Maximum Element After Decreasing and Rearranging

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ArraysGreedy Algorithms

You are given an array of positive integers arr. Perform some operations (possibly none) on arr so that it satisfies these conditions:

The value of the first element in arr must be 1.
The absolute difference between any 2 adjacent elements must be less than or equal to 1. In other words, abs(arr[i] - arr[i - 1]) <= 1 for each i where 1 <= i < arr.length (0-indexed). abs(x) is the absolute value of x.

There are 2 types of operations that you can perform any number of times:

Decrease the value of any element of arr to a smaller positive integer.
Rearrange the elements of arr to be in any order.

Return the maximum possible value of an element in arr after performing the operations to satisfy the conditions.

Example 1:

arr = [2, 2, 1, 2, 1]

Output: 2

Explanation: We can satisfy the conditions by rearranging arr so it becomes [1, 2, 2, 2, 1]. The largest element in arr is 2.

Example 2:

arr = [100, 1, 1000]

Output: 3

Explanation: One possible way to satisfy the conditions is by doing the following:

Rearrange arr so it becomes [1, 100, 1000].
Decrease the value of the second element to 2.
Decrease the value of the third element to 3.

Now arr = [1, 2, 3], which satisfies the conditions. The largest element in arr is 3.

Example 3:

arr = [1, 2, 3, 4, 5]

Output: 5

Explanation: The array already satisfies the conditions, and the largest element is 5.

Maximum Element After Decreasing and Rearranging

Medium

Amazon

2 views

Topics:

ArraysGreedy Algorithms

You are given an array of positive integers arr. Perform some operations (possibly none) on arr so that it satisfies these conditions:

The value of the first element in arr must be 1.
The absolute difference between any 2 adjacent elements must be less than or equal to 1. In other words, abs(arr[i] - arr[i - 1]) <= 1 for each i where 1 <= i < arr.length (0-indexed). abs(x) is the absolute value of x.

There are 2 types of operations that you can perform any number of times:

Decrease the value of any element of arr to a smaller positive integer.
Rearrange the elements of arr to be in any order.

Return the maximum possible value of an element in arr after performing the operations to satisfy the conditions.

Example 1:

arr = [2, 2, 1, 2, 1]

Output: 2

Explanation: We can satisfy the conditions by rearranging arr so it becomes [1, 2, 2, 2, 1]. The largest element in arr is 2.

Example 2:

arr = [100, 1, 1000]

Output: 3

Explanation: One possible way to satisfy the conditions is by doing the following:

Rearrange arr so it becomes [1, 100, 1000].
Decrease the value of the second element to 2.
Decrease the value of the third element to 3.

Now arr = [1, 2, 3], which satisfies the conditions. The largest element in arr is 3.

Example 3:

arr = [1, 2, 3, 4, 5]

Output: 5

Explanation: The array already satisfies the conditions, and the largest element is 5.

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 values for the integers in the input array?
Can the input array be empty or null?
Are there any constraints on the length of the input array?
Are duplicate numbers allowed in the input array, and how should they be handled during the decreasing and rearranging process?
If after decreasing and rearranging, multiple arrays result in the same maximum element, is any such maximum element acceptable?

Brute Force Solution

Approach

The brute force approach to finding the maximum element after decreasing and rearranging is like trying every possible combination. We explore all ways to modify the numbers and their positions, checking if each way gives us a valid arrangement.

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

Consider all possible rearrangements of the original numbers.
For each arrangement, consider all possible ways to decrease the numbers so that the first number is 1, and the difference between adjacent numbers is at most 1.
After performing the decreasing and rearranging steps, check the last element of the arrangement.
Repeat steps 1-3 for every possible combination of rearrangements and decreasing.
Keep track of the largest last element (the maximum element) found across all these combinations.
After checking every possibility, return the largest last element you found. This is the overall maximum.

Code Implementation

import itertools

def maximum_element_after_decreasing_and_rearranging_brute_force(numbers):
    maximum_element = 0
    for permutation in itertools.permutations(numbers):
        # Iterate through all possible permutations of the input list
        for i in range(1 << (len(numbers) - 1)): 
            decreased_numbers = list(permutation)
            
            binary_representation = bin(i)[2:].zfill(len(numbers) - 1)
            
            for index in range(len(numbers)):                
                if index == 0:
                    decreased_numbers[index] = 1
                else:
                    # Determine if decreasing is needed
                    if binary_representation[index-1] == '1':
                        decreased_numbers[index] = min(decreased_numbers[index], decreased_numbers[index-1] + 1)
                    else:
                        decreased_numbers[index] = min(decreased_numbers[index], decreased_numbers[index-1] - 1 if decreased_numbers[index-1] > 1 else decreased_numbers[index-1] + 1)                    
                
            is_valid = True
            for index in range(1, len(decreased_numbers)):
                if abs(decreased_numbers[index] - decreased_numbers[index - 1]) > 1:
                    is_valid = False
                    break
            
            if is_valid:
                maximum_element = max(maximum_element, decreased_numbers[-1])
                
    return maximum_element

def maximum_element_after_decreasing_and_rearranging(numbers):
    numbers.sort()
    numbers[0] = 1
    for i in range(1, len(numbers)):
        # Ensure adjacent elements differ by at most 1
        numbers[i] = min(numbers[i], numbers[i - 1] + 1)

    # The maximum element is the last element after processing
    return numbers[-1]

Big(O) Analysis

Time Complexity

O(n! * n^n) – The brute force approach first considers all possible rearrangements of the input array of size n, which takes O(n!) time. For each of these rearrangements, it explores all possible ways to decrease the numbers to satisfy the given conditions. In the worst case, each number could be decreased significantly, and since there are n numbers, each with potentially n different decreasing values (down to 1), this could contribute O(n^n) complexity. Therefore, the total time complexity is approximately O(n! * n^n).

Space Complexity

O(N!) – The brute force approach considers all possible rearrangements of the input array of size N. Generating each permutation requires storing N elements in a temporary array. Because all N! rearrangements are considered and presumably processed one at a time, the space used by an individual permutation array is reused, however the need to generate these permutations implies a call stack potentially growing with the number of permutations. This results in a space complexity that is at least O(N!).

Optimal Solution

Approach

The trick to solving this quickly is to first organize the numbers and then make sure they fit the rules. We sort them to easily find the right sequence, and then adjust them in a specific way to guarantee we get the biggest possible final number.

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

First, organize the numbers from smallest to largest.
Then, make sure the first number is a 1. If it's not, change it to 1.
After that, go through the rest of the numbers one by one. For each number, check if it's too much bigger than the number before it. If it is, reduce it to be just one bigger than the previous number.
When you're done going through all the numbers, the last number in the sequence will be the biggest number you can possibly get following the rules.

Code Implementation

def maximum_element_after_decreasing_and_rearranging(array_of_numbers):
    array_of_numbers.sort()

    # Ensure the first element is 1, per the problem statement.
    array_of_numbers[0] = 1

    for index in range(1, len(array_of_numbers)):
        # If the current element is too large,
        # decrease it to be at most 1 greater than the previous.
        if array_of_numbers[index] > array_of_numbers[index - 1] + 1:

            array_of_numbers[index] = array_of_numbers[index - 1] + 1

    # The last element is the maximum possible value
    return array_of_numbers[-1]

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this solution is sorting the input array of size n, which typically takes O(n log n) time using efficient sorting algorithms like merge sort or quicksort. The subsequent steps involve iterating through the sorted array once to adjust the elements. This linear iteration takes O(n) time. Since O(n log n) grows faster than O(n) as n increases, the overall time complexity is determined by the sorting step. Therefore, the time complexity of the entire algorithm is O(n log n).

Space Complexity

O(1) – The algorithm sorts the input array in place, meaning no new array of size N is created for sorting. Only a few variables are used to iterate through the sorted array and compare adjacent elements. Therefore, the space used is constant and independent of the input size N.

Edge Cases

Case	How to Handle
Empty input array	Return 0 since there are no elements to rearrange and the maximum element would be non-existent.
Array with a single element	Return 1, because after decrementing (if needed) and rearranging, it will always be 1.
Array with all elements being the same value	The solution should still produce a valid rearranged array; after sorting and applying the decrementing and rearranging rules, it correctly results in a sequence 1 to N.
Array already sorted in ascending order	The solution will still ensure arr[0] = 1 and the difference between adjacent elements is at most 1.
Array contains very large integers leading to potential integer overflow during calculations	The given problem constraints specify that the integers are within a reasonable range, but consider using a larger data type (e.g., long) if the input integers are extremely large, or alternatively using modular arithmetic if applicable.
Array with a highly skewed distribution (e.g., many small numbers and a few very large numbers)	Sorting the array as the initial step ensures that we can effectively decrement and rearrange the elements to maximize the last element, regardless of the initial distribution.
Maximum size input array (n = 10^5 as stated in the constraints) with worst-case values	Sorting-based approach (O(n log n)) is efficient enough and the algorithm scales well within the specified constraints.
Array contains duplicate numbers	Sorting allows the algorithm to consider each element only once in the context of establishing the maximum element after rearrangement.