Maximum Gap (Pigeonhole / Sort + Scan)

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Steps

setup

Check array length

The algorithm first checks if the input array has fewer than two elements. Since the input has 4 elements, it proceeds.

💡 If the array is too short, no gap exists, so returning 0 immediately saves time.

Line:

if len(nums) < 2:
    return 0

💡 The algorithm only processes arrays where a gap can exist.

setup

Find minimum and maximum values

The algorithm finds the minimum and maximum values in the array: min_num=1 and max_num=9.

💡 Knowing the range helps detect if all elements are equal and is useful for gap calculations.

Line:

min_num, max_num = min(nums), max(nums)

💡 The min and max define the bounds of the array values.

setup

Check if all elements are equal

The algorithm checks if min_num equals max_num. Since 1 != 9, it continues.

💡 If all elements are equal, the maximum gap is zero, so returning early saves time.

Line:

if min_num == max_num:
    return 0

💡 The array has distinct values, so a positive gap may exist.

expand

Sort the array

The array is sorted in ascending order, resulting in [1, 3, 6, 9].

💡 Sorting arranges elements so adjacent differences can be checked linearly.

Line:

nums.sort()

💡 Sorting enables efficient scanning for maximum gap between neighbors.

setup

Initialize max_gap to zero

The variable max_gap is initialized to 0 before scanning begins.

💡 max_gap will track the largest gap found during scanning.

Line:

max_gap = 0

💡 Starting with zero ensures any positive gap found will update max_gap.

traverse

Start scanning from second element (i=1)

Set pointer i to index 1 to compare nums[1] and nums[0].

💡 We compare adjacent pairs to find gaps.

Line:

for i in range(1, len(nums)):

💡 Iteration starts at second element to compare with previous.

compare

Calculate gap between nums[1] and nums[0]

Calculate gap = nums[1] - nums[0] = 3 - 1 = 2. max_gap is updated from 0 to 2.

💡 This is the first gap found and sets a baseline for max_gap.

Line:

max_gap = max(max_gap, nums[i] - nums[i-1])

💡 max_gap tracks the largest gap found so far.

traverse

Move pointer i to index 2

Pointer i moves to index 2 to compare nums[2] and nums[1].

💡 We continue scanning adjacent pairs to find larger gaps.

Line:

for i in range(1, len(nums)):

💡 Each adjacent pair is checked sequentially.

compare

Calculate gap between nums[2] and nums[1]

Calculate gap = nums[2] - nums[1] = 6 - 3 = 3. max_gap updated from 2 to 3.

💡 A larger gap is found, so max_gap is updated.

Line:

max_gap = max(max_gap, nums[i] - nums[i-1])

💡 max_gap always holds the largest gap found so far.

traverse

Move pointer i to index 3

Pointer i moves to index 3 to compare nums[3] and nums[2].

💡 Continue scanning adjacent pairs for maximum gap.

Line:

for i in range(1, len(nums)):

💡 Sequential scanning covers all adjacent pairs.

compare

Calculate gap between nums[3] and nums[2]

Calculate gap = nums[3] - nums[2] = 9 - 6 = 3. max_gap remains 3.

💡 Gap equals current max_gap, so max_gap stays the same.

Line:

max_gap = max(max_gap, nums[i] - nums[i-1])

💡 max_gap only updates when a strictly larger gap is found.

reconstruct

Return the maximum gap

The algorithm returns max_gap = 3 as the maximum gap found.

💡 The final answer is the largest gap between adjacent sorted elements.

Line:

return max_gap

💡 The maximum gap is the final output of the algorithm.

from typing import List

def maximumGap(nums: List[int]) -> int:
    # STEP 1: Check if array length < 2
    if len(nums) < 2:
        return 0
    # STEP 2: Find min and max
    min_num, max_num = min(nums), max(nums)
    # STEP 3: Check if all elements equal
    if min_num == max_num:
        return 0
    # STEP 4: Sort the array
    nums.sort()
    # STEP 5: Initialize max_gap
    max_gap = 0
    # STEP 6-11: Scan array to find max gap
    for i in range(1, len(nums)):
        max_gap = max(max_gap, nums[i] - nums[i-1])
    # STEP 12: Return max_gap
    return max_gap

if __name__ == '__main__':
    print(maximumGap([3,6,9,1]))  # Output: 3

📊

Maximum Gap (Pigeonhole / Sort + Scan) - Watch the Algorithm Execute, Step by Step

Watching each step reveals how sorting enables a simple linear scan to find the maximum gap, making the algorithm efficient and intuitive.

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expand

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Result: 0

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Result: 3

record

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Result: 3

Key Takeaways

✓ Sorting the array transforms the problem into a simple linear scan for maximum adjacent gap.

This insight is hard to see from code alone because the power of sorting enabling linear scanning is conceptual.

✓ Early checks for array length and equal elements optimize performance by avoiding unnecessary work.

Understanding these early exits helps appreciate efficiency improvements beyond the main loop.

✓ The max_gap variable updates only when a strictly larger gap is found, ensuring the final answer is correct.

Seeing each update visually clarifies how the algorithm tracks the maximum gap step-by-step.

Practice

1. You are given an array of integers and need to rearrange it so that no two adjacent elements are the same. Which approach guarantees an optimal solution for this problem?

easy

A. Count frequencies, then use a max heap to greedily place the most frequent elements first, ensuring spacing.

B. Use backtracking to generate all permutations and check for valid arrangements.

C. Sort the array and then swap adjacent elements to avoid duplicates.

D. Use dynamic programming to track valid sequences avoiding adjacent duplicates.

Solution

1. Step 1: Understand problem constraints

   The problem requires rearranging elements so no two adjacent are equal, which demands careful placement of frequent elements.

2. Step 2: Identify approach that handles frequency and spacing

   Using a max heap to always pick the most frequent element available and placing it ensures no two identical elements are adjacent, which brute force or simple sorting cannot guarantee efficiently.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Max heap approach handles frequency and spacing correctly [OK]

Hint: Max heap ensures spacing of frequent elements [OK]

Common Mistakes:

• Thinking sorting alone avoids adjacent duplicates
• Assuming DP solves this without frequency tracking

2. Consider the following code snippet implementing the optimal solution to rearrange an array so no two adjacent elements are equal. What is the output of the function call make_array_beautiful_heap([1,1,2])?

easy

A. []

B. [1, 1, 2]

C. [1, 2, 1]

D. [2, 1, 1]

Solution

1. Step 1: Check max frequency condition

   Frequency counts: 1 -> 2, 2 -> 1; max_freq=2, n=3, (n+1)//2=2, so condition passes.

2. Step 2: Trace heap operations and result construction

   Heap: [(-2,1), (-1,2)] Pop (-2,1): res=[1], prev_count=-1, prev_num=1 Pop (-1,2): res=[1,2], push back (-1,1) since prev_count<0 Pop (-1,1): res=[1,2,1] Return [1,2,1]

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Output matches expected rearrangement with no adjacent duplicates [OK]

Hint: Check max frequency and simulate heap pops [OK]

Common Mistakes:

• Returning original array without rearrangement
• Confusing heap push/pop order

3. Suppose the citation counts can be negative or zero, and you want to compute the h-index considering only non-negative citations. Which modification to the optimal algorithm is correct?

hard

A. Filter out negative citations before sorting, then apply the same binary search algorithm on the filtered list.

B. Keep all citations including negatives, but treat negative values as zero during counting.

C. Use the brute force approach to handle negative values correctly without filtering.

D. Sort citations ascending and perform binary search on h values including negatives.

Solution

1. Step 1: Understand problem with negatives

   Negative citations do not contribute to h-index; they should be excluded.

2. Step 2: Modify input before algorithm

   Filter out negatives, then sort descending and apply binary search as usual.

3. Step 3: Why other options fail

   Treating negatives as zero inflates counts incorrectly; brute force is inefficient; ascending sort breaks counting logic.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Filtering negatives preserves correctness and efficiency [OK]

Hint: Exclude negatives before sorting and searching [OK]

Common Mistakes:

• Counting negatives as zero
• Not filtering negatives
• Sorting ascending

4. Suppose the problem is modified so that elements can be reused unlimited times to form an infinitely long array with no two adjacent elements equal. Which modification to the heap-based algorithm correctly handles this variant?

hard

A. Remove the max frequency check and keep pushing elements back with updated counts indefinitely.

B. Use a queue instead of a heap to cycle through elements ensuring no two adjacent are equal.

C. Use a max heap but reset counts to original frequencies after each full pass through the heap.

D. Modify the algorithm to never decrement counts and always push elements back to allow infinite reuse.

Solution

1. Step 1: Understand infinite reuse implication

   Since elements can be reused unlimited times, counts do not decrease; the algorithm must allow repeated placement without decrementing counts.

2. Step 2: Modify heap usage accordingly

   Never decrement counts and always push elements back to heap to maintain availability, ensuring no two adjacent elements are equal.

3. Step 3: Evaluate other options

   Removing max frequency check or resetting counts breaks correctness or efficiency; queue does not prioritize frequency.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Maintaining counts without decrement allows infinite reuse without adjacency violation [OK]

Hint: Infinite reuse means counts stay constant [OK]

Common Mistakes:

• Removing checks without adjusting counts
• Resetting counts causing incorrect order

5. Suppose the problem is modified so that time points can repeat multiple times and you want to find the minimum difference between any two occurrences, but the input size is very large (up to 10^7). Which approach is best suited to handle this efficiently?

hard

A. Use the brute force nested loops approach to check all pairs for minimum difference.

B. Use the bucket sort approach with a boolean array to detect duplicates and find minimum difference in O(n) time.

C. Use a hash map to count occurrences and then sort unique times to find minimum difference.

D. Use a balanced binary search tree to insert times and find minimum difference on the fly.

Solution

1. Step 1: Consider input size and duplicates

   With up to 10^7 times and possible duplicates, brute force is infeasible.

2. Step 2: Bucket sort with boolean array is optimal

   Using a fixed-size 1440-length boolean array detects duplicates instantly and scans once for minimum difference in O(n) time.

3. Step 3: Other approaches are less efficient

   Sorting or balanced trees add overhead and cannot beat fixed-size bucket scanning for this problem.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Bucket sort approach handles large inputs and duplicates efficiently [OK]

Hint: Fixed 1440 bucket array handles duplicates and large inputs efficiently [OK]

Common Mistakes:

• Trying brute force on large inputs
• Sorting large inputs with duplicates is costly
• Using trees adds unnecessary overhead