2229. Check if an Array Is Consecutive

Problem Description

The given problem presents us with a challenge to determine whether an array of integers, named nums, is consecutive or not. An array is considered consecutive if it includes all the numbers from x to x + n - 1 without any gaps or repetitions, where x is the smallest number in the array, and n is the total count of numbers in the array. To reiterate, our task is to analyze the integer array and return true if the array meets this consecutive criteria or false otherwise.

Intuition

To intuitively solve this problem, we can break down the definition of a consecutive array into two main conditions:

1. The array must not contain any duplicate values, which is essential for ensuring that the array includes a complete range of numbers without repetition.
2. The difference between the maximum and minimum number in the array should be exactly n - 1, where n is the size of the array. This condition guarantees that the range spans enough numbers to fill the array without gaps.

So, the straightforward approach is to first identify the smallest and largest numbers in the array. Once these are identified, we can check whether all array elements are unique and whether the difference between the maximum and minimum values equals n - 1. If both conditions are satisfied, we can confidently determine that the array is consecutive.

Here's a step by step reasoning of the implementation:

• Find the minimum value mi in the array nums using the min() function. This represents x.
• Find the maximum value mx in the array nums using the max() function. This represents x + n - 1.
• Calculate the array size n using the len() function.
• Check if all elements in the array are unique by converting the array into a set and comparing its length to n.
• Finally, check if the maximum value equals the minimum value plus n minus 1, i.e., mx == mi + n - 1.
• If both checks pass, return true. Otherwise, return false.

By checking each of these conditions with simple operations, we can solve the problem in an efficient and understandable manner.

Solution Approach

The solution is implemented using simple Python constructs and relies on the unique properties of sets in Python, and mathematical reasoning. Here's the breakdown of the steps involved in checking if the list of integers is consecutive:

1. Minimum and Maximum Values: We begin by using Python's built-in min() and max() functions to find the smallest number mi and the largest number mx in the nums array. These functions iterate through the list and deliver the minimum and maximum values efficiently.

2. Unique Elements Check: We then convert the list nums into a set. Sets in Python are unordered collections of unique elements. By comparing the length of this set with the length of the original list (using len(set(nums)) == n), we determine if all elements in the array are unique. If there are any duplicates, the set's length will be smaller than that of the list because sets automatically remove duplicate items.

3. Consecutive Numbers Check: Finally, we check if the numbers are consecutive by verifying that the maximum element mx equals the minimum element mi plus n - 1. This check ensures that the numbers span a range which corresponds to the size of the list. If the length of the set is equal to the length of the list and the maximum value is as expected, the list is consecutive.

The solution code combines these three checks into one line:

1return len(set(nums)) == n and mx == mi + n - 1

This line of code first asserts that each number in the array is unique (since the size of the set should be equal to the length of the array) and then ensures that the range (mx - mi) is exactly one less than the length of the array (n - 1). If both conditions are satisfied, this indicates that the array contains exactly n consecutive numbers starting at the minimum value found, thus making it a consecutive array, and the function returns True. Otherwise, the function returns False.

In terms of efficiency, this is a very concise approach since it involves a single pass over the array to check for all the required conditions, making it an O(n) time complexity algorithm, with O(n) space complexity due to set creation where n is the number of elements in the nums array.

Example Walkthrough

To illustrate the solution approach, consider a small example where the nums array is [4, 5, 6, 7].

1. Finding Minimum and Maximum Values: First, the smallest number mi is found using min(nums), which is 4 in this case. The largest number mx is found using max(nums), which is 7 in our example.

2. Unique Elements Check: We convert the nums array into a set {4, 5, 6, 7}. The original list's length n is 4 and the set's length is also 4. As the lengths are equal, we can conclude that all elements in the array are unique.

3. Consecutive Numbers Check: We now verify if the numbers are consecutive. We do this by checking if mx = mi + n - 1. In our example, 7 = 4 + 4 - 1 simplifies to 7 = 7, which is true.

The array [4, 5, 6, 7] passed the unique elements check and the consecutive numbers check. Thus, by implementing the solution approach:

1mi = min(nums)
2mx = max(nums)
3n = len(nums)
4return len(set(nums)) == n and mx == mi + n - 1

The conditions are satisfied and the function would return True, confirming that the array is consecutive.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n) where n is the length of the nums list. This is because the min() and max() functions will each take O(n) to find the minimum and maximum elements in the list, and len() takes O(1) time. Furthermore, converting the list to a set also takes O(n) time in average case to eliminate duplicates and check if all the elements are unique.

The space complexity of the code is O(n) since we are creating a new set from the list of numbers, which in the worst case will contain a unique value for every element in the list if there are no duplicates.

Learn more about how to find time and space complexity quickly using problem constraints.