IPO (Maximize Capital, K Projects)

📋

IPO (Maximize Capital, K Projects)

Edge cases:

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IDE

def findMaxCapital(K: int, W: int, Profits: list[int], Capital: list[int]) -> int:public int findMaxCapital(int K, int W, int[] Profits, int[] Capital)int findMaxCapital(int K, int W, vector<int>& Profits, vector<int>& Capital)function findMaxCapital(K, W, Profits, Capital)

def findMaxCapital(K, W, Profits, Capital):
    # Write your solution here
    pass

class Solution {
    public int findMaxCapital(int K, int W, int[] Profits, int[] Capital) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int findMaxCapital(int K, int W, vector<int>& Profits, vector<int>& Capital) {
    // Write your solution here
    return 0;
}

function findMaxCapital(K, W, Profits, Capital) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: Returns initial capital W even when projects are affordable and profitable.Fails to add projects to max-heap when capital allows, or ignores profits.✅ Push all projects with capital requirement <= current capital into max-heap before picking.

Wrong: Selects projects ignoring capital constraints, resulting in inflated capital.Greedy approach picking highest profit projects without checking capital requirements.✅ Check capital requirement before adding projects to max-heap and before selection.

Wrong: Allows selecting the same project multiple times, inflating capital beyond correct maximum.No tracking of used projects or removal after selection.✅ Mark projects as used or remove them from heaps after selection to enforce 0/1 constraint.

Wrong: Fails on empty input arrays or K=0, throwing errors or returning wrong values.No checks for empty arrays or zero K before processing.✅ Return initial capital W immediately if K=0 or projects list is empty.

Wrong: Time limit exceeded on large inputs.Using brute force or exponential time approach.✅ Implement sorting and two heaps approach for O(n log n + K log n) complexity.

✓

Test Cases

Focus on handling edge cases like zero projects or no affordable projects.

Watch out for greedy traps and 0/1 selection constraints.

Optimize your solution with sorting and two heaps to avoid timeouts.

t1_01basic

Input{"K":2,"W":0,"Profits":[1,2,3],"Capital":[0,1,1]}

Expected4

⏱ Performance - must finish in 2000ms

Start with capital 0, pick project 0 (profit 1), capital becomes 1. Now capital is 1, pick project 2 (profit 3), capital becomes 4. Maximum capital after 2 projects is 4.

💡 Consider which projects are affordable at each step and pick the most profitable one.

💡 Use two heaps: one min-heap for capital requirements and one max-heap for profits of affordable projects.

💡 Push all projects into min-heap by capital, then move affordable projects to max-heap by profit and pick the max profit project.

Why it failed: Failed to pick projects in order of affordability and profit. Fix by using two heaps: min-heap for capital and max-heap for profits, pushing projects to max-heap only when capital allows.

✓ Correctly selects projects maximizing capital using two heaps.

t1_02basic

Input{"K":3,"W":1,"Profits":[5,10,15],"Capital":[0,1,2]}

Expected31

⏱ Performance - must finish in 2000ms

Start with capital 1, pick project 1 (profit 10), capital becomes 11. Now capital 11 allows picking project 2 (profit 15), capital becomes 26. Then pick project 0 (profit 5), capital becomes 31.

💡 Track projects affordable with current capital and pick the highest profit one.

💡 Sort projects by capital and use a max-heap for profits of affordable projects.

💡 After each pick, update capital and add newly affordable projects to max-heap.

Why it failed: Incorrectly ignores projects that become affordable after capital increases. Fix by adding all newly affordable projects to max-heap after each pick.

✓ Correctly updates capital and project availability after each pick.

t2_01edge

Input{"K":0,"W":10,"Profits":[1,2,3],"Capital":[0,1,1]}

Expected10

⏱ Performance - must finish in 2000ms

No projects can be completed (K=0), so capital remains initial W=10.

💡 Check the base case when K=0; no projects should be selected.

💡 Return initial capital immediately if K=0.

💡 Add a condition to return W if K is zero before processing projects.

Why it failed: Fails to handle K=0 case, returns incorrect capital. Fix by returning W immediately when K=0.

✓ Correctly returns initial capital when no projects can be done.

t2_02edge

Input{"K":3,"W":0,"Profits":[1,2,3],"Capital":[5,6,7]}

Expected0

⏱ Performance - must finish in 2000ms

All projects require more capital than available (W=0), so no projects can be started; capital remains 0.

💡 Check if any project is affordable with initial capital.

💡 If no projects are affordable, capital remains unchanged.

💡 Ensure projects are only added to max-heap if capital >= project capital requirement.

Why it failed: Incorrectly picks projects that require more capital than available. Fix by filtering projects with capital requirement <= current capital before selection.

✓ Correctly returns initial capital when no projects are affordable.

t2_03edge

Input{"K":2,"W":5,"Profits":[0,0,0],"Capital":[0,1,2]}

Expected5

⏱ Performance - must finish in 2000ms

All projects have zero profit, so capital remains unchanged after any number of projects.

💡 Consider projects with zero profit and their effect on capital.

💡 Selecting zero profit projects should not increase capital.

💡 Ensure profit addition is correctly applied and zero profits do not change capital.

Why it failed: Adds profits incorrectly or modifies capital even when profits are zero. Fix by adding profits only as given, no extra increments.

✓ Correctly handles zero profit projects without changing capital.

t2_04edge

Input{"K":0,"W":0,"Profits":[],"Capital":[]}

Expected0

⏱ Performance - must finish in 2000ms

Empty projects arrays and zero K means capital remains initial W=0.

💡 Handle empty input arrays gracefully.

💡 Return initial capital if no projects exist or K=0.

💡 Check for empty Profits and Capital arrays before processing.

Why it failed: Fails on empty input, throws error or returns wrong value. Fix by returning W if projects list is empty or K=0.

✓ Correctly handles empty input returning initial capital.

t3_01corner

Input{"K":3,"W":0,"Profits":[1,2,3],"Capital":[0,0,0]}

Expected6

⏱ Performance - must finish in 2000ms

All projects are affordable initially; picking the top 3 profits (3,2,1) sums to 6.

💡 When all projects are affordable at start, pick the top K profits.

💡 Use a max-heap for profits to pick highest profit projects first.

💡 Ensure all projects with capital <= W are pushed to max-heap before picking.

Why it failed: Fails to consider all initially affordable projects or picks fewer than K projects. Fix by pushing all projects with capital <= W into max-heap before selection.

✓ Correctly picks top K profitable projects when all are affordable.

t3_02corner

Input{"K":2,"W":1,"Profits":[1,2,3],"Capital":[0,1,2]}

Expected6

⏱ Performance - must finish in 2000ms

Greedy by profit alone may fail if capital requirements are ignored. Correct max capital is 4 by picking projects 0 and 1 in order.

💡 Greedy by profit alone may fail if capital requirements are ignored.

💡 Ensure projects are only picked if current capital meets their capital requirement.

💡 Use two heaps to manage capital constraints and profit maximization.

Why it failed: Greedy approach picking highest profit regardless of capital requirement leads to wrong answer. Fix by enforcing capital constraints before picking projects.

✓ Correctly respects capital constraints while maximizing profit.

t3_03corner

Input{"K":3,"W":0,"Profits":[1,2,3],"Capital":[0,1,1]}

Expected6

⏱ Performance - must finish in 2000ms

Confirms 0/1 project selection: each project can be selected at most once, so max capital after 2 picks is 4, not more.

💡 Do not reuse projects multiple times; each project can be selected once.

💡 Track visited projects or remove them after selection.

💡 Ensure no project is selected more than once in the algorithm.

Why it failed: Incorrectly allows selecting the same project multiple times, inflating capital. Fix by marking projects as used or removing them after selection.

✓ Correctly enforces 0/1 project selection constraint.

t4_01performance

Input{"K":100,"W":500000000,"Profits":[1000000000,999999999,999999998,999999997,999999996,999999995,999999994,999999993,999999992,999999991,999999990,999999989,999999988,999999987,999999986,999999985,999999984,999999983,999999982,999999981,999999980,999999979,999999978,999999977,999999976,999999975,999999974,999999973,999999972,999999971,999999970,999999969,999999968,999999967,999999966,999999965,999999964,999999963,999999962,999999961,999999960,999999959,999999958,999999957,999999956,999999955,999999954,999999953,999999952,999999951,999999950,999999949,999999948,999999947,999999946,999999945,999999944,999999943,999999942,999999941,999999940,999999939,999999938,999999937,999999936,999999935,999999934,999999933,999999932,999999931,999999930,999999929,999999928,999999927,999999926,999999925,999999924,999999923,999999922,999999921,999999920,999999919,999999918,999999917,999999916,999999915,999999914,999999913,999999912,999999911,999999910,999999909,999999908,999999907,999999906,999999905,999999904,999999903,999999902,999999901,999999900],"Capital":[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, K=100, O(n log n + K log n) expected to complete within 2s.

💡 Use two heaps to efficiently select projects within capital constraints.

💡 Avoid brute force exponential approaches.

💡 Sorting projects by capital and using heaps for profits achieves O(n log n + K log n).

Why it failed: Brute force or naive solutions will TLE due to exponential complexity. Fix by implementing two heaps approach with sorting and priority queues.

✓ Efficient two heaps approach confirmed to run within time limits.

Practice

1. Given the following Python code implementing the optimal approach for minimizing deviation, what is the final returned value when the input is [1, 2, 3, 4]?

import heapq

def minimumDeviation(nums):
    heap = []
    min_val = float('inf')
    for num in nums:
        if num % 2 == 1:
            num *= 2
        min_val = min(min_val, num)
        heapq.heappush(heap, -num)  # max heap by pushing negative

    min_dev = float('inf')
    while True:
        max_val = -heapq.heappop(heap)
        min_dev = min(min_dev, max_val - min_val)
        if max_val % 2 == 0:
            max_val //= 2
            min_val = min(min_val, max_val)
            heapq.heappush(heap, -max_val)
        else:
            break
    return min_dev

print(minimumDeviation([1,2,3,4]))

easy

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Normalize input and build max-heap

   Input [1,2,3,4] becomes [2,2,6,4] after doubling odds. min_val = 2. Heap = [-6,-4,-2,-2].

2. Step 2: Iteratively reduce max and update min_dev

   Pop max=6, min_dev=6-2=4. Halve 6 -> 3, min_val=2, push -3. Pop max=4, min_dev=4-2=2. Halve 4 -> 2, min_val=2, push -2. Pop max=3, min_dev=2-2=0. 3 is odd, stop. Minimum deviation found is 1 (from earlier step).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Final returned value matches expected minimal deviation [OK]

Hint: Doubling odds first and halving max even values reduces deviation [OK]

Common Mistakes:

• Forgetting to double odd numbers initially
• Stopping too early without halving max even values
• Miscomputing min_val after halving

2. Consider the following Python code implementing the optimal rearrangement algorithm. What is the output of rearrangeString("aabc", 2)?

easy

A. "abac"

B. "abca"

C. "aabc"

D. "" (empty string)

Solution

1. Step 1: Trace first iterations of the loop

   Initial freq: {a:2, b:1, c:1}, max_heap=[(-2,'a'),(-1,'b'),(-1,'c')]. Pop 'a', res=['a'], queue=[('a',-1)]. Queue size < k=2, continue.

2. Step 2: Next iterations and final string

   Pop 'b', res=['a','b'], queue=[('a',-1),('b',0)]. Queue size == k, pop front ('a',-1), since count<0, push back (-1,'a'). Next pop 'a', res=['a','b','a'], queue=[('b',0),('a',0)]. Pop front ('b',0), count not <0, no push. Pop 'c', res=['a','b','a','c'], queue=[('a',0),('c',0)]. Final string length matches input length.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Output "abac" respects k=2 spacing and uses all chars [OK]

Hint: Track heap pops and queue size carefully [OK]

Common Mistakes:

• Confusing when to push back characters into heap
• Off-by-one in decrementing counts
• Returning partial strings without length check

3. Consider the following buggy code snippet for sliding window median using a balanced sorted list. Which line contains the subtle bug that causes incorrect median results for even-sized windows?

medium

A. Line with 'return float(window[k // 2 - 1] + window[k // 2])'

B. Line with 'bisect.insort_left(window, nums[i])'

C. Line with 'idx = bisect.bisect_left(window, nums[i - k])'

D. Line with 'window = sorted(nums[:k])'

Solution

1. Step 1: Analyze median calculation for even k

   Median for even k is average of two middle elements, so sum must be divided by 2.

2. Step 2: Identify bug

   Line returning sum without division causes median to be twice the correct value.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Missing division by 2 in even median calculation [OK]

Hint: Even median requires averaging two middle elements [OK]

Common Mistakes:

• Forgetting to divide sum by 2 for even k median
• Incorrect index calculation for median
• Wrong removal or insertion causing imbalance

4. What is the time complexity of the optimal two heaps approach for finding the next interval for each of n intervals, and why?

medium

A. O(n log n) because building and processing two heaps each take O(n log n) time.

B. O(n²) because each interval may be compared with all others in the worst case.

C. O(n) because each interval is pushed and popped from heaps only once.

D. O(n log k) where k is the number of intervals with start ≥ end for each interval.

Solution

1. Step 1: Analyze heap operations

   Building two heaps of size n takes O(n) each, but heap push/pop operations during processing take O(log n) each.

2. Step 2: Count total operations

   Each interval is pushed and popped at most once from each heap, resulting in O(n log n) total time.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Heap operations dominate complexity, confirming O(n log n) [OK]

Hint: Heap push/pop cost dominates, not linear scans [OK]

Common Mistakes:

• Assuming linear time due to single pass
• Confusing n and k in complexity
• Mistaking nested loops for heap operations

5. Suppose the cooldown constraint is relaxed so that tasks can be scheduled again immediately (space = 0), but each task can only be executed a limited number of times given by a frequency map. Which modification to the optimized scheduler is correct?

hard

A. Ignore cooldown and schedule tasks in any order until frequencies are exhausted

B. Simulate day-by-day with a queue to track cooldown and frequency simultaneously

C. Continue using the hash map for cooldown but ignore frequency limits

D. Use a max-heap to always pick the task with highest remaining frequency, decrement frequency after scheduling

Solution

1. Step 1: Understand the new constraints

   Cooldown is zero, so tasks can be scheduled consecutively, but frequency limits restrict how many times each task can run.

2. Step 2: Identify correct scheduling approach

   Using a max-heap to pick the task with highest remaining frequency ensures tasks are scheduled optimally without cooldown constraints.

3. Step 3: Why other options fail

   Ignoring frequency leads to over-scheduling; hash map cooldown tracking is unnecessary; simulating day-by-day is inefficient.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Max-heap prioritizes tasks by frequency -> [OK]

Hint: Max-heap for frequency scheduling when cooldown is zero [OK]

Common Mistakes:

• Ignoring frequency constraints
• Using cooldown tracking when space=0
• Simulating day-by-day unnecessarily