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Maximum Value at a Given Index in a Bounded Array

Medium

Amazon

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Topics:

Binary SearchGreedy Algorithms

You are given three positive integers: n, index, and maxSum. You want to construct an array nums (0-indexed) that satisfies the following conditions:

nums.length == n
nums[i] is a positive integer where 0 <= i < n.
abs(nums[i] - nums[i+1]) <= 1 where 0 <= i < n-1.
The sum of all the elements of nums does not exceed maxSum.
nums[index] is maximized.

Return nums[index] of the constructed array.

Note that abs(x) equals x if x >= 0, and -x otherwise.

Example 1:

Input: n = 4, index = 2,  maxSum = 6
Output: 2
Explanation: nums = [1,2,2,1] is one array that satisfies all the conditions.
There are no arrays that satisfy all the conditions and have nums[2] == 3, so 2 is the maximum nums[2].

Example 2:

Input: n = 6, index = 1,  maxSum = 10
Output: 3

Constraints:

1 <= n <= maxSum <= 10⁹
0 <= index < n

Maximum Value at a Given Index in a Bounded Array

Medium

Amazon

1 view

Topics:

Binary SearchGreedy Algorithms

You are given three positive integers: n, index, and maxSum. You want to construct an array nums (0-indexed) that satisfies the following conditions:

nums.length == n
nums[i] is a positive integer where 0 <= i < n.
abs(nums[i] - nums[i+1]) <= 1 where 0 <= i < n-1.
The sum of all the elements of nums does not exceed maxSum.
nums[index] is maximized.

Return nums[index] of the constructed array.

Note that abs(x) equals x if x >= 0, and -x otherwise.

Example 1:

Input: n = 4, index = 2,  maxSum = 6
Output: 2
Explanation: nums = [1,2,2,1] is one array that satisfies all the conditions.
There are no arrays that satisfy all the conditions and have nums[2] == 3, so 2 is the maximum nums[2].

Example 2:

Input: n = 6, index = 1,  maxSum = 10
Output: 3

Constraints:

1 <= n <= maxSum <= 10⁹
0 <= index < n

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 are the constraints on `n`, `index`, and `maxSum`? Specifically, what are the maximum possible values for each?
Is `index` guaranteed to be within the bounds of the array (0 <= index < n)?
Are all inputs positive integers, and is `maxSum` guaranteed to be greater than or equal to `n`?
If it's impossible to construct an array that satisfies the conditions, what value should I return?
Could you provide a few examples of input and expected output to ensure I understand the problem correctly?

Brute Force Solution

Approach

The brute force strategy here is about trying every possible value at a specific spot in our collection. We'll check if each value, one at a time, works with the given constraints for the entire collection. If a value works, we found a solution; if it doesn't, we move on and try the next value.

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

Start by guessing a potential maximum value for the special spot.
Then, construct the rest of the collection by filling the spots to the left and right of the special spot, making sure the numbers decrease as we move away from it.
Check if the total of all the numbers in our constructed collection meets the required total.
If the total is too small, try a larger value for the special spot and repeat the process.
If the total is too big, try a smaller value for the special spot and repeat the process.
Keep adjusting the value at the special spot until the total of the collection matches the requirement, while also respecting the decreasing pattern away from the special spot.
If no value works under these constraints, there is no suitable solution using the brute force approach.

Code Implementation

def find_maximum_value_brute_force(number_of_elements, index, maximum_sum):
    for potential_maximum_value in range(1, maximum_sum + 1):
        array_sum = potential_maximum_value
        left_value = potential_maximum_value - 1
        right_value = potential_maximum_value - 1

        # Build left side of the array
        for left_index in range(index - 1, -1, -1):
            if left_value > 0:
                array_sum += left_value
                left_value -= 1
            else:
                array_sum += 1

        # Build right side of the array
        for right_index in range(index + 1, number_of_elements):
            if right_value > 0:
                array_sum += right_value
                right_value -= 1
            else:
                # Ensure we contribute to the final sum
                array_sum += 1

        # We found the maximum possible value
        if array_sum == maximum_sum:
            return potential_maximum_value
        elif array_sum > maximum_sum:
            break

    return -1

Big(O) Analysis

Time Complexity

O(n*target) – The brute force approach iterates through each possible value (up to target) for the index. For each of these possible values, the algorithm constructs the array by decrementing values to the left and right of the index. Constructing the array involves at most n operations to fill the array from the chosen index outwards. Thus, for each possible value at the index, it takes O(n) time. Since we iterate up to the target value, the overall time complexity is O(n * target).

Space Complexity

O(1) – The brute force approach described constructs the collection by filling spots to the left and right of the special spot to check if a given maximum value is feasible. However, it doesn't explicitly mention storing this entire collection in memory. The algorithm seems to only calculate the sum of the numbers and compare it with the required total. Therefore, it only uses a few constant space variables for the current guess, the running total, and potentially some loop counters, all of which are independent of the input size n. Thus the auxiliary space complexity is O(1).

Optimal Solution

Approach

The most efficient way to solve this puzzle involves using a technique similar to guessing a number in a number guessing game. Instead of trying every possible value one by one, we start with a reasonable guess and adjust it up or down based on the outcome to quickly converge to the answer.

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

Start by guessing a possible value for the number at the specific location.
Imagine building a shape around that location, where the numbers get smaller as you move away from it. The shape will look like a pyramid or a plateau.
Calculate the total sum of all the numbers in this shape based on your guessed value and the location.
Compare the total sum with the given limit. If it's too high, you need to lower your guess; if it's too low, you need to increase your guess.
Repeat this process, adjusting your guess each time, until you find the value that creates a shape where the sum is as close as possible to the limit without exceeding it.
The optimal value is the one that lets you build the largest possible shape without breaking the total sum rule.

Code Implementation

def maximum_value_at_a_given_index(number_of_elements, index_value, maximum_sum):
    left_bound = 1
    right_bound = maximum_sum

    while left_bound <= right_bound:
        center_value = (left_bound + right_bound) // 2
        sum_of_array = calculate_array_sum(number_of_elements, index_value, center_value)

        # Adjust search space based on current sum
        if sum_of_array <= maximum_sum:
            result = center_value
            left_bound = center_value + 1
        else:
            right_bound = center_value - 1

    return result

def calculate_array_sum(number_of_elements, index_value, peak_value):
    left_count = index_value
    right_count = number_of_elements - index_value - 1

    total_sum = 0

    # Calculate sum of left side
    if left_count >= peak_value - 1:
        total_sum += (peak_value - 1 + 1) * (peak_value - 1) // 2 + (left_count - (peak_value - 1))

    else:
        total_sum += (peak_value - 1 + peak_value - 1 - left_count) * (left_count + 1) // 2

    # Calculate sum of right side
    if right_count >= peak_value - 1:
        total_sum += (peak_value - 1 + 1) * (peak_value - 1) // 2 + (right_count - (peak_value - 1))

    else:
        total_sum += (peak_value - 1 + peak_value - 1 - right_count) * (right_count + 1) // 2

    # Adding peak_value since it was ommitted in both sides
    total_sum += peak_value
    return total_sum - peak_value

Big(O) Analysis

Time Complexity

O(log(maxSum)) – The solution employs a binary search approach to determine the maximum possible value at the specified index. The search space ranges from 1 to maxSum. Within each iteration of the binary search, the algorithm calculates the sum of the bounded array based on the current mid-value, which takes constant time O(1) since it uses a formula. The binary search continues until the optimal value is found. Therefore, the time complexity is determined by the binary search iterations, which is logarithmic with respect to the maximum possible sum, maxSum, resulting in O(log(maxSum)).

Space Complexity

O(1) – The algorithm described primarily involves iterative guessing and calculation of sums. It does not mention the use of auxiliary data structures like arrays, lists, or hash maps. The variables used for guessing and calculating sums consume a constant amount of space, irrespective of the input size n (where n could relate to the given limit or the index). Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
n = 1, index = 0, maxSum = 1	Should return 1 as the single element can be 1.
index = 0, maxSum is very small compared to n	The peak value will be small, and we need to correctly calculate the distribution.
index = n-1, maxSum is very small compared to n	The peak will be on the right edge and we need to adjust for that.
maxSum is very large compared to n	The peak will be the dominant factor, and the sides will be close to 1.
n is extremely large, leading to potential integer overflow during calculations of sums of arithmetic series.	Use long data types for intermediate calculations to prevent integer overflow.
index is close to either 0 or n-1 and maxSum is also small.	The sum calculation needs to correctly handle the case where the sides of the 'V' shape don't fully extend to 1.
maxSum is less than n	No valid answer can exist as the minimum sum is n if all elements are 1 and you should return a value indicating no solution.
index is in the middle, maxSum is relatively small	The height is likely to not reach a high value and must be handled accurately using binary search, and the total sum should never exceed maxSum

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 are the constraints on `n`, `index`, and `maxSum`? Specifically, what are the maximum possible values for each?
Is `index` guaranteed to be within the bounds of the array (0 <= index < n)?
Are all inputs positive integers, and is `maxSum` guaranteed to be greater than or equal to `n`?
If it's impossible to construct an array that satisfies the conditions, what value should I return?
Could you provide a few examples of input and expected output to ensure I understand the problem correctly?

Brute Force Solution

Approach

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

Start by guessing a potential maximum value for the special spot.
Then, construct the rest of the collection by filling the spots to the left and right of the special spot, making sure the numbers decrease as we move away from it.
Check if the total of all the numbers in our constructed collection meets the required total.
If the total is too small, try a larger value for the special spot and repeat the process.
If the total is too big, try a smaller value for the special spot and repeat the process.
Keep adjusting the value at the special spot until the total of the collection matches the requirement, while also respecting the decreasing pattern away from the special spot.
If no value works under these constraints, there is no suitable solution using the brute force approach.

Code Implementation

def find_maximum_value_brute_force(number_of_elements, index, maximum_sum):
    for potential_maximum_value in range(1, maximum_sum + 1):
        array_sum = potential_maximum_value
        left_value = potential_maximum_value - 1
        right_value = potential_maximum_value - 1

        # Build left side of the array
        for left_index in range(index - 1, -1, -1):
            if left_value > 0:
                array_sum += left_value
                left_value -= 1
            else:
                array_sum += 1

        # Build right side of the array
        for right_index in range(index + 1, number_of_elements):
            if right_value > 0:
                array_sum += right_value
                right_value -= 1
            else:
                # Ensure we contribute to the final sum
                array_sum += 1

        # We found the maximum possible value
        if array_sum == maximum_sum:
            return potential_maximum_value
        elif array_sum > maximum_sum:
            break

    return -1

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by guessing a possible value for the number at the specific location.
Imagine building a shape around that location, where the numbers get smaller as you move away from it. The shape will look like a pyramid or a plateau.
Calculate the total sum of all the numbers in this shape based on your guessed value and the location.
Compare the total sum with the given limit. If it's too high, you need to lower your guess; if it's too low, you need to increase your guess.
Repeat this process, adjusting your guess each time, until you find the value that creates a shape where the sum is as close as possible to the limit without exceeding it.
The optimal value is the one that lets you build the largest possible shape without breaking the total sum rule.

Code Implementation

def maximum_value_at_a_given_index(number_of_elements, index_value, maximum_sum):
    left_bound = 1
    right_bound = maximum_sum

    while left_bound <= right_bound:
        center_value = (left_bound + right_bound) // 2
        sum_of_array = calculate_array_sum(number_of_elements, index_value, center_value)

        # Adjust search space based on current sum
        if sum_of_array <= maximum_sum:
            result = center_value
            left_bound = center_value + 1
        else:
            right_bound = center_value - 1

    return result

def calculate_array_sum(number_of_elements, index_value, peak_value):
    left_count = index_value
    right_count = number_of_elements - index_value - 1

    total_sum = 0

    # Calculate sum of left side
    if left_count >= peak_value - 1:
        total_sum += (peak_value - 1 + 1) * (peak_value - 1) // 2 + (left_count - (peak_value - 1))

    else:
        total_sum += (peak_value - 1 + peak_value - 1 - left_count) * (left_count + 1) // 2

    # Calculate sum of right side
    if right_count >= peak_value - 1:
        total_sum += (peak_value - 1 + 1) * (peak_value - 1) // 2 + (right_count - (peak_value - 1))

    else:
        total_sum += (peak_value - 1 + peak_value - 1 - right_count) * (right_count + 1) // 2

    # Adding peak_value since it was ommitted in both sides
    total_sum += peak_value
    return total_sum - peak_value

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 1, index = 0, maxSum = 1	Should return 1 as the single element can be 1.
index = 0, maxSum is very small compared to n	The peak value will be small, and we need to correctly calculate the distribution.
index = n-1, maxSum is very small compared to n	The peak will be on the right edge and we need to adjust for that.
maxSum is very large compared to n	The peak will be the dominant factor, and the sides will be close to 1.
n is extremely large, leading to potential integer overflow during calculations of sums of arithmetic series.	Use long data types for intermediate calculations to prevent integer overflow.
index is close to either 0 or n-1 and maxSum is also small.	The sum calculation needs to correctly handle the case where the sides of the 'V' shape don't fully extend to 1.
maxSum is less than n	No valid answer can exist as the minimum sum is n if all elements are 1 and you should return a value indicating no solution.
index is in the middle, maxSum is relatively small	The height is likely to not reach a high value and must be handled accurately using binary search, and the total sum should never exceed maxSum