Problem Description

Given an array of distinct integers nums, you need to find the largest subarray of length k.

A subarray is a contiguous sequence of elements from the array. For example, if nums = [1,2,3,4], then [2,3] is a subarray but [1,3] is not (elements are not contiguous).

The comparison rule for determining which array is "larger" works as follows:

• Compare the arrays element by element from left to right
• At the first position i where the elements differ (i.e., A[i] != B[i]), the array with the larger element at that position is considered larger
• For example:
  • [1,3,2,4] > [1,2,2,4] because at index 1, we have 3 > 2
  • [1,4,4,4] < [2,1,1,1] because at index 0, we have 1 < 2

Your task is to find and return the subarray of length k that is the largest among all possible subarrays of length k in the given array.

The solution leverages the fact that all integers are distinct. Since we want the lexicographically largest subarray, we need to find the subarray that starts with the maximum possible value. We search for the maximum element within the range where a valid subarray of length k can start (positions 0 through n-k), then return the subarray starting from that position.

Intuition

Since we need the "largest" subarray based on lexicographic comparison, the key insight is that the subarray starting with the biggest possible first element will always be the largest.

Think about it this way: when comparing two subarrays, we look at elements from left to right and stop at the first difference. If one subarray starts with element x and another starts with element y where x > y, then the first subarray is automatically larger regardless of what comes after.

Because all integers in the array are distinct (no duplicates), we know that:

1. Each element appears exactly once
2. There's a unique maximum element in any range

So our strategy becomes simple: find the largest element that can be the starting point of a valid subarray of length k.

Where can a subarray of length k start? It can start at any index from 0 to n-k (inclusive), because we need k elements from the starting position. If we start any later than index n-k, we won't have enough elements to form a subarray of length k.

Therefore, we just need to:

1. Look at all possible starting positions: indices 0 through n-k
2. Find which position has the maximum element
3. Return the subarray of length k starting from that position

This greedy approach works because once we maximize the first element, we've guaranteed the lexicographically largest subarray possible.

Learn more about Greedy patterns.

Solution Approach

The implementation is straightforward once we understand the intuition:

1. Find the valid range for starting positions: We can only start a subarray of length k from indices 0 to n-k (inclusive). This is calculated as len(nums) - k + 1, which gives us the number of valid starting positions.

2. Find the maximum element in the valid range: We use Python's max() function to find the largest element among all possible starting positions:

   max(nums[: len(nums) - k + 1])

   This slices the array from index 0 to n-k and finds the maximum value.

3. Get the index of that maximum element: We use the index() method to find where this maximum element is located:

   i = nums.index(max(nums[: len(nums) - k + 1]))

   Since all elements are distinct, there's exactly one occurrence of this maximum value.

4. Return the subarray: Once we have the starting index i, we simply slice the array from i to i+k:

   return nums[i : i + k]

The entire solution is accomplished in just two lines:

• First line finds the optimal starting position
• Second line extracts and returns the subarray

Time Complexity: O(n) where n is the length of the array, as we need to scan through the valid range to find the maximum and then find its index.

Space Complexity: O(k) for the output array (or O(1) if we don't count the output).

Example Walkthrough

Let's walk through an example with nums = [3, 5, 2, 8, 6, 1, 9, 4] and k = 3.

Step 1: Identify valid starting positions

• Array length n = 8, we need subarrays of length k = 3
• Valid starting positions are from index 0 to n-k = 8-3 = 5
• So we can start at indices: 0, 1, 2, 3, 4, 5

Step 2: Find the maximum element in the valid range

• Look at elements at valid starting positions: nums[0:6] = [3, 5, 2, 8, 6, 1]
• The maximum element is 8 at index 3

Step 3: Extract the subarray

• Start from index 3 where we found 8
• Take k = 3 elements: nums[3:6] = [8, 6, 1]

Why this is the largest subarray: Let's compare with other possible subarrays of length 3:

• Starting at index 0: [3, 5, 2] → First element is 3
• Starting at index 1: [5, 2, 8] → First element is 5
• Starting at index 2: [2, 8, 6] → First element is 2
• Starting at index 3: [8, 6, 1] → First element is 8 ✓
• Starting at index 4: [6, 1, 9] → First element is 6
• Starting at index 5: [1, 9, 4] → First element is 1

Since [8, 6, 1] starts with 8, which is the largest possible first element among all valid subarrays, it's lexicographically the largest. When comparing any other subarray with [8, 6, 1], the comparison stops at the first position where 8 beats any other starting element.

Code execution:

i = nums.index(max(nums[:6]))  # i = 3 (index of 8)
return nums[3:6]                # returns [8, 6, 1]

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the array.

The time complexity breaks down as follows:

• nums[: len(nums) - k + 1] creates a slice, which takes O(n - k + 1) time, simplifying to O(n)
• max() function iterates through the slice to find the maximum element, taking O(n - k + 1) time, which is O(n)
• nums.index() searches for the index of the maximum element in the original array, taking O(n) time in the worst case
• nums[i : i + k] creates the final subarray slice, which takes O(k) time

Overall, the dominant operations are O(n), resulting in a total time complexity of O(n).

Space Complexity: O(k) for the output array, or O(1) if we ignore the space consumption of the answer.

The space complexity analysis:

• nums[: len(nums) - k + 1] creates a temporary slice that uses O(n - k + 1) space, which is O(n) in the worst case
• The output array nums[i : i + k] requires O(k) space
• Other variables like i use constant space O(1)

If we count all space used including temporary slices, the space complexity is O(n). However, following the reference answer's convention of ignoring the space consumption of the answer itself, and considering that Python's slicing for the max() operation could be optimized by the interpreter to avoid creating a full copy, the auxiliary space complexity is O(1).

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

Common Pitfalls

1. Off-by-One Error in Valid Range Calculation

A common mistake is incorrectly calculating the valid range for starting positions. Developers might write:

• max(nums[:len(nums) - k]) (missing the +1)
• max(nums[:len(nums) - k - 1]) (subtracting an extra 1)

Why this happens: Confusion about whether we need inclusive or exclusive bounds when calculating valid starting positions.

Correct approach: The valid starting positions are from index 0 to n-k (inclusive). Since Python slicing is exclusive of the end index, we need nums[:len(nums) - k + 1].

Example to illustrate:

nums = [1, 2, 3, 4, 5], k = 3
# Valid starting indices: 0, 1, 2 (can form [1,2,3], [2,3,4], [3,4,5])
# len(nums) - k = 5 - 3 = 2
# We need to check indices 0, 1, 2, so slice should be nums[:3] = nums[:len(nums) - k + 1]

2. Not Handling Edge Cases

The code assumes valid input but doesn't explicitly handle edge cases:

• What if k > len(nums)? The code would look for max in an empty slice
• What if k == 0? Should return an empty array
• What if the array itself is empty?

Solution with edge case handling:

def largestSubarray(self, nums: List[int], k: int) -> List[int]:
    # Handle edge cases
    if not nums or k == 0:
        return []
    if k > len(nums):
        return []  # or raise ValueError("k cannot be larger than array length")
  
    # Original solution
    max_start_index = nums.index(max(nums[:len(nums) - k + 1]))
    return nums[max_start_index:max_start_index + k]

3. Assuming Multiple Occurrences of Maximum

While the problem states integers are distinct, developers might forget this constraint and worry about multiple occurrences of the maximum value. The index() method returns the first occurrence, which is perfect here since there's only one.

If the constraint didn't exist, we'd need to handle ties by finding all occurrences and comparing subsequent elements:

# This would be needed if elements weren't distinct
def find_best_start(nums, k):
    valid_range = nums[:len(nums) - k + 1]
    max_val = max(valid_range)
    # Find all positions with max value
    positions = [i for i, val in enumerate(valid_range) if val == max_val]
    if len(positions) == 1:
        return positions[0]
    # Compare subarrays starting at each position
    # (Additional logic needed here)

4. Performance Consideration for Large Arrays

The current solution makes two passes through the valid range:

1. max() to find the maximum value
2. index() to find its position

More efficient single-pass solution:

def largestSubarray(self, nums: List[int], k: int) -> List[int]:
    max_val = float('-inf')
    max_idx = 0
  
    # Single pass to find both max value and its index
    for i in range(len(nums) - k + 1):
        if nums[i] > max_val:
            max_val = nums[i]
            max_idx = i
  
    return nums[max_idx:max_idx + k]

This reduces the constant factor in the O(n) time complexity, though both approaches are O(n).