Maximum Gap
#833 Most AskedMedium
Maximum Gap #833 Most Asked
Medium
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Topics:
ArraysGreedy Algorithms
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.
Example 1:
Input: nums = [3,6,9,1] Output: 3 Explanation: The sorted form of the array is [1,3,6,9], either (3,6) or (6,9) has the maximum difference 3.
Example 2:
Input: nums = [10] Output: 0 Explanation: The array contains less than 2 elements, therefore return 0.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 109
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What is the range of possible values for the integers in the input array? Can they be negative, zero, or only positive?
- What should be returned if the input array `nums` is empty or contains only one element?
- Are duplicate values allowed in the input array, and if so, how should they be handled when determining the maximum gap?
- Could you please clarify what you mean by 'successive elements in its sorted form'? Does it mean adjacent after sorting, or should I consider other possibilities?
- What is the expected size of the input array `nums`? Should I be concerned about integer overflow when calculating differences?
Brute Force Solution
Approach
The brute force approach to finding the maximum gap involves exhaustively comparing every possible pair of numbers. We're essentially checking every single combination to see which pair has the biggest difference. It's like looking at every single student pairing in a class to find the two tallest students, without skipping anyone.
Here's how the algorithm would work step-by-step:
- Take the first number from the list.
- Compare this number with every other number in the list one by one.
- Calculate the difference between the first number and each of the other numbers.
- Keep track of the largest difference you've found so far.
- Now, take the second number from the list.
- Compare the second number with every other number in the list (including the first number).
- Calculate the difference between the second number and each of the other numbers.
- If any of these new differences are bigger than the largest difference you've kept track of so far, update it.
- Repeat this process for every number in the list, comparing each one to all the others.
- After you've compared all possible pairs of numbers, the largest difference you've kept track of is the maximum gap.
Code Implementation
def maximum_gap_brute_force(numbers):
if not numbers or len(numbers) < 2:
return 0
max_difference = 0
for first_number_index in range(len(numbers)):
for second_number_index in range(len(numbers)):
# Avoid comparing a number with itself
if first_number_index == second_number_index:
continue
# Calculate the absolute difference between numbers
current_difference = abs(numbers[first_number_index] - numbers[second_number_index])
# Update max_difference if a larger gap is found
if current_difference > max_difference:
max_difference = current_difference
return max_differenceBig(O) Analysis
Time Complexity
O(n²) – The brute force approach iterates through each of the n numbers in the input array. For each number, it compares it against every other number in the array to find the difference. This means that for each of the n numbers, we perform approximately n comparisons, leading to nested iterations. Therefore, the total number of operations is proportional to n multiplied by n, resulting in a time complexity of O(n²).
Space Complexity
O(1) – The provided brute force approach only involves iterating through the input list and comparing elements. We maintain a single variable to store the largest difference found so far. No additional data structures, such as auxiliary arrays or hash maps, are created. Therefore, the auxiliary space required remains constant regardless of the input size N, and the space complexity is O(1).
Optimal Solution
Approach
The core idea is to avoid comparing every pair of numbers directly. We want to group similar numbers and then only check the gaps between these groups. This avoids a lot of unnecessary comparisons and makes finding the largest gap much faster.
Here's how the algorithm would work step-by-step:
- First, divide the range of numbers into equal-sized groups (think buckets).
- For each group, keep track of the smallest and largest numbers that fall into it.
- Now, find the biggest gap. The biggest gap will *never* occur between numbers in the same group; it must occur between different groups.
- To find it, look at the difference between the largest number in one group and the smallest number in the *next non-empty* group.
- Repeat this process by moving across the range to find the largest such difference. This is your maximum gap.
Code Implementation
def maximum_gap(numbers):
if len(numbers) < 2:
return 0
minimum_value = min(numbers)
maximum_value = max(numbers)
# If min and max are the same, all elements are identical.
if minimum_value == maximum_value:
return 0
number_of_elements = len(numbers)
bucket_size = max(1, (maximum_value - minimum_value) // (number_of_elements - 1))
number_of_buckets = (maximum_value - minimum_value) // bucket_size + 1
buckets_minimum = [float('inf')] * number_of_buckets
buckets_maximum = [float('-inf')] * number_of_buckets
# Populate the buckets with min/max values
for number in numbers:
bucket_index = (number - minimum_value) // bucket_size
buckets_minimum[bucket_index] = min(buckets_minimum[bucket_index], number)
buckets_maximum[bucket_index] = max(buckets_maximum[bucket_index], number)
maximum_gap_value = 0
previous_maximum = buckets_maximum[0]
# Find the largest gap across buckets
for i in range(1, number_of_buckets):
if buckets_minimum[i] == float('inf'):
continue
maximum_gap_value = max(maximum_gap_value, buckets_minimum[i] - previous_maximum)
previous_maximum = buckets_maximum[i]
return maximum_gap_valueBig(O) Analysis
Time Complexity
O(n) – The algorithm first iterates through the input array of size n to determine the minimum and maximum values, which takes O(n) time. Then, it creates buckets and assigns each element to its corresponding bucket, taking O(n) time. Finally, it iterates through the buckets to find the maximum gap between the largest element of one non-empty bucket and the smallest element of the next non-empty bucket, which also takes O(n) time. Since these operations are performed sequentially and each takes O(n) time, the overall time complexity is O(n).
Space Complexity
O(N) – The algorithm divides the range into buckets, creating two arrays (or lists) of size N, where N is the number of elements in the input. These arrays, representing the minimum and maximum values for each bucket, constitute the auxiliary space. Therefore, the space complexity is directly proportional to the input size N, resulting in O(N) space complexity.
Edge Cases
| Case | How to Handle |
|---|---|
| Empty or null input array | Return 0 immediately as no gap can be calculated. |
| Array with fewer than 2 elements | Return 0 since a gap requires at least two elements. |
| Array with all identical elements | After sorting, the difference between consecutive elements will be 0, which is correctly handled. |
| Array with a wide range of integer values (Integer.MAX_VALUE and Integer.MIN_VALUE) | The sorting algorithm must handle this efficiently; using bucket sort requires careful size calculation. |
| Array containing duplicate elements | Duplicates do not affect the maximum gap calculation after sorting. |
| Array with negative numbers | The sorting algorithm must correctly handle negative numbers (should sort correctly). |
| Large input size, potential memory constraints | Bucket sort or radix sort variations may be preferred over comparison-based sorting to improve performance and potentially reduce memory footprint. |
| Integer overflow during calculations (e.g., difference between max and min values) | Use long data type to store differences to avoid potential integer overflow issues when calculating the gap between elements. |
Empty or null input array
How to Handle:
Return 0 immediately as no gap can be calculated.
Array with fewer than 2 elements
How to Handle:
Return 0 since a gap requires at least two elements.
Array with all identical elements
How to Handle:
After sorting, the difference between consecutive elements will be 0, which is correctly handled.
Array with a wide range of integer values (Integer.MAX_VALUE and Integer.MIN_VALUE)
How to Handle:
The sorting algorithm must handle this efficiently; using bucket sort requires careful size calculation.
Array containing duplicate elements
How to Handle:
Duplicates do not affect the maximum gap calculation after sorting.
Array with negative numbers
How to Handle:
The sorting algorithm must correctly handle negative numbers (should sort correctly).
Large input size, potential memory constraints
How to Handle:
Bucket sort or radix sort variations may be preferred over comparison-based sorting to improve performance and potentially reduce memory footprint.
Integer overflow during calculations (e.g., difference between max and min values)
How to Handle:
Use long data type to store differences to avoid potential integer overflow issues when calculating the gap between elements.
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