Magnetic Force Between Two Balls

🎯

Magnetic Force Between Two Balls

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Imagine placing magnetic balls on a metal rod so that the minimum magnetic force between any two balls is as large as possible, ensuring they don't clump together.

💡 This problem is a classic example of binary search on the answer space, which can be confusing because instead of searching through the input array, we search through possible answers. Beginners often struggle to identify when and how to apply this technique.

📋

Problem Statement

You are given an array position where position[i] represents the position of the ith ball on a 1D line. You want to place m balls into these positions such that the minimum magnetic force between any two balls is maximized. The magnetic force between two balls at positions x and y is |x - y|. Return the maximum possible minimum magnetic force between any two balls.

1 ≤ position.length ≤ 10^51 ≤ position[i] ≤ 10^9All integers in position are distinct.2 ≤ m ≤ position.length

💡

Example

Input"position = [1,2,3,4,7], m = 3"

Output3

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.

• Minimum number of balls (m=2) → output is max distance between two positions
• All positions are consecutive integers → minimum distance is 1
• Positions are very spread out → binary search range is large
• m equals the number of positions → minimum distance is the smallest gap between any two positions

⚠️

Common Mistakes

Not sorting the positions array before greedy check

Greedy placement fails, resulting in incorrect feasibility checks and wrong answers

✅ Always sort the positions array before any placement logic

Using linear search over distances instead of binary search

Solution times out on large inputs due to excessive iterations

✅ Use binary search on the answer space to reduce time complexity

Not stopping early in the feasibility check after placing all balls

Unnecessary iterations increase runtime, especially for large arrays

✅ Add early stopping once all m balls are placed

Incorrect binary search boundaries (e.g., low=0 instead of 1)

May cause infinite loops or incorrect results

✅ Set low = 1 because minimum distance cannot be zero

Using integer division incorrectly causing overflow or wrong mid calculation

Incorrect mid value leads to wrong binary search behavior

✅ Calculate mid as low + (high - low) // 2 to avoid overflow

🧠

Brute Force (Check All Distances)

💡 This approach tries every possible minimum distance and checks if placing m balls is possible. It is slow but helps understand the problem deeply.

Intuition

Try every possible minimum distance from 1 up to the maximum distance between the smallest and largest positions, and check if we can place all balls with at least that distance apart.

Algorithm

1. Sort the positions array.
2. Calculate the maximum possible distance between the first and last position.
3. For each distance d from 1 to max distance:
4. Check if it is possible to place m balls such that the minimum distance between any two balls is at least d.
5. Return the largest d for which placement is possible.

💡 This approach is intuitive but inefficient because it checks all distances without pruning.

</>

Code

def maxDistanceBruteForce(position, m):
    position.sort()
    max_dist = position[-1] - position[0]
    def can_place(distance):
        count = 1
        last_pos = position[0]
        for pos in position[1:]:
            if pos - last_pos >= distance:
                count += 1
                last_pos = pos
                if count == m:
                    return True
        return False
    result = 0
    for dist in range(1, max_dist + 1):
        if can_place(dist):
            result = dist
    return result

# Driver code
if __name__ == '__main__':
    pos = [1,2,3,4,7]
    m = 3
    print(maxDistanceBruteForce(pos, m))  # Output: 3

Line Notes

position.sort()Sort positions to check distances in order, enabling greedy placement

max_dist = position[-1] - position[0]Calculate the maximum possible distance between the first and last ball positions

def can_place(distance):Helper function to check if m balls can be placed with at least 'distance' apart

for pos in position[1:]:Iterate through positions to greedily place balls

if pos - last_pos >= distance:Place a ball only if the distance condition is met to maintain minimum spacing

if count == m:Return True early if all balls have been placed successfully

for dist in range(1, max_dist + 1):Try all possible distances from 1 up to max_dist

if can_place(dist):Update result if placement is possible for the current distance

import java.util.*;
public class Solution {
    public static boolean canPlace(int[] position, int m, int distance) {
        int count = 1, lastPos = position[0];
        for (int i = 1; i < position.length; i++) {
            if (position[i] - lastPos >= distance) {
                count++;
                lastPos = position[i];
                if (count == m) return true;
            }
        }
        return false;
    }
    public static int maxDistanceBruteForce(int[] position, int m) {
        Arrays.sort(position);
        int maxDist = position[position.length - 1] - position[0];
        int result = 0;
        for (int dist = 1; dist <= maxDist; dist++) {
            if (canPlace(position, m, dist)) {
                result = dist;
            }
        }
        return result;
    }
    public static void main(String[] args) {
        int[] pos = {1,2,3,4,7};
        int m = 3;
        System.out.println(maxDistanceBruteForce(pos, m)); // Output: 3
    }
}

Line Notes

Arrays.sort(position);Sort positions to enable greedy placement in order

int maxDist = position[position.length - 1] - position[0];Calculate maximum possible distance between balls

if (position[i] - lastPos >= distance)Check if current position can place a ball maintaining minimum distance

count++Increment count when a ball is placed

if (count == m) return true;Return early if all balls have been placed

for (int dist = 1; dist <= maxDist; dist++)Try all distances from 1 to maxDist

if (canPlace(position, m, dist))Update result if placement is possible for current distance

#include <bits/stdc++.h>
using namespace std;

bool canPlace(vector<int>& position, int m, int distance) {
    int count = 1, lastPos = position[0];
    for (int i = 1; i < (int)position.size(); i++) {
        if (position[i] - lastPos >= distance) {
            count++;
            lastPos = position[i];
            if (count == m) return true;
        }
    }
    return false;
}

int maxDistanceBruteForce(vector<int>& position, int m) {
    sort(position.begin(), position.end());
    int maxDist = position.back() - position.front();
    int result = 0;
    for (int dist = 1; dist <= maxDist; dist++) {
        if (canPlace(position, m, dist)) {
            result = dist;
        }
    }
    return result;
}

int main() {
    vector<int> pos = {1,2,3,4,7};
    int m = 3;
    cout << maxDistanceBruteForce(pos, m) << "\n"; // Output: 3
    return 0;
}

Line Notes

sort(position.begin(), position.end());Sort positions to enable greedy placement in order

int maxDist = position.back() - position.front();Calculate maximum possible distance between balls

if (position[i] - lastPos >= distance)Check if current position can place a ball maintaining minimum distance

count++Increment count when a ball is placed

if (count == m) return true;Return early if all balls have been placed

for (int dist = 1; dist <= maxDist; dist++)Try all distances from 1 to maxDist

if (canPlace(position, m, dist))Update result if placement is possible for current distance

function canPlace(position, m, distance) {
    let count = 1;
    let lastPos = position[0];
    for (let i = 1; i < position.length; i++) {
        if (position[i] - lastPos >= distance) {
            count++;
            lastPos = position[i];
            if (count === m) return true;
        }
    }
    return false;
}

function maxDistanceBruteForce(position, m) {
    position.sort((a,b) => a - b);
    let maxDist = position[position.length - 1] - position[0];
    let result = 0;
    for (let dist = 1; dist <= maxDist; dist++) {
        if (canPlace(position, m, dist)) {
            result = dist;
        }
    }
    return result;
}

// Test
const pos = [1,2,3,4,7];
const m = 3;
console.log(maxDistanceBruteForce(pos, m)); // Output: 3

Line Notes

position.sort((a,b) => a - b);Sort positions to enable greedy placement in order

let maxDist = position[position.length - 1] - position[0];Calculate maximum possible distance between balls

if (position[i] - lastPos >= distance)Check if current position can place a ball maintaining minimum distance

count++Increment count when a ball is placed

if (count === m) return true;Return early if all balls have been placed

for (let dist = 1; dist <= maxDist; dist++)Try all distances from 1 to maxDist

if (canPlace(position, m, dist))Update result if placement is possible for current distance

⏱

Complexity

TimeO(n * d) where d is max distance between positions

SpaceO(1) additional space

For each distance from 1 to maxDist, we scan the array once to check feasibility, resulting in O(n * d) time.

💡 If n=10^5 and maxDist=10^9, this approach is impractical due to huge operations.

Interview Verdict: TLE

This approach is too slow for large inputs but helps understand the problem and correctness.

🧠

Binary Search on Answer Space with Greedy Check

💡 This approach uses binary search on the minimum distance (answer space) and a greedy check to verify feasibility, drastically improving efficiency.

Intuition

Instead of checking all distances, use binary search between 1 and max distance. For each mid distance, greedily check if m balls can be placed with at least mid distance apart.

Algorithm

1. Sort the positions array.
2. Set low = 1 and high = max distance between first and last position.
3. While low <= high:
4. Calculate mid = (low + high) // 2.
5. Check if it is possible to place m balls with minimum distance mid.
6. If possible, update result = mid and set low = mid + 1 to try larger distances.
7. Else, set high = mid - 1 to try smaller distances.
8. Return result.

💡 Binary search narrows down the answer space efficiently, and greedy check quickly verifies feasibility.

</>

Code

def maxDistanceBinarySearch(position, m):
    position.sort()
    def can_place(distance):
        count = 1
        last_pos = position[0]
        for pos in position[1:]:
            if pos - last_pos >= distance:
                count += 1
                last_pos = pos
                if count == m:
                    return True
        return False
    low, high = 1, position[-1] - position[0]
    result = 0
    while low <= high:
        mid = (low + high) // 2
        if can_place(mid):
            result = mid
            low = mid + 1
        else:
            high = mid - 1
    return result

# Driver code
if __name__ == '__main__':
    pos = [1,2,3,4,7]
    m = 3
    print(maxDistanceBinarySearch(pos, m))  # Output: 3

Line Notes

position.sort()Sort positions to enable greedy placement in order

def can_place(distance):Helper function to check feasibility of placing balls with given minimum distance

for pos in position[1:]:Iterate through positions to greedily place balls

if pos - last_pos >= distance:Place ball only if distance condition is met to maintain minimum spacing

if count == m:Return True early if all balls have been placed successfully

low, high = 1, position[-1] - position[0]Initialize binary search boundaries for minimum distance

while low <= high:Binary search loop over possible minimum distances

mid = (low + high) // 2Calculate mid distance to check feasibility

if can_place(mid):If placement is possible, try larger distances

else:If placement is not possible, try smaller distances

import java.util.*;
public class Solution {
    public static boolean canPlace(int[] position, int m, int distance) {
        int count = 1, lastPos = position[0];
        for (int i = 1; i < position.length; i++) {
            if (position[i] - lastPos >= distance) {
                count++;
                lastPos = position[i];
                if (count == m) return true;
            }
        }
        return false;
    }
    public static int maxDistanceBinarySearch(int[] position, int m) {
        Arrays.sort(position);
        int low = 1, high = position[position.length - 1] - position[0];
        int result = 0;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (canPlace(position, m, mid)) {
                result = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }
    public static void main(String[] args) {
        int[] pos = {1,2,3,4,7};
        int m = 3;
        System.out.println(maxDistanceBinarySearch(pos, m)); // Output: 3
    }
}

Line Notes

Arrays.sort(position);Sort positions to enable greedy placement in order

int low = 1, high = position[position.length - 1] - position[0];Initialize binary search boundaries for minimum distance

while (low <= high)Binary search loop over possible minimum distances

int mid = low + (high - low) / 2;Calculate mid distance safely to avoid overflow

if (canPlace(position, m, mid))If placement is feasible, try larger distances

elseIf placement is not feasible, try smaller distances

return result;Return the maximum feasible minimum distance

#include <bits/stdc++.h>
using namespace std;

bool canPlace(vector<int>& position, int m, int distance) {
    int count = 1, lastPos = position[0];
    for (int i = 1; i < (int)position.size(); i++) {
        if (position[i] - lastPos >= distance) {
            count++;
            lastPos = position[i];
            if (count == m) return true;
        }
    }
    return false;
}

int maxDistanceBinarySearch(vector<int>& position, int m) {
    sort(position.begin(), position.end());
    int low = 1, high = position.back() - position.front();
    int result = 0;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (canPlace(position, m, mid)) {
            result = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

int main() {
    vector<int> pos = {1,2,3,4,7};
    int m = 3;
    cout << maxDistanceBinarySearch(pos, m) << "\n"; // Output: 3
    return 0;
}

Line Notes

sort(position.begin(), position.end());Sort positions to enable greedy placement in order

int low = 1, high = position.back() - position.front();Initialize binary search boundaries for minimum distance

while (low <= high)Binary search loop over possible minimum distances

int mid = low + (high - low) / 2;Calculate mid distance safely to avoid overflow

if (canPlace(position, m, mid))If placement is feasible, try larger distances

elseIf placement is not feasible, try smaller distances

return result;Return the maximum feasible minimum distance

function canPlace(position, m, distance) {
    let count = 1;
    let lastPos = position[0];
    for (let i = 1; i < position.length; i++) {
        if (position[i] - lastPos >= distance) {
            count++;
            lastPos = position[i];
            if (count === m) return true;
        }
    }
    return false;
}

function maxDistanceBinarySearch(position, m) {
    position.sort((a,b) => a - b);
    let low = 1;
    let high = position[position.length - 1] - position[0];
    let result = 0;
    while (low <= high) {
        let mid = Math.floor((low + high) / 2);
        if (canPlace(position, m, mid)) {
            result = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

// Test
const pos = [1,2,3,4,7];
const m = 3;
console.log(maxDistanceBinarySearch(pos, m)); // Output: 3

Line Notes

position.sort((a,b) => a - b);Sort positions to enable greedy placement in order

let low = 1;Initialize lower bound for binary search

let high = position[position.length - 1] - position[0];Initialize upper bound for binary search

while (low <= high)Binary search loop over possible minimum distances

let mid = Math.floor((low + high) / 2);Calculate mid distance to check feasibility

if (canPlace(position, m, mid))If placement is feasible, try larger distances

elseIf placement is not feasible, try smaller distances

⏱

Complexity

TimeO(n log d) where d is max distance between positions

SpaceO(1) additional space

Binary search runs in log d iterations, each checking feasibility in O(n) time.

💡 For n=10^5 and maxDist=10^9, this approach is efficient and practical.

Interview Verdict: Accepted

This is the optimal approach for this problem and is expected in interviews.

🧠

Optimized Binary Search with Early Stopping in Greedy Check

💡 This approach is a minor optimization of the previous one by stopping the feasibility check early once all balls are placed, improving runtime in practice.

Intuition

During the greedy check, as soon as we place all m balls, we stop checking further positions to save time.

Algorithm

1. Sort the positions array.
2. Set low = 1 and high = max distance between first and last position.
3. While low <= high:
4. Calculate mid = (low + high) // 2.
5. Check feasibility with early stopping if m balls placed.
6. If feasible, update result and try larger distances.
7. Else, try smaller distances.
8. Return result.

💡 Early stopping reduces unnecessary iterations in the feasibility check, improving efficiency.

</>

Code

def maxDistanceOptimized(position, m):
    position.sort()
    def can_place(distance):
        count = 1
        last_pos = position[0]
        for pos in position[1:]:
            if pos - last_pos >= distance:
                count += 1
                last_pos = pos
                if count == m:
                    return True  # Early stop
        return False
    low, high = 1, position[-1] - position[0]
    result = 0
    while low <= high:
        mid = (low + high) // 2
        if can_place(mid):
            result = mid
            low = mid + 1
        else:
            high = mid - 1
    return result

# Driver code
if __name__ == '__main__':
    pos = [1,2,3,4,7]
    m = 3
    print(maxDistanceOptimized(pos, m))  # Output: 3

Line Notes

position.sort()Sort positions to enable greedy placement in order

def can_place(distance):Feasibility check with early stopping once all balls placed

for pos in position[1:]:Iterate through positions to greedily place balls

if pos - last_pos >= distance:Place ball only if distance condition is met

if count == m:Early stop once all balls are placed to save time

low, high = 1, position[-1] - position[0]Initialize binary search boundaries

while low <= high:Binary search loop over possible minimum distances

mid = (low + high) // 2Calculate mid distance to check feasibility

if can_place(mid):If placement is feasible, try larger distances

else:If placement is not feasible, try smaller distances

import java.util.*;
public class Solution {
    public static boolean canPlace(int[] position, int m, int distance) {
        int count = 1, lastPos = position[0];
        for (int i = 1; i < position.length; i++) {
            if (position[i] - lastPos >= distance) {
                count++;
                lastPos = position[i];
                if (count == m) return true; // Early stop
            }
        }
        return false;
    }
    public static int maxDistanceOptimized(int[] position, int m) {
        Arrays.sort(position);
        int low = 1, high = position[position.length - 1] - position[0];
        int result = 0;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (canPlace(position, m, mid)) {
                result = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }
    public static void main(String[] args) {
        int[] pos = {1,2,3,4,7};
        int m = 3;
        System.out.println(maxDistanceOptimized(pos, m)); // Output: 3
    }
}

Line Notes

Arrays.sort(position);Sort positions to enable greedy placement in order

if (count == m) return true;Early stop once all balls placed to save time

while (low <= high)Binary search loop over possible minimum distances

int mid = low + (high - low) / 2;Calculate mid distance safely to avoid overflow

if (canPlace(position, m, mid))If placement is feasible, try larger distances

elseIf placement is not feasible, try smaller distances

return result;Return the maximum feasible minimum distance

#include <bits/stdc++.h>
using namespace std;

bool canPlace(vector<int>& position, int m, int distance) {
    int count = 1, lastPos = position[0];
    for (int i = 1; i < (int)position.size(); i++) {
        if (position[i] - lastPos >= distance) {
            count++;
            lastPos = position[i];
            if (count == m) return true; // Early stop
        }
    }
    return false;
}

int maxDistanceOptimized(vector<int>& position, int m) {
    sort(position.begin(), position.end());
    int low = 1, high = position.back() - position.front();
    int result = 0;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (canPlace(position, m, mid)) {
            result = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

int main() {
    vector<int> pos = {1,2,3,4,7};
    int m = 3;
    cout << maxDistanceOptimized(pos, m) << "\n"; // Output: 3
    return 0;
}

Line Notes

sort(position.begin(), position.end());Sort positions to enable greedy placement in order

if (count == m) return true;Early stop once all balls placed to save time

while (low <= high)Binary search loop over possible minimum distances

int mid = low + (high - low) / 2;Calculate mid distance safely to avoid overflow

if (canPlace(position, m, mid))If placement is feasible, try larger distances

elseIf placement is not feasible, try smaller distances

return result;Return the maximum feasible minimum distance

function canPlace(position, m, distance) {
    let count = 1;
    let lastPos = position[0];
    for (let i = 1; i < position.length; i++) {
        if (position[i] - lastPos >= distance) {
            count++;
            lastPos = position[i];
            if (count === m) return true; // Early stop
        }
    }
    return false;
}

function maxDistanceOptimized(position, m) {
    position.sort((a,b) => a - b);
    let low = 1;
    let high = position[position.length - 1] - position[0];
    let result = 0;
    while (low <= high) {
        let mid = Math.floor((low + high) / 2);
        if (canPlace(position, m, mid)) {
            result = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return result;
}

// Test
const pos = [1,2,3,4,7];
const m = 3;
console.log(maxDistanceOptimized(pos, m)); // Output: 3

Line Notes

position.sort((a,b) => a - b);Sort positions to enable greedy placement in order

if (count === m) return true;Early stop once all balls placed to save time

while (low <= high)Binary search loop over possible minimum distances

let mid = Math.floor((low + high) / 2);Calculate mid distance to check feasibility

if (canPlace(position, m, mid))If placement is feasible, try larger distances

elseIf placement is not feasible, try smaller distances

⏱

Complexity

TimeO(n log d) where d is max distance between positions

SpaceO(1) additional space

Early stopping reduces constant factors but asymptotic complexity remains the same.

💡 This optimization can speed up runtime in practice, especially when m is small.

Interview Verdict: Accepted

This is a practical optimization that can improve runtime without changing the algorithm.

📊

All Approaches - One-Glance Tradeoffs

💡 In interviews, always aim to code the binary search on answer space with greedy check (Approach 2 or 3) as it balances clarity and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n * d) where d is max distance	O(1)	No	N/A	Mention only - never code due to inefficiency
2. Binary Search on Answer Space with Greedy Check	O(n log d)	O(1)	No	N/A	Code this approach for optimal solution
3. Optimized Binary Search with Early Stopping	O(n log d) with better constants	O(1)	No	N/A	Code this if time permits for practical optimization

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding the optimal approach, and prepare to explain your reasoning clearly in interviews.

How to Present

Step 1: Clarify the problem and constraints with the interviewer.Step 2: Describe the brute force approach to show understanding.Step 3: Explain why brute force is inefficient and introduce binary search on answer space.Step 4: Detail the greedy feasibility check and binary search steps.Step 5: Code the binary search solution with early stopping.Step 6: Test with sample and edge cases.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test: 5min. Total ~23min

What the Interviewer Tests

Interviewer tests your ability to identify binary search on answer space, implement greedy checks, and optimize code for efficiency.

Common Follow-ups

• What if positions are not sorted? → Sort first, as sorting is essential.
• Can you solve it without sorting? → No, sorting is necessary for greedy placement.
• What if m equals the number of positions? → Minimum distance is the smallest gap between consecutive positions.
• How to handle very large input efficiently? → Use binary search on answer space with early stopping.

💡 These follow-ups test your understanding of problem constraints, edge cases, and efficiency improvements.

🔍

Pattern Recognition

When to Use

1) Asked to maximize or minimize a distance or value that depends on pairs or groups, 2) Input array is large and sorted or can be sorted, 3) Checking feasibility for a guessed answer is possible in linear time, 4) Problem hints at binary search on answer space.

Signature Phrases

'maximize the minimum magnetic force''place m balls into positions'

NOT This Pattern When

Problems that require binary search on sorted arrays for element lookup or DP problems with different state definitions.

Similar Problems

Aggressive cows - maximize minimum distance between cows in stallsCapacity To Ship Packages Within D Days - binary search on shipping capacitySplit Array Largest Sum - binary search on largest sum allowed

Practice

1. You are given a sorted array and a target number x. The task is to find k elements closest to x in the array, with ties broken by choosing the smaller elements first. Which of the following approaches guarantees an optimal solution with logarithmic search complexity on the array?

easy

A. Use a greedy approach by iteratively picking the closest element to x until k elements are chosen.

B. Use dynamic programming to compute the minimum distance sum for all windows of size k and pick the minimum.

C. Perform a binary search to find the left boundary of the window of size k that contains the closest elements to x.

D. Use a two-pointer sliding window starting from both ends of the array and shrink towards the center until k elements remain.

Solution

1. Step 1: Understand problem constraints

   The array is sorted, and we want k closest elements to x with tie-breakers favoring smaller elements.

2. Step 2: Identify approach with logarithmic complexity

   Binary search on the left boundary of the window of size k efficiently narrows down the candidate window in O(log(n-k)) time, guaranteeing optimality.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Binary search exploits sorted order and window size constraints [OK]

Hint: Binary search on window boundaries exploits sorted array [OK]

Common Mistakes:

• Assuming greedy selection always yields optimal window
• Using DP unnecessarily increasing complexity
• Two-pointer approach is linear, not logarithmic

2. You are given an unsorted array of integers and need to find an index of an element that is greater than its neighbors (or boundary elements if at edges). Which algorithmic approach guarantees an optimal O(log n) time solution for this problem?

easy

A. Linear scan checking each element and its neighbors sequentially

B. Dynamic programming to store and compare local maxima

C. Modified binary search that compares middle element with its right neighbor to decide search direction

D. Greedy approach selecting the first element larger than the previous one

Solution

1. Step 1: Understand problem constraints

   The problem requires finding a peak element efficiently, which suggests a logarithmic time approach.

2. Step 2: Identify algorithm matching O(log n) time

   Modified binary search that compares mid element with its neighbor to decide which half to search next achieves O(log n) time.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Binary search halves search space each step [OK]

Hint: Peak element search uses modified binary search [OK]

Common Mistakes:

• Assuming greedy or DP solves this optimally
• Confusing linear scan as optimal
• Using DP unnecessarily

3. Given the following code for finding a peak element, what is the final output when the input array is [1, 3, 2]?

def findPeakElement(nums):
    def binarySearch(low, high):
        if low == high:
            return low
        mid = (low + high) // 2
        if nums[mid] < nums[mid + 1]:
            return binarySearch(mid + 1, high)
        else:
            return binarySearch(low, mid)
    return binarySearch(0, len(nums) - 1)

print(findPeakElement([1, 3, 2]))

easy

A. None of the above

B. 0

C. 2

D. 1

Solution

1. Step 1: Trace first call binarySearch(0, 2)

   mid = (0+2)//2 = 1; nums[1]=3, nums[2]=2; since 3 > 2, recurse binarySearch(0,1)

2. Step 2: Trace second call binarySearch(0, 1)

   mid = (0+1)//2 = 0; nums[0]=1, nums[1]=3; since 1 < 3, recurse binarySearch(1,1)

3. Step 3: Trace third call binarySearch(1, 1)

   low == high, return 1

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Returned index 1 corresponds to peak element 3 [OK]

Hint: Check mid and mid+1 comparison carefully [OK]

Common Mistakes:

• Returning mid instead of low/high
• Off-by-one in mid calculation
• Confusing which half to recurse

4. What is the time complexity of the optimal approach that uses a single binary search followed by expanding boundaries to find the first and last positions of the target in a sorted array?

medium

A. O(n)

B. O(log n + k), where k is the number of occurrences of the target

C. O(log n * k), where k is the number of occurrences of the target

D. O(log n)

Solution

1. Step 1: Analyze binary search phase

   Binary search runs in O(log n) to find one occurrence of target.

2. Step 2: Analyze boundary expansion phase

   Expanding left and right to find all occurrences takes O(k), where k is count of target.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Combined complexity is O(log n + k), not multiplicative [OK]

Hint: Binary search plus linear expansion over occurrences [OK]

Common Mistakes:

• Assuming O(log n) only
• Multiplying log n by k incorrectly
• Ignoring expansion cost

5. Suppose the problem is modified so that packages can be shipped in any order (not necessarily the given order) within the given days. Which of the following approaches correctly finds the minimal ship capacity under this new constraint?

hard

A. Use dynamic programming to assign packages to days minimizing maximum load per day, combined with binary search on capacity

B. Sort packages in descending order and use a greedy bin-packing heuristic combined with binary search on capacity

C. Use a simple greedy approach assigning packages to days in any order without binary search

D. Use the same binary search on capacity with the original feasibility check since order doesn't matter

Solution

1. Step 1: Understand the relaxed constraint

   Packages can be shipped in any order, so the problem becomes a variant of multi-way partition or bin packing minimizing max load.

2. Step 2: Identify correct approach

   Binary search on capacity still applies, but feasibility check requires DP or backtracking to assign packages optimally to days, as greedy heuristics may fail.

3. Step 3: Why other options fail

   Original feasibility check assumes fixed order; greedy heuristics are approximate and may not find minimal capacity.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   DP feasibility check with binary search needed for unordered assignment [OK]

Hint: Order relaxation requires DP feasibility check, not simple greedy [OK]

Common Mistakes:

• Using original feasibility check ignoring order change
• Assuming greedy suffices
• Ignoring complexity increase