_heap may not have the smallest capital project at the top, breaking the affordability filtering logic.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Min-heap must be built from sorted projects or pushed in sorted order [OK]

Hint: Min-heap top must be smallest capital project [OK]

Common Mistakes:

• Assuming heapq.heappush sorts automatically
• Confusing max-heap and min-heap roles
• Ignoring early stopping condition

5. Suppose projects can be selected multiple times (reused) in the IPO problem. Which modification to the two-heaps algorithm correctly handles this scenario?

hard

A. Use a max-heap only and pick the top K profitable projects ignoring capital constraints

B. Do not remove projects from min-capital heap; instead, keep all projects always available

C. After selecting a project, reinsert it into the min-capital heap to allow reuse

D. After selecting a project, do not remove it from max-profit heap to allow repeated selection

Solution

1. Step 1: Understand reuse implication

   Allowing reuse means projects remain available after selection, so they must be reinserted to be considered again.

2. Step 2: Modify algorithm accordingly

   After popping a project from min-capital heap and selecting it, reinsert it to allow future reuse; otherwise, it disappears from consideration.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Reinserting projects enables multiple selections [OK]

Hint: Reinsert projects after selection to allow reuse [OK]

Common Mistakes:

• Ignoring reinsertion, losing projects after first pick
• Keeping all projects always available breaks capital filtering
• Ignoring capital constraints when reusing projects