algoadvance

Given an integer array nums, return the maximum difference between two successive elements in its sorted form. If the array contains less than two elements, return 0.

You must write an algorithm that runs in linear time and uses linear extra space.

Clarifying Questions

  1. Q: Can the array contain negative numbers?
    • A: Yes, the array can contain negative numbers.
  2. Q: What should be returned if the array is empty or contains only one element?
    • A: Return 0 in such cases.
  3. Q: Do we need to sort the array, and is there an efficient way to do this in linear time?
    • A: Yes, we need to find the maximum gap after the array is sorted. However, traditional sorting algorithms like quicksort or mergesort would take O(n log n) time, which is not sufficient. We can use a bucket-based solution (a variation of Radix Sort) to accomplish this in linear time.

Strategy

  1. Base Case Handling: If the array contains less than two elements, return 0.
  2. Find Maximum and Minimum: Compute the maximum and minimum values in the array.
  3. Determine Bucket Size and Count:
    • The ideal bucket size can be calculated using the formula: [ \text{bucket_size} = \max(1, \frac{\text{max_val} - \text{min_val}}{N - 1}) ]
    • Here, N is the number of elements in the array.
  4. Initialize Buckets: Create buckets to hold elements. Each bucket will store:
    • min_val: the minimum value in the bucket
    • max_val: the maximum value in the bucket
  5. Distribute Elements into Buckets:
    • For each element, calculate the appropriate bucket index and update the min_val and max_val of that bucket accordingly.
  6. Compute Maximum Gap:
    • Iterate through the buckets and compute the gap between the maximum of the previous bucket and the minimum of the current bucket.
  7. Return Results: Finally, return the maximum gap found.

Code

def maximumGap(nums):
    if len(nums) < 2:
        return 0

    min_val, max_val = min(nums), max(nums)
    if min_val == max_val:
        return 0

    n = len(nums)
    bucket_size = max(1, (max_val - min_val) // (n - 1))
    bucket_count = (max_val - min_val) // bucket_size + 1

    buckets = [{'min': float('inf'), 'max': float('-inf')} for _ in range(bucket_count)]

    for num in nums:
        bucket_index = (num - min_val) // bucket_size
        buckets[bucket_index]['min'] = min(buckets[bucket_index]['min'], num)
        buckets[bucket_index]['max'] = max(buckets[bucket_index]['max'], num)

    max_gap = 0
    prev_max = min_val  # Initialize with min_val

    for bucket in buckets:
        if bucket['min'] == float('inf'):  # Empty bucket
            continue
        max_gap = max(max_gap, bucket['min'] - prev_max)
        prev_max = bucket['max']

    return max_gap

# Example usage:
nums = [3, 6, 9, 1]
print(maximumGap(nums))  # Output should be 3

Time Complexity

This solution ensures we achieve the desired linear time complexity while using linear space, meeting the problem’s constraints.

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