Count the Number of K-Big Indices

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Arrays

You are given a 0-indexed array nums of n integers and a positive integer k.

Let's define an index i of the array nums as K-big if the following conditions hold:

There exist at least k elements in nums that are strictly smaller than nums[i].
There exist at least k elements in nums that are strictly greater than nums[i].

Return the number of K-big indices in nums.

Example 1:

Input: nums = [2, 3, 6, 5, 2, 3], k = 2
Output: 2
Explanation: For index 2:
- There exist four elements that are strictly smaller than nums[2] = 6, which are nums[0], nums[1], nums[4], and nums[5].
- There exist zero elements that are strictly greater than nums[2] = 6.
Therefore, index 2 is not a 2-big index.

For index 3:
- There exist four elements that are strictly smaller than nums[3] = 5, which are nums[0], nums[1], nums[4], and nums[5].
- There exists one element that is strictly greater than nums[3] = 5, which is nums[2].
Therefore, index 3 is not a 2-big index.

For index 0:
- There exist zero elements that are strictly smaller than nums[0] = 2.
- There exist four elements that are strictly greater than nums[0] = 2, which are nums[1], nums[2], nums[3], and nums[5].
Therefore, index 0 is not a 2-big index.

The only two 2-big indices are 1 and 5 because:
- For index 1: nums[1] = 3 > 2 and nums[1] > 2. There are two elements greater than 3 and two elements smaller than 3.
- For index 5: nums[5] = 3 > 2 and nums[5] > 2. There are two elements greater than 3 and two elements smaller than 3.

Example 2:

Input: nums = [1, 1, 1, 1], k = 0
Output: 4
Explanation: For each index i in the array, there exist zero elements that are strictly smaller than nums[i] and zero elements that are strictly greater than nums[i].
Therefore, each index is a 0-big index.

Example 3:

Input: nums = [5, 6, 7, 8], k = 2
Output: 0
Explanation: There are no index that satisfies the conditions.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 1000
0 <= k <= 49

Try coding it in LeetCode

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 expected range of values within the input array? Can I assume they are all non-negative integers?
Can the input array be empty or null? If so, what should I return?
Could you please clarify what is expected as a return value if no K-Big indices exist in the array?
Are there any constraints on the value of 'k'? For example, is it always a positive integer, and will it always be smaller than the length of the input array?
If an index satisfies the K-Big condition multiple times (i.e., has multiple sub-arrays of size k where it is the largest), should it be counted multiple times or only once?

Brute Force Solution

Approach

We need to find specific spots in a lineup of numbers that satisfy a special condition involving other numbers around them. A brute force strategy means we will check every single spot, one by one, to see if it meets the requirement. We'll do this by comparing the number at that spot to all the numbers before and after it, checking if enough of them are bigger.

Here's how the algorithm would work step-by-step:

Look at the first number in the lineup.
Check how many numbers before it are bigger than a certain amount, we will call it K.
Check how many numbers after it are bigger than K.
If the number of bigger numbers before AND after it are both at least K, then we have found a match! Increase our count.
Now, move to the second number in the lineup.
Repeat the same checking process: counting how many numbers before and after are bigger than K.
Again, if both counts are at least K, increase our match count.
Continue this process, checking each number in the lineup one at a time, until we have checked every single number.
The final count of matches is the answer.

Code Implementation

def count_k_big_indices_brute_force(numbers, k_value):
    number_of_k_big_indices = 0
    list_length = len(numbers)

    for index in range(list_length):
        # Reset counters for each index
        number_bigger_before = 0
        number_bigger_after = 0

        # Count elements before current index
        for before_index in range(index):
            if numbers[before_index] > numbers[index]:
                number_bigger_before += 1

        # Count elements after current index
        for after_index in range(index + 1, list_length):
            if numbers[after_index] > numbers[index]:
                number_bigger_after += 1

        # Key condition: Check if index is K-big
        if number_bigger_before >= k_value and number_bigger_after >= k_value:
            # Increment the counter
            number_of_k_big_indices += 1

    return number_of_k_big_indices

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the input array. For each element, it then iterates through the elements before and after it to count how many are greater than the given value k. In the worst case, this inner loop iterates through nearly all the other n elements for each of the n elements in the outer loop. Therefore, the number of operations is proportional to n * n, resulting in a time complexity of O(n²).

Space Complexity

O(1) – The provided algorithm iterates through the input array, performing comparisons and counting values. It only uses a few integer variables to store the current count of k-big indices, the index of the current element being examined, and temporary counts for the number of larger elements before and after. The number of these variables remains constant regardless of the input size N. Thus, the space complexity is O(1).

Optimal Solution

Approach

The goal is to efficiently find positions in a list where the number of larger values to the left and right both meet a certain minimum count. Instead of checking each position separately, we can precompute some information to speed things up. This is achieved by preparing counts from both directions simultaneously and combining them.

Here's how the algorithm would work step-by-step:

First, go through the list from left to right. At each position, count how many numbers to the left are bigger than the number at that position. Store these counts.
Next, go through the list from right to left. At each position, count how many numbers to the right are bigger than the number at that position. Store these counts.
Now, for each position in the list, look up the two counts we precomputed: the number of larger values to the left and the number of larger values to the right.
If both of these counts are at least as big as the given minimum count, then we know that position is a 'K-Big Index'.
Keep track of how many positions are 'K-Big Indices'.
The final count is the answer.

Code Implementation

def count_k_big_indices(numbers, k_value):
    list_length = len(numbers)
    left_count = [0] * list_length
    right_count = [0] * list_length

    # Populate left counts
    for i in range(1, list_length):
        for j in range(i):
            if numbers[j] > numbers[i]:
                left_count[i] += 1

    # Populate right counts
    for i in range(list_length - 2, -1, -1):
        for j in range(i + 1, list_length):
            if numbers[j] > numbers[i]:
                right_count[i] += 1

    k_big_index_count = 0
    # Count indices where both left and right counts are >= k
    for i in range(list_length):
        if left_count[i] >= k_value and right_count[i] >= k_value:
            # Only increment if both conditions are met
            k_big_index_count += 1

    return k_big_index_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input array of size n twice: once from left to right and once from right to left, to compute counts of larger elements. Within each of these loops, for each element, it compares with all the preceding (or succeeding) elements. Thus, for each of the n elements, there's potentially another loop of size n in the worst case (specifically n-1 comparisons). This results in approximately n * n total operations, which simplifies to O(n²).

Space Complexity

O(N) – The algorithm uses two lists, `left_counts` and `right_counts`, to store the counts of larger elements to the left and right of each element in the input list. Both `left_counts` and `right_counts` have the same size as the input list, which is N. Therefore, the auxiliary space required is proportional to the size of the input list, N. The space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no indices to evaluate.
Array with only one element	Return 0 as a single element cannot satisfy the K-Big condition.
k = 0	Handle k=0 by returning the count of indices that have larger elements on both sides.
k is larger than the array size	If k exceeds the valid range for comparison, ensure the code handles it gracefully (e.g. return 0 or throw an exception with appropriate message).
Array contains all identical values	No index can be K-Big; return 0.
Large array size exceeding memory constraints	Optimize the space complexity to avoid memory overflow (e.g. use iterative approaches with constant space).
Integer overflow when summing elements on either side of the index	Use appropriate data types (e.g., long) to store the sum of elements to prevent overflow.
The sums of elements on the left and right are identical to K	The condition requires the sums to be strictly greater than K, so these indices should not be counted.

Empty input array

How to Handle:

Return 0 as there are no indices to evaluate.

Array with only one element

How to Handle:

Return 0 as a single element cannot satisfy the K-Big condition.

k = 0

How to Handle:

Handle k=0 by returning the count of indices that have larger elements on both sides.

k is larger than the array size

How to Handle:

If k exceeds the valid range for comparison, ensure the code handles it gracefully (e.g. return 0 or throw an exception with appropriate message).

Array contains all identical values

How to Handle:

No index can be K-Big; return 0.

Large array size exceeding memory constraints

How to Handle:

Optimize the space complexity to avoid memory overflow (e.g. use iterative approaches with constant space).

Integer overflow when summing elements on either side of the index

How to Handle:

Use appropriate data types (e.g., long) to store the sum of elements to prevent overflow.

The sums of elements on the left and right are identical to K

How to Handle:

The condition requires the sums to be strictly greater than K, so these indices should not be counted.