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Get Maximum in Generated Array

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Arrays

You are given an integer n. A 0-indexed integer array nums of length n + 1 is generated in the following way:

nums[0] = 0
nums[1] = 1
nums[2 * i] = nums[i] when 2 <= 2 * i <= n
nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n

Return the maximum integer in the array nums.

Example 1:

Input: n = 7
Output: 3
Explanation: According to the given rules:
  nums[0] = 0
  nums[1] = 1
  nums[(1 * 2) = 2] = nums[1] = 1
  nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2
  nums[(2 * 2) = 4] = nums[2] = 1
  nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3
  nums[(3 * 2) = 6] = nums[3] = 2
  nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3
Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is max(0,1,1,2,1,3,2,3) = 3.

Example 2:

Input: n = 2
Output: 1
Explanation: According to the given rules, nums = [0,1,1]. The maximum is max(0,1,1) = 1.

Example 3:

Input: n = 3
Output: 2
Explanation: According to the given rules, nums = [0,1,1,2]. The maximum is max(0,1,1,2) = 2.

Constraints:

0 <= n <= 100

Try coding it in LeetCode

Get Maximum in Generated Array

Easy

Asked by:

10 views

Topics:

Arrays

You are given an integer n. A 0-indexed integer array nums of length n + 1 is generated in the following way:

nums[0] = 0
nums[1] = 1
nums[2 * i] = nums[i] when 2 <= 2 * i <= n
nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n

Return the maximum integer in the array nums.

Example 1:

Input: n = 7
Output: 3
Explanation: According to the given rules:
  nums[0] = 0
  nums[1] = 1
  nums[(1 * 2) = 2] = nums[1] = 1
  nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2
  nums[(2 * 2) = 4] = nums[2] = 1
  nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3
  nums[(3 * 2) = 6] = nums[3] = 2
  nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3
Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is max(0,1,1,2,1,3,2,3) = 3.

Example 2:

Input: n = 2
Output: 1
Explanation: According to the given rules, nums = [0,1,1]. The maximum is max(0,1,1) = 1.

Example 3:

Input: n = 3
Output: 2
Explanation: According to the given rules, nums = [0,1,1,2]. The maximum is max(0,1,1,2) = 2.

Constraints:

0 <= n <= 100

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 maximum possible value for the integer 'n'?
Can 'n' be 0 or 1? If so, what are the expected outputs for these cases?
The problem states 'nums[i/2]' and 'nums[i/2 + 1]'. Does integer division for 'i/2' always round down as per standard integer division rules in programming languages?
Are there any constraints on the data types for the elements in the generated array, or can they grow arbitrarily large?
If the maximum value appears multiple times, should I return any one of them, or is there a specific criterion (e.g., the first occurrence)?

Brute Force Solution

Approach

To find the largest number in a specially generated sequence, we can simulate the entire generation process step-by-step. This means we'll calculate each number in the sequence based on the rules, keeping track of every number as we go, and then simply pick the biggest one we've recorded.

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

Start by knowing the first few numbers in the sequence.
For each subsequent number, figure out what it should be by looking at the numbers that came before it, according to the specific generation rules.
As you determine each new number, add it to your growing collection of numbers.
Continue this process until you have generated all the numbers up to the specified limit.
Once you have your complete collection of generated numbers, go through all of them.
Compare each number to find the one with the highest value.
The number with the highest value you found is your answer.

Code Implementation

def getMaximumGenerated(n):
    if n == 0:
        return 0
    if n == 1:
        return 1

    generatedNumbers = [0] * (n + 1)
    generatedNumbers[0] = 0
    generatedNumbers[1] = 1
    maximumValueFound = 1

    # This loop simulates the generation process to fill the array
    for currentIndex in range(2, n + 1):
        # Determine if the current index is even or odd to apply the correct rule
        if currentIndex % 2 == 0:
            # Even indices are generated by doubling the value at half the index
            generatedNumbers[currentIndex] = generatedNumbers[currentIndex // 2]
        else:
            # Odd indices are generated by summing values at floor(i/2) and ceil(i/2)
            generatedNumbers[currentIndex] = generatedNumbers[currentIndex // 2] + generatedNumbers[currentIndex // 2 + 1]

        # Update the maximum value found so far as we generate new numbers
        if generatedNumbers[currentIndex] > maximumValueFound:
            maximumValueFound = generatedNumbers[currentIndex]

    # After generating all numbers, the maximumValueFound holds our answer
    return maximumValueFound

Big(O) Analysis

Time Complexity

O(n) – The solution involves simulating the generation of the array up to the given number n. We will iterate from 0 up to n to calculate each element of the generated array. For each step in this iteration, we perform a constant number of operations (checks and assignments) based on previous array elements. Finally, finding the maximum element in the generated array of size n takes at most n comparisons. Therefore, the total number of operations is directly proportional to n, resulting in a linear time complexity of O(n).

Space Complexity

O(n) – The solution requires an auxiliary array to store all the generated numbers up to the specified limit 'n'. This array will hold 'n+1' elements. Therefore, the auxiliary space complexity is directly proportional to the input size 'n', resulting in O(n) space.

Optimal Solution

Approach

This problem asks us to find the largest number in a specially generated sequence. The clever way to solve this is to build the sequence itself in an organized manner, keeping track of the biggest number encountered as we go, rather than regenerating it multiple times.

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

Imagine you have a blank space to write down the numbers of our special sequence.
You start by knowing the very first two numbers.
For subsequent numbers, their value depends on previous numbers in the sequence based on whether their position is even or odd.
If a position is even, the number there is the same as the number at half that position.
If a position is odd, the number there is the sum of the number at half that position and the number at the next position after half.
As you generate each new number, compare it with the largest number you've seen so far and update your record if the new number is bigger.
Continue this process until you have generated all the numbers up to the required size.
The largest number you recorded throughout this process is your answer.

Code Implementation

def getMaximumGenerated(number_of_elements):
    if number_of_elements < 1:
        return 0
    if number_of_elements == 1:
        return 0

    generated_sequence = [0] * (number_of_elements + 1)
    generated_sequence[1] = 1
    maximum_generated_value = 1

    # We need to build the sequence up to the specified number of elements.
    for current_index in range(2, number_of_elements + 1):
        # Even indexed elements are derived from their half-index.
        if current_index % 2 == 0:
            half_index = current_index // 2
            generated_sequence[current_index] = generated_sequence[half_index]
        # Odd indexed elements are derived from their half-index and the next element.
        else:
            half_index = current_index // 2
            generated_sequence[current_index] = generated_sequence[half_index] + generated_sequence[half_index + 1]

        # Track the largest value seen so far during generation.
        if generated_sequence[current_index] > maximum_generated_value:
            maximum_generated_value = generated_sequence[current_index]

    return maximum_generated_value

Big(O) Analysis

Time Complexity

O(n) – The solution involves generating an array of size n. Each element in this array is computed once based on previously computed elements. We iterate through the array from index 2 up to n to fill it. Inside this loop, operations are constant time (accessing array elements, addition, comparison). Therefore, the total number of operations is directly proportional to n, resulting in a time complexity of O(n).

Space Complexity

O(n) – The solution requires an auxiliary array to store the generated sequence up to the specified size 'n'. This array directly stores 'n' integer values, meaning the additional memory used is proportional to 'n'. The variables used to track the maximum value and loop indices are constant space. Therefore, the dominant factor in auxiliary space complexity is the array used to store the sequence, leading to O(n) space complexity.

Edge Cases

Case	How to Handle
Input n = 0	The array will have size 1, containing only nums[0] = 0, and the maximum will be 0.
Input n = 1	The array will have size 2, containing nums[0] = 0 and nums[1] = 1, and the maximum will be 1.
Input n is very large	The solution uses an array of size n+1, so memory constraints and potential integer overflow for extremely large n should be considered.
The generated numbers grow very quickly	The maximum value could potentially exceed standard integer limits if n is large enough, requiring consideration of larger integer types.
All generated numbers are the same	This scenario is not possible given the generation rules, as nums[1] is always 1 and nums[2] is derived from nums[1].
No valid solution exists	A valid solution always exists for non-negative n as the array is deterministically generated.
Integer overflow during calculation of nums[i / 2] + nums[i / 2 + 1]	If n is sufficiently large, the sum could overflow standard integer types, necessitating the use of larger integer types or checks for overflow.
Constraints on n not specified (e.g., negative n)	The problem statement implies non-negative n; a robust solution might handle negative inputs by returning an error or an empty array.

Input n = 0

How to Handle:

The array will have size 1, containing only nums[0] = 0, and the maximum will be 0.

Input n = 1

How to Handle:

The array will have size 2, containing nums[0] = 0 and nums[1] = 1, and the maximum will be 1.

Input n is very large

How to Handle:

The solution uses an array of size n+1, so memory constraints and potential integer overflow for extremely large n should be considered.

The generated numbers grow very quickly

How to Handle:

The maximum value could potentially exceed standard integer limits if n is large enough, requiring consideration of larger integer types.

All generated numbers are the same

How to Handle:

This scenario is not possible given the generation rules, as nums[1] is always 1 and nums[2] is derived from nums[1].

No valid solution exists

How to Handle:

A valid solution always exists for non-negative n as the array is deterministically generated.

Integer overflow during calculation of nums[i / 2] + nums[i / 2 + 1]

How to Handle:

If n is sufficiently large, the sum could overflow standard integer types, necessitating the use of larger integer types or checks for overflow.

Constraints on n not specified (e.g., negative n)

How to Handle:

The problem statement implies non-negative n; a robust solution might handle negative inputs by returning an error or an empty array.

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 maximum possible value for the integer 'n'?
Can 'n' be 0 or 1? If so, what are the expected outputs for these cases?
The problem states 'nums[i/2]' and 'nums[i/2 + 1]'. Does integer division for 'i/2' always round down as per standard integer division rules in programming languages?
Are there any constraints on the data types for the elements in the generated array, or can they grow arbitrarily large?
If the maximum value appears multiple times, should I return any one of them, or is there a specific criterion (e.g., the first occurrence)?

Brute Force Solution

Approach

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

Start by knowing the first few numbers in the sequence.
For each subsequent number, figure out what it should be by looking at the numbers that came before it, according to the specific generation rules.
As you determine each new number, add it to your growing collection of numbers.
Continue this process until you have generated all the numbers up to the specified limit.
Once you have your complete collection of generated numbers, go through all of them.
Compare each number to find the one with the highest value.
The number with the highest value you found is your answer.

Code Implementation

def getMaximumGenerated(n):
    if n == 0:
        return 0
    if n == 1:
        return 1

    generatedNumbers = [0] * (n + 1)
    generatedNumbers[0] = 0
    generatedNumbers[1] = 1
    maximumValueFound = 1

    # This loop simulates the generation process to fill the array
    for currentIndex in range(2, n + 1):
        # Determine if the current index is even or odd to apply the correct rule
        if currentIndex % 2 == 0:
            # Even indices are generated by doubling the value at half the index
            generatedNumbers[currentIndex] = generatedNumbers[currentIndex // 2]
        else:
            # Odd indices are generated by summing values at floor(i/2) and ceil(i/2)
            generatedNumbers[currentIndex] = generatedNumbers[currentIndex // 2] + generatedNumbers[currentIndex // 2 + 1]

        # Update the maximum value found so far as we generate new numbers
        if generatedNumbers[currentIndex] > maximumValueFound:
            maximumValueFound = generatedNumbers[currentIndex]

    # After generating all numbers, the maximumValueFound holds our answer
    return maximumValueFound

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine you have a blank space to write down the numbers of our special sequence.
You start by knowing the very first two numbers.
For subsequent numbers, their value depends on previous numbers in the sequence based on whether their position is even or odd.
If a position is even, the number there is the same as the number at half that position.
If a position is odd, the number there is the sum of the number at half that position and the number at the next position after half.
As you generate each new number, compare it with the largest number you've seen so far and update your record if the new number is bigger.
Continue this process until you have generated all the numbers up to the required size.
The largest number you recorded throughout this process is your answer.

Code Implementation

def getMaximumGenerated(number_of_elements):
    if number_of_elements < 1:
        return 0
    if number_of_elements == 1:
        return 0

    generated_sequence = [0] * (number_of_elements + 1)
    generated_sequence[1] = 1
    maximum_generated_value = 1

    # We need to build the sequence up to the specified number of elements.
    for current_index in range(2, number_of_elements + 1):
        # Even indexed elements are derived from their half-index.
        if current_index % 2 == 0:
            half_index = current_index // 2
            generated_sequence[current_index] = generated_sequence[half_index]
        # Odd indexed elements are derived from their half-index and the next element.
        else:
            half_index = current_index // 2
            generated_sequence[current_index] = generated_sequence[half_index] + generated_sequence[half_index + 1]

        # Track the largest value seen so far during generation.
        if generated_sequence[current_index] > maximum_generated_value:
            maximum_generated_value = generated_sequence[current_index]

    return maximum_generated_value

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n = 0	The array will have size 1, containing only nums[0] = 0, and the maximum will be 0.
Input n = 1	The array will have size 2, containing nums[0] = 0 and nums[1] = 1, and the maximum will be 1.
Input n is very large	The solution uses an array of size n+1, so memory constraints and potential integer overflow for extremely large n should be considered.
The generated numbers grow very quickly	The maximum value could potentially exceed standard integer limits if n is large enough, requiring consideration of larger integer types.
All generated numbers are the same	This scenario is not possible given the generation rules, as nums[1] is always 1 and nums[2] is derived from nums[1].
No valid solution exists	A valid solution always exists for non-negative n as the array is deterministically generated.
Integer overflow during calculation of nums[i / 2] + nums[i / 2 + 1]	If n is sufficiently large, the sum could overflow standard integer types, necessitating the use of larger integer types or checks for overflow.
Constraints on n not specified (e.g., negative n)	The problem statement implies non-negative n; a robust solution might handle negative inputs by returning an error or an empty array.

Input n = 0

How to Handle:

The array will have size 1, containing only nums[0] = 0, and the maximum will be 0.

Input n = 1

How to Handle:

The array will have size 2, containing nums[0] = 0 and nums[1] = 1, and the maximum will be 1.

Input n is very large

How to Handle:

The solution uses an array of size n+1, so memory constraints and potential integer overflow for extremely large n should be considered.

The generated numbers grow very quickly

How to Handle:

The maximum value could potentially exceed standard integer limits if n is large enough, requiring consideration of larger integer types.

All generated numbers are the same

How to Handle:

This scenario is not possible given the generation rules, as nums[1] is always 1 and nums[2] is derived from nums[1].

No valid solution exists

How to Handle:

A valid solution always exists for non-negative n as the array is deterministically generated.

Integer overflow during calculation of nums[i / 2] + nums[i / 2 + 1]

How to Handle:

If n is sufficiently large, the sum could overflow standard integer types, necessitating the use of larger integer types or checks for overflow.

Constraints on n not specified (e.g., negative n)

How to Handle:

The problem statement implies non-negative n; a robust solution might handle negative inputs by returning an error or an empty array.