Magnetic Force Between Two Balls

📋

Magnetic Force Between Two Balls

Edge cases:

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IDE

def maxDistance(position: list[int], m: int) -> int:public int maxDistance(int[] position, int m)int maxDistance(vector<int>& position, int m)function maxDistance(position, m)

def maxDistance(position, m):
    # Write your solution here
    pass

class Solution {
    public int maxDistance(int[] position, int m) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int maxDistance(vector<int>& position, int m) {
    // Write your solution here
    return 0;
}

function maxDistance(position, m) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: Too large minimum distance (e.g., 4 instead of 3)Greedy check incorrectly assumes balls can be placed with larger distance without verifying placement count.✅ Implement greedy check that counts how many balls can be placed with given distance and adjust binary search accordingly.

Wrong: Too small minimum distance (e.g., 1 instead of 3)Binary search lower bound or greedy check logic is incorrect, causing premature termination.✅ Ensure binary search updates low and high bounds correctly based on greedy check results.

Wrong: Off-by-one errors in output (e.g., 3 instead of 2)Incorrect mid calculation or comparison in binary search loop.✅ Use mid = low + (high - low + 1) // 2 and adjust low/high based on feasibility to avoid infinite loops and off-by-one errors.

Wrong: Fails on large inputs or times outUsing brute force approach with O(n*d) complexity instead of binary search on answer space.✅ Implement binary search on answer space with O(n log d) complexity and greedy feasibility check.

Wrong: Incorrect output when m equals number of positionsGreedy check skips positions or does not handle placing balls at every position.✅ Ensure greedy check places balls at every position if needed and binary search range includes minimum gap.

✓

Test Cases

Focus on handling smallest and largest inputs correctly, including boundary values.

Watch out for common pitfalls like greedy traps and off-by-one errors in binary search.

Optimize your solution to O(n log d) using binary search on answer space and greedy feasibility checks.

t1_01basic

Input{"position":[1,2,3,4,7],"m":3}

Expected3

⏱ Performance - must finish in 2000ms

Place balls at positions 1, 4, and 7. The minimum distance is min(4-1=3,7-4=3) = 3, which is the largest possible.

💡 Think about binary searching the minimum distance between balls.

💡 Use a greedy check to verify if balls can be placed with at least a given distance.

💡 Binary search over distance range and use greedy placement to find max minimum distance.

Why it failed: Returned incorrect max minimum distance; likely failed to implement binary search or greedy check correctly. Fix by ensuring binary search mid is tested with a proper greedy placement function.

✓ Correctly finds the maximum minimum distance using binary search and greedy check.

t1_02basic

Input{"position":[5,10,15,20,25],"m":2}

Expected20

⏱ Performance - must finish in 2000ms

With 2 balls, placing at positions 5 and 25 yields the maximum minimum distance of 20.

💡 For m=2, the answer is the max distance between any two positions.

💡 Binary search range is from 1 to max(position) - min(position).

💡 Check if you can place 2 balls with at least mid distance by placing first ball at min position and next at earliest position >= last + mid.

Why it failed: Incorrect output likely due to not considering max distance as initial upper bound or incorrect greedy placement logic. Fix by sorting positions and correctly implementing binary search boundaries and greedy check.

✓ Correctly handles minimum number of balls and finds max distance.

t2_01edge

Input{"position":[1,2],"m":2}

Expected1

⏱ Performance - must finish in 2000ms

Only two positions and two balls; max minimum distance is the difference 1.

💡 Check the smallest input size with minimum balls.

💡 Binary search range is from 1 to position[-1] - position[0].

💡 Greedy check should place balls at positions 1 and 2 successfully with distance 1.

Why it failed: Fails on minimum input size; likely does not handle smallest input or boundary conditions correctly. Fix by ensuring binary search handles low range and greedy check places all balls.

✓ Handles minimum input size and balls correctly.

t2_02edge

Input{"position":[1,2,3,4,5],"m":5}

Expected1

⏱ Performance - must finish in 2000ms

All positions consecutive and m equals number of positions; minimum distance is 1.

💡 When m equals number of positions, minimum distance is smallest gap between consecutive positions.

💡 Check that greedy placement does not skip any positions.

💡 Binary search should find minimum distance 1 as maximum possible.

Why it failed: Fails to handle case where m equals number of positions; likely greedy check skips positions or binary search range incorrect. Fix by ensuring greedy check places balls at every position if needed.

✓ Correctly handles placing balls at every position with minimum distance 1.

t2_03edge

Input{"position":[1,1000000000],"m":2}

Expected999999999

⏱ Performance - must finish in 2000ms

Positions are very spread out; max minimum distance is the difference 999999999.

💡 Test large distance range for binary search.

💡 Ensure no integer overflow in distance calculations.

💡 Greedy check should place balls at positions 1 and 1000000000 with distance 999999999.

Why it failed: Fails on large distance range; likely integer overflow or incorrect binary search bounds. Fix by using appropriate data types and correct binary search implementation.

✓ Handles large distance ranges without overflow.

t2_04edge

Input{"position":[1],"m":1}

Expected0

⏱ Performance - must finish in 2000ms

Single position and one ball; minimum distance is 0 since only one ball placed.

💡 Check behavior when only one ball is to be placed.

💡 Minimum distance is zero because no pair exists.

💡 Ensure code handles m=1 gracefully.

Why it failed: Fails on single element input or m=1; likely does not handle trivial cases. Fix by returning 0 when m=1 or only one position.

✓ Correctly handles single ball or single position cases.

t3_01corner

Input{"position":[1,2,4,8,9],"m":3}

Expected3

⏱ Performance - must finish in 2000ms

Greedy trap test: placing balls greedily with too large distance may fail; correct max minimum distance is 3.

💡 Beware of greedy trap: placing balls too far apart may miss valid placements.

💡 Binary search with greedy check must verify feasibility for each mid distance.

💡 Adjust binary search boundaries carefully and verify placement count.

Why it failed: Fails greedy trap test by returning too large distance without verifying placement feasibility. Fix by implementing correct greedy check that counts balls placed and adjusts binary search accordingly.

✓ Correctly handles greedy trap by verifying placement feasibility.

t3_02corner

Input{"position":[1,3,5,7,9],"m":4}

Expected2

⏱ Performance - must finish in 2000ms

Off-by-one error test: correct max minimum distance is 2, not 3.

💡 Check off-by-one errors in binary search mid calculation.

💡 Ensure greedy check places exactly m balls or fewer to adjust search.

💡 Binary search should narrow down to correct max minimum distance.

Why it failed: Returns incorrect distance due to off-by-one error in binary search or greedy check. Fix by carefully handling mid calculation and placement count comparison.

✓ Correctly handles off-by-one in binary search and greedy check.

t3_03corner

Input{"position":[1,2,4,7,11],"m":3}

Expected4

⏱ Performance - must finish in 2000ms

Test for confusion between 0/1 knapsack and binary search: max minimum distance is 5.

💡 This problem requires binary search on answer space, not knapsack.

💡 Greedy check places balls at earliest positions satisfying distance.

💡 Do not confuse with subset sum or knapsack approaches.

Why it failed: Fails by using incorrect approach like knapsack or subset sum instead of binary search. Fix by implementing binary search on minimum distance with greedy feasibility check.

✓ Correctly uses binary search on answer space with greedy check.

t4_01performance

Input{"position":[1,10001,20001,30001,40001,50001,60001,70001,80001,90001,100001,110001,120001,130001,140001,150001,160001,170001,180001,190001,200001,210001,220001,230001,240001,250001,260001,270001,280001,290001,300001,310001,320001,330001,340001,350001,360001,370001,380001,390001,400001,410001,420001,430001,440001,450001,460001,470001,480001,490001,500001,510001,520001,530001,540001,550001,560001,570001,580001,590001,600001,610001,620001,630001,640001,650001,660001,670001,680001,690001,700001,710001,720001,730001,740001,750001,760001,770001,780001,790001,800001,810001,820001,830001,840001,850001,860001,870001,880001,890001,900001,910001,920001,930001,940001,950001,960001,970001,980001,990001,1000001],"m":50}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=100 and spaced positions to test O(n log d) complexity within 2 seconds.

💡 Binary search on answer space runs in O(n log d) time.

💡 Greedy check runs in O(n) time per binary search iteration.

💡 Brute force O(n*d) will time out; optimize with binary search.

Why it failed: TLE due to brute force or inefficient approach; fix by implementing binary search on answer space with O(n log d) complexity.

✓ Efficient binary search and greedy check pass within time limit.

Practice

1. Consider the following Python code that finds the smallest divisor for nums = [1, 2, 5, 9] and threshold = 6. What is the value of the variable mid during the first iteration of the while loop inside smallest_divisor?

easy

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Initialize low and high

   low = 1, high = max(nums) = 9

2. Step 2: Calculate mid in first iteration

   mid = (1 + 9) // 2 = 5; integer division yields 5.

3. Step 3: Confirm mid value

   The first mid value is 5, matching 5.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Integer division midpoint between 1 and 9 is 5 -> [OK]

Hint: Mid is integer division of low and high boundaries [OK]

Common Mistakes:

• Off-by-one error in calculating mid
• Confusing mid with low or high values
• Misunderstanding integer division behavior

2. What is the time complexity of the recursive modified binary search algorithm for finding a peak element in an array of size n, and why might some candidates mistakenly think it is worse?

medium

A. O(n) because each recursive call processes a subarray of size n-1

B. O(log n) because each recursive call halves the search space

C. O(n log n) because recursion adds a log n factor to linear scanning

D. O(log n) time but O(n) space due to recursion stack

Solution

1. Step 1: Analyze recursive calls

   Each call halves the search space, so number of calls is O(log n).

2. Step 2: Confirm work per call

   Each call does O(1) work (simple comparisons), so total time is O(log n).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Halving search space leads to logarithmic time [OK]

Hint: Halving input size -> O(log n) time [OK]

Common Mistakes:

• Confusing recursion stack space with time
• Assuming linear scan inside recursion
• Thinking recursion multiplies complexity

3. What is the time complexity of the optimal binary search solution for finding the minimum speed to arrive on time, given n distances and a precision epsilon for floating point speed search?

medium

A. O(log n * (maxSpeed / epsilon))

B. O(n * maxSpeed)

C. O(n^2 * log(maxSpeed))

D. O(n * log((maxSpeed - minSpeed) / epsilon))

Solution

1. Step 1: Identify the binary search iterations

   Binary search on speed range [minSpeed, maxSpeed] with precision epsilon takes O(log((maxSpeed - minSpeed)/epsilon)) steps.

2. Step 2: Calculate per-iteration cost

   Each feasibility check sums over n distances, so O(n) per iteration.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Combining steps yields O(n log((maxSpeed - minSpeed)/epsilon)) [OK]

Hint: Binary search iterations times linear feasibility check [OK]

Common Mistakes:

• Confusing maxSpeed with n
• Ignoring epsilon precision in log factor

4. Suppose the problem is modified so that each number in the array can be divided multiple times (reuse allowed), and the sum of all ceiling divisions must still be ≤ threshold. Which modification to the algorithm is necessary to correctly find the smallest divisor?

hard

A. Use the same binary search with feasibility function, but modify feasibility to count repeated divisions per element

B. Switch to a greedy approach that picks divisors based on the largest elements first

C. Use dynamic programming to track all possible divisor sums with reuse

D. Binary search is no longer applicable; use brute force over all divisors

Solution

1. Step 1: Understand the reuse modification

   Allowing multiple divisions per element means feasibility must consider repeated divisions, but the divisor search space remains monotonic.

2. Step 2: Adapt feasibility function

   Modify feasibility to count how many times each element can be divided by the divisor (e.g., sum of ceiling divisions with reuse).

3. Step 3: Retain binary search on answer space

   Binary search still applies because the feasibility condition remains monotonic with divisor.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Binary search with adapted feasibility handles reuse -> [OK]

Hint: Reuse changes feasibility, not binary search applicability [OK]

Common Mistakes:

• Discarding binary search prematurely
• Using greedy which breaks global constraints
• Overcomplicating with DP when not needed

5. Suppose the problem is modified so that the matrix rows are sorted but columns are not guaranteed to be sorted. Which approach would correctly find the kth smallest element efficiently?

hard

A. Use the min-heap approach merging rows as before, since rows remain sorted.

B. Flatten the matrix and sort it, then return the kth element.

C. Use binary search on the value range with a counting function scanning the entire matrix.

D. Use a min-heap merging columns instead of rows.

Solution

1. Step 1: Analyze matrix properties

   Rows are sorted but columns are not, so min-heap merging rows only partially orders elements.

2. Step 2: Evaluate approaches

   Min-heap merging rows assumes columns sorted for correctness; binary search counting requires sorted rows and columns for efficient counting; flatten and sort works regardless.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Flatten and sort is the fallback when only rows sorted [OK]

Hint: Partial sorting breaks heap and binary search assumptions [OK]

Common Mistakes:

• Assuming min-heap merging rows still works
• Using binary search counting without full sorting
• Ignoring matrix sorting properties