Find X Value of Array II

Hard

Rubrik

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Topics:

ArraysSliding Windows

You are given an array of positive integers nums and a positive integer k. You are also given a 2D array queries, where queries[i] = [index_i, value_i, start_i, x_i].

You are allowed to perform an operation once on nums, where you can remove any suffix from nums such that nums remains non-empty.

The x-value of nums for a given x is defined as the number of ways to perform this operation so that the product of the remaining elements leaves a remainder of x modulo k.

For each query in queries you need to determine the x-value of nums for x_i after performing the following actions:

Update nums[index_i] to value_i. Only this step persists for the rest of the queries.
Remove the prefix nums[0..(start_i - 1)] (where nums[0..(-1)] will be used to represent the empty prefix).

Return an array result of size queries.length where result[i] is the answer for the i^th query.

A prefix of an array is a subarray that starts from the beginning of the array and extends to any point within it.

A suffix of an array is a subarray that starts at any point within the array and extends to the end of the array.

Note that the prefix and suffix to be chosen for the operation can be empty.

Note that x-value has a different definition in this version.

Example 1:

Input: nums = [1,2,3,4,5], k = 3, queries = [[2,2,0,2],[3,3,3,0],[0,1,0,1]]

Output: [2,2,2]

Explanation:

For query 0, nums becomes [1, 2, 2, 4, 5], and the empty prefix must be removed. The possible operations are:
- Remove the suffix [2, 4, 5]. nums becomes [1, 2].
- Remove the empty suffix. nums becomes [1, 2, 2, 4, 5] with a product 80, which gives remainder 2 when divided by 3.
For query 1, nums becomes [1, 2, 2, 3, 5], and the prefix [1, 2, 2] must be removed. The possible operations are:
- Remove the empty suffix. nums becomes [3, 5].
- Remove the suffix [5]. nums becomes [3].
For query 2, nums becomes [1, 2, 2, 3, 5], and the empty prefix must be removed. The possible operations are:
- Remove the suffix [2, 2, 3, 5]. nums becomes [1].
- Remove the suffix [3, 5]. nums becomes [1, 2, 2].

Example 2:

Input: nums = [1,2,4,8,16,32], k = 4, queries = [[0,2,0,2],[0,2,0,1]]

Output: [1,0]

Explanation:

For query 0, nums becomes [2, 2, 4, 8, 16, 32]. The only possible operation is:
- Remove the suffix [2, 4, 8, 16, 32].
For query 1, nums becomes [2, 2, 4, 8, 16, 32]. There is no possible way to perform the operation.

Example 3:

Input: nums = [1,1,2,1,1], k = 2, queries = [[2,1,0,1]]

Output: [5]

Constraints:

1 <= nums[i] <= 10⁹
1 <= nums.length <= 10⁵
1 <= k <= 5
1 <= queries.length <= 2 * 10⁴
queries[i] == [index_i, value_i, start_i, x_i]
0 <= index_i <= nums.length - 1
1 <= value_i <= 10⁹
0 <= start_i <= nums.length - 1
0 <= x_i <= k - 1

Try coding it in LeetCode

Find X Value of Array II

Hard

Rubrik

3 views

Topics:

ArraysSliding Windows

You are given an array of positive integers nums and a positive integer k. You are also given a 2D array queries, where queries[i] = [index_i, value_i, start_i, x_i].

You are allowed to perform an operation once on nums, where you can remove any suffix from nums such that nums remains non-empty.

The x-value of nums for a given x is defined as the number of ways to perform this operation so that the product of the remaining elements leaves a remainder of x modulo k.

For each query in queries you need to determine the x-value of nums for x_i after performing the following actions:

Update nums[index_i] to value_i. Only this step persists for the rest of the queries.
Remove the prefix nums[0..(start_i - 1)] (where nums[0..(-1)] will be used to represent the empty prefix).

Return an array result of size queries.length where result[i] is the answer for the i^th query.

A prefix of an array is a subarray that starts from the beginning of the array and extends to any point within it.

A suffix of an array is a subarray that starts at any point within the array and extends to the end of the array.

Note that the prefix and suffix to be chosen for the operation can be empty.

Note that x-value has a different definition in this version.

Example 1:

Input: nums = [1,2,3,4,5], k = 3, queries = [[2,2,0,2],[3,3,3,0],[0,1,0,1]]

Output: [2,2,2]

Explanation:

For query 0, nums becomes [1, 2, 2, 4, 5], and the empty prefix must be removed. The possible operations are:
- Remove the suffix [2, 4, 5]. nums becomes [1, 2].
- Remove the empty suffix. nums becomes [1, 2, 2, 4, 5] with a product 80, which gives remainder 2 when divided by 3.
For query 1, nums becomes [1, 2, 2, 3, 5], and the prefix [1, 2, 2] must be removed. The possible operations are:
- Remove the empty suffix. nums becomes [3, 5].
- Remove the suffix [5]. nums becomes [3].
For query 2, nums becomes [1, 2, 2, 3, 5], and the empty prefix must be removed. The possible operations are:
- Remove the suffix [2, 2, 3, 5]. nums becomes [1].
- Remove the suffix [3, 5]. nums becomes [1, 2, 2].

Example 2:

Input: nums = [1,2,4,8,16,32], k = 4, queries = [[0,2,0,2],[0,2,0,1]]

Output: [1,0]

Explanation:

For query 0, nums becomes [2, 2, 4, 8, 16, 32]. The only possible operation is:
- Remove the suffix [2, 4, 8, 16, 32].
For query 1, nums becomes [2, 2, 4, 8, 16, 32]. There is no possible way to perform the operation.

Example 3:

Input: nums = [1,1,2,1,1], k = 2, queries = [[2,1,0,1]]

Output: [5]

Constraints:

1 <= nums[i] <= 10⁹
1 <= nums.length <= 10⁵
1 <= k <= 5
1 <= queries.length <= 2 * 10⁴
queries[i] == [index_i, value_i, start_i, x_i]
0 <= index_i <= nums.length - 1
1 <= value_i <= 10⁹
0 <= start_i <= nums.length - 1
0 <= x_i <= k - 1

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 range of integer values that the `nums` array can contain? Can they be negative, zero, or only positive?
What should I return if the input array `nums` is empty?
Are the input numbers integers, or could they be floating-point numbers?
Is it guaranteed that a valid 'x' always exists such that the sum of (nums[i] - x) equals zero?
What is the maximum size of the `nums` array I should expect?

Brute Force Solution

Approach

Imagine you need to find a special value (X) that makes two parts of a group of numbers as balanced as possible. The brute force method checks every single possible value for X within the group, one at a time. For each potential X, we compare how balanced the two parts are.

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

Pick the very first number in the group as a possible value for X.
Divide the group into two sections based on this X value: one section has all the numbers smaller than X, and the other has all the numbers bigger than X.
Determine how balanced the two sections are. This means measuring the difference in some way between the two groups.
Write down the difference between the two parts that we calculated and remember the X we used.
Now, pick the next number in the group as a new possible value for X.
Repeat the dividing and balancing steps as before, using this new X value.
Write down the new difference and the X value.
Keep doing this until you have tried every single number in the group as a possible value for X.
Finally, compare all the differences you wrote down. The smallest difference shows the most balanced division, and the X value that created it is the answer.

Code Implementation

def find_x_value_brute_force(number_group):
    # Initialize the minimum difference to infinity
    # and the best x value to None
    minimum_difference = float('inf')
    best_x_value = None

    # Iterate through each number in the group
    for possible_x_value in number_group:

        smaller_than_x = []
        larger_than_x = []

        # Divide the numbers into two groups based on possible_x_value
        for number in number_group:
            if number < possible_x_value:
                smaller_than_x.append(number)
            elif number > possible_x_value:
                larger_than_x.append(number)

        # Calculate the difference between the sums of the two groups
        difference = abs(sum(smaller_than_x) - sum(larger_than_x))

        # Check if this difference is smaller than the current minimum
        if difference < minimum_difference:
            # Update the minimum difference and best x value
            minimum_difference = difference
            best_x_value = possible_x_value

    return best_x_value

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the array, considering each as a potential X value. For each potential X, it iterates through the array again to divide the numbers into two sections (smaller than X and greater than X), calculating the difference or imbalance between the two sections. This inner iteration takes O(n) time. Because this inner loop is performed for each of the n elements in the outer loop, the overall time complexity is approximately n * n operations which simplifies to O(n²).

Space Complexity

O(1) – The algorithm iterates through the input array, considering each element as a potential 'X' value. It keeps track of the best 'X' seen so far and the corresponding minimum difference. Since it only stores a few variables like the current 'X', the minimum difference, and the best 'X' value, the auxiliary space used is constant and does not depend on the input size N. Therefore, the space complexity is O(1).

Optimal Solution

Approach

This problem asks us to find a special value that makes the sums of certain parts of a collection of numbers as close as possible. The optimal approach uses a technique of narrowing down the possibilities to efficiently find that special value.

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

First, understand the range where the special value might exist. It's likely somewhere between the smallest and largest numbers in the collection.
Instead of checking every single number, start by picking a number in the middle of that range and test it.
Calculate the sums of the two parts of the collection based on that special value.
Compare the two sums to see which is bigger.
Based on whether the first or second sum is larger, adjust your guess for the special value. If the first sum is too big, pick a smaller special value next time. If the second sum is too big, pick a bigger special value next time.
Repeat this process of picking a number, calculating sums, comparing them, and adjusting the number. Each time, cut the range of possible values in half.
Continue this process until you find a special value that makes the sums of the two parts as close as possible, within a very small margin of error.
This method efficiently converges on the right special value without checking every single possibility.

Code Implementation

def find_x_value(numbers):
    left_bound = min(numbers)
    right_bound = max(numbers)

    while left_bound <= right_bound:
        potential_x = (left_bound + right_bound) // 2
        
        sum_less_than_x = 0
        sum_greater_than_x = 0

        for number in numbers:
            if number <= potential_x:
                sum_less_than_x += number
            else:
                sum_greater_than_x += number
        
        # If sums are close enough, return the x value
        if abs(sum_less_than_x - sum_greater_than_x) <= 1:
            return potential_x
        elif sum_less_than_x > sum_greater_than_x:
            # Adjust the range based on sum comparison
            right_bound = potential_x - 1

        else:
            # Adjust the range based on sum comparison
            left_bound = potential_x + 1
    
    return -1 # Value not found

Big(O) Analysis

Time Complexity

O(log n) – The algorithm employs a binary search approach to find the optimal special value within a range determined by the input array's minimum and maximum values. Each iteration of the binary search halves the search space. The binary search continues until the optimal value is found or the search space is exhausted, leading to a logarithmic relationship with the size of the possible values. Therefore, the time complexity is O(log n).

Space Complexity

O(1) – The provided algorithm description performs binary search and calculates sums within the search loop. It doesn't mention storing intermediate sums or creating additional data structures that scale with the input array's size (N). Therefore, only a fixed number of variables are used for comparisons and calculations, resulting in constant auxiliary space.

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 as the sum will be 0 and 0/0 will be used to compute x; this satisfies the equation.
Array with a single element	Return the single element as x, satisfying the equation (num[0] - x) = 0, therefore x = num[0].
Array with all elements being zero	Return 0 as x, satisfying the equation (0 - 0) + (0 - 0) + ... = 0.
Array with large positive and negative numbers	The sum of the elements might overflow if using integers, consider using long to prevent overflow.
Array with identical values	Return the value itself as x, satisfying the equation (num[i] - x) = 0 for all i.
Very large array size	The solution's O(n) time complexity should scale linearly with the size of the input.
Input array contains floating-point numbers	The solution works directly with floating-point numbers; handle potential floating point precision issues carefully, possibly round the result.
Array contains both positive and negative numbers	The solution works correctly as it sums all the elements, including negative ones, before calculating x.