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Decrease Elements To Make Array Zigzag

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ArraysGreedy Algorithms

Given an array nums of integers, a move consists of choosing any element and decreasing it by 1.

An array A is a zigzag array if either:

Every even-indexed element is greater than adjacent elements, ie. A[0] > A[1] < A[2] > A[3] < A[4] > ...
OR, every odd-indexed element is greater than adjacent elements, ie. A[0] < A[1] > A[2] < A[3] > A[4] < ...

Return the minimum number of moves to transform the given array nums into a zigzag array.

Example 1:

Input: nums = [1,2,3]
Output: 2
Explanation: We can decrease 2 to 0 or 3 to 1.

Example 2:

Input: nums = [9,6,1,6,2]
Output: 4

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 1000

Try coding it in LeetCode

Decrease Elements To Make Array Zigzag

Medium

Asked by:

4 views

Topics:

ArraysGreedy Algorithms

Given an array nums of integers, a move consists of choosing any element and decreasing it by 1.

An array A is a zigzag array if either:

Every even-indexed element is greater than adjacent elements, ie. A[0] > A[1] < A[2] > A[3] < A[4] > ...
OR, every odd-indexed element is greater than adjacent elements, ie. A[0] < A[1] > A[2] < A[3] > A[4] < ...

Return the minimum number of moves to transform the given array nums into a zigzag array.

Example 1:

Input: nums = [1,2,3]
Output: 2
Explanation: We can decrease 2 to 0 or 3 to 1.

Example 2:

Input: nums = [9,6,1,6,2]
Output: 4

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 1000

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 are the possible values for the elements in the input array? Could they be negative, zero, or very large?
What should I return if the input array is empty or contains only one element?
If multiple zigzag arrangements are possible with the same minimum number of decrements, is any one of them acceptable, or is there a specific zigzag arrangement that should be preferred?
Are there any constraints on modifying the original input array in place, or should I create a copy?
By 'zigzag', do you strictly mean `nums[i] < nums[i+1] > nums[i+2] < ...` or `nums[i] > nums[i+1] < nums[i+2] > ...`, or are both valid zigzag patterns?

Brute Force Solution

Approach

The brute force method for this puzzle involves checking every possible adjustment we can make to the numbers in the list. We want to find an arrangement where the numbers go up and down like a zigzag, and this approach guarantees we will try everything.

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

Consider the first position in the list. We could change the number in that position or leave it as it is.
For each possible change we make in the first position, explore all options for the second position: change it or don't.
Continue this process for every position in the list. At each position, consider making all possible changes to the number at that location.
After making a complete set of changes (possibly none at all), check if the list now satisfies the zigzag requirement.
If it does, keep track of how many adjustments we had to make.
Repeat the entire process from the beginning, trying a different set of adjustments each time.
Once we have tried all possible combinations of adjustments, select the set of changes that resulted in a zigzag pattern with the fewest number of adjustments.

Code Implementation

def decrease_elements_to_make_array_zigzag_brute_force(numbers):
    array_length = len(numbers)
    minimum_moves = float('inf')

    def is_zigzag(array):
        for index in range(array_length):
            if index % 2 == 0:
                # Check if even indexed element is greater than its neighbors
                if index > 0 and array[index] >= array[index - 1]:
                    return False
                if index < array_length - 1 and array[index] >= array[index + 1]:
                    return False
            else:
                # Check if odd indexed element is greater than its neighbors
                if index > 0 and array[index] >= array[index - 1]:
                    return False
                if index < array_length - 1 and array[index] >= array[index + 1]:
                    return False
        return True

    def solve(index, current_array, moves):
        nonlocal minimum_moves

        if index == array_length:
            if is_zigzag(current_array):
                minimum_moves = min(minimum_moves, moves)
            return

        # Option 1: Don't change the current element
        solve(index + 1, current_array[:], moves)

        # Option 2: Try decreasing the current element
        original_value = current_array[index]
        for new_value in range(original_value):
            current_array[index] = new_value
            solve(index + 1, current_array[:], moves + (original_value - new_value))

        # Reset the current array. Necessary for backtracking.
        current_array[index] = original_value

    solve(0, numbers[:], 0)

    return minimum_moves

Big(O) Analysis

Time Complexity

O(Infinity) – The described brute force method involves considering all possible modifications to each element in the array. While not explicitly stated how, the approach implies exploring all possible values for each element. Without constraints on possible element values, the number of possible modifications for each element approaches infinity. Since this process is repeated for each element in the array of size n, the overall time complexity is effectively O(Infinity) due to the unbounded number of modifications possible at each position. Trying all these modifications will result in infinite computations and can never finish. The number of modifications explode combinatorially, therefore this approach doesn't yield a Big O complexity that depends on n.

Space Complexity

O(N) – The brute force method explores all possible combinations of adjustments by implicitly using recursion. In the worst-case scenario, the depth of the recursion can go as deep as the number of elements in the list (N), where N is the length of the input array. Each recursive call adds a new frame to the call stack, holding the current state of the array and the index being considered. Thus, the auxiliary space used by the call stack scales linearly with the input size N, resulting in O(N) space complexity.

Optimal Solution

Approach

To make the array zigzag, we want each number to be smaller than its neighbors. We can achieve this by focusing on making either the even-positioned numbers or the odd-positioned numbers smaller, and choosing the option that requires the fewest changes overall.

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

Imagine you want all the numbers at even positions (first, third, fifth, etc.) to be smaller than their neighbors.
Go through the numbers at even positions. If a number is bigger than either of its neighbors, reduce it to be just one smaller than the smaller neighbor.
Keep track of how many times you had to change a number to make the even positions smaller than their neighbors.
Now, imagine you want all the numbers at odd positions (second, fourth, sixth, etc.) to be smaller than their neighbors.
Go through the numbers at odd positions. If a number is bigger than either of its neighbors, reduce it to be just one smaller than the smaller neighbor.
Keep track of how many times you had to change a number to make the odd positions smaller than their neighbors.
Compare the number of changes you made for even positions versus odd positions.
Choose the zigzag pattern (even or odd smaller) that required fewer changes. The number of changes is your answer.

Code Implementation

def decrease_elements_to_make_array_zigzag(numbers):
    array_length = len(numbers)

    def calculate_moves(start_index):
        temp_numbers = numbers[:]
        moves_needed = 0

        for i in range(start_index, array_length, 2):
            # Determine neighbors to compare with
            left_neighbor = temp_numbers[i - 1] if i > 0 else float('inf')
            right_neighbor = temp_numbers[i + 1] if i < array_length - 1 else float('inf')

            # Find the smaller neighbor
            smaller_neighbor = min(left_neighbor, right_neighbor)

            # If current element is not smaller, reduce it
            if temp_numbers[i] >= smaller_neighbor:
                moves_needed += temp_numbers[i] - smaller_neighbor + 1
                temp_numbers[i] = smaller_neighbor - 1
        return moves_needed

    # Calculate moves starting from even indices.
    even_start_moves = calculate_moves(0)

    # Calculate moves starting from odd indices.
    odd_start_moves = calculate_moves(1)

    # Return the minimum moves needed.
    return min(even_start_moves, odd_start_moves)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums twice. The first pass focuses on making even-indexed elements smaller than their neighbors, and the second pass focuses on making odd-indexed elements smaller than their neighbors. In each pass, for each element, it only performs a constant number of comparisons (at most two neighbor comparisons) and potentially one subtraction. Therefore, the time complexity is directly proportional to the size of the input array, n, making it O(n).

Space Complexity

O(1) – The algorithm uses a constant number of variables to store the counts of changes needed for both even and odd positions, irrespective of the input array size N. No auxiliary data structures like lists or maps that scale with the input are utilized. The space used is independent of the input size. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as an empty array is trivially zigzag.
Array with a single element	Return 0, as a single element array is trivially zigzag.
Array with two elements	Calculate the cost of making each element smaller than the other, and return the minimum.
Array with all identical elements	Calculate the cost of making all elements at even indices smaller and all elements at odd indices smaller, returning the minimum.
Array with alternating small and large elements, already zigzag	The algorithm should correctly compute that no changes are needed in this case, resulting in a cost of 0.
Very large array (scalability)	Ensure the solution has a time complexity of O(n) to handle large arrays efficiently by avoiding nested loops or unnecessary sorting.
Array with negative numbers	The algorithm should correctly handle negative numbers by taking the absolute difference when calculating the cost of decrementing elements.
Integer overflow when calculating costs	Use appropriate data types (e.g., long) or modulo operations if the problem specifies a maximum cost to prevent integer overflow.

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 possible values for the elements in the input array? Could they be negative, zero, or very large?
What should I return if the input array is empty or contains only one element?
If multiple zigzag arrangements are possible with the same minimum number of decrements, is any one of them acceptable, or is there a specific zigzag arrangement that should be preferred?
Are there any constraints on modifying the original input array in place, or should I create a copy?
By 'zigzag', do you strictly mean `nums[i] < nums[i+1] > nums[i+2] < ...` or `nums[i] > nums[i+1] < nums[i+2] > ...`, or are both valid zigzag patterns?

Brute Force Solution

Approach

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

Consider the first position in the list. We could change the number in that position or leave it as it is.
For each possible change we make in the first position, explore all options for the second position: change it or don't.
Continue this process for every position in the list. At each position, consider making all possible changes to the number at that location.
After making a complete set of changes (possibly none at all), check if the list now satisfies the zigzag requirement.
If it does, keep track of how many adjustments we had to make.
Repeat the entire process from the beginning, trying a different set of adjustments each time.
Once we have tried all possible combinations of adjustments, select the set of changes that resulted in a zigzag pattern with the fewest number of adjustments.

Code Implementation

def decrease_elements_to_make_array_zigzag_brute_force(numbers):
    array_length = len(numbers)
    minimum_moves = float('inf')

    def is_zigzag(array):
        for index in range(array_length):
            if index % 2 == 0:
                # Check if even indexed element is greater than its neighbors
                if index > 0 and array[index] >= array[index - 1]:
                    return False
                if index < array_length - 1 and array[index] >= array[index + 1]:
                    return False
            else:
                # Check if odd indexed element is greater than its neighbors
                if index > 0 and array[index] >= array[index - 1]:
                    return False
                if index < array_length - 1 and array[index] >= array[index + 1]:
                    return False
        return True

    def solve(index, current_array, moves):
        nonlocal minimum_moves

        if index == array_length:
            if is_zigzag(current_array):
                minimum_moves = min(minimum_moves, moves)
            return

        # Option 1: Don't change the current element
        solve(index + 1, current_array[:], moves)

        # Option 2: Try decreasing the current element
        original_value = current_array[index]
        for new_value in range(original_value):
            current_array[index] = new_value
            solve(index + 1, current_array[:], moves + (original_value - new_value))

        # Reset the current array. Necessary for backtracking.
        current_array[index] = original_value

    solve(0, numbers[:], 0)

    return minimum_moves

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine you want all the numbers at even positions (first, third, fifth, etc.) to be smaller than their neighbors.
Go through the numbers at even positions. If a number is bigger than either of its neighbors, reduce it to be just one smaller than the smaller neighbor.
Keep track of how many times you had to change a number to make the even positions smaller than their neighbors.
Now, imagine you want all the numbers at odd positions (second, fourth, sixth, etc.) to be smaller than their neighbors.
Go through the numbers at odd positions. If a number is bigger than either of its neighbors, reduce it to be just one smaller than the smaller neighbor.
Keep track of how many times you had to change a number to make the odd positions smaller than their neighbors.
Compare the number of changes you made for even positions versus odd positions.
Choose the zigzag pattern (even or odd smaller) that required fewer changes. The number of changes is your answer.

Code Implementation

def decrease_elements_to_make_array_zigzag(numbers):
    array_length = len(numbers)

    def calculate_moves(start_index):
        temp_numbers = numbers[:]
        moves_needed = 0

        for i in range(start_index, array_length, 2):
            # Determine neighbors to compare with
            left_neighbor = temp_numbers[i - 1] if i > 0 else float('inf')
            right_neighbor = temp_numbers[i + 1] if i < array_length - 1 else float('inf')

            # Find the smaller neighbor
            smaller_neighbor = min(left_neighbor, right_neighbor)

            # If current element is not smaller, reduce it
            if temp_numbers[i] >= smaller_neighbor:
                moves_needed += temp_numbers[i] - smaller_neighbor + 1
                temp_numbers[i] = smaller_neighbor - 1
        return moves_needed

    # Calculate moves starting from even indices.
    even_start_moves = calculate_moves(0)

    # Calculate moves starting from odd indices.
    odd_start_moves = calculate_moves(1)

    # Return the minimum moves needed.
    return min(even_start_moves, odd_start_moves)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as an empty array is trivially zigzag.
Array with a single element	Return 0, as a single element array is trivially zigzag.
Array with two elements	Calculate the cost of making each element smaller than the other, and return the minimum.
Array with all identical elements	Calculate the cost of making all elements at even indices smaller and all elements at odd indices smaller, returning the minimum.
Array with alternating small and large elements, already zigzag	The algorithm should correctly compute that no changes are needed in this case, resulting in a cost of 0.
Very large array (scalability)	Ensure the solution has a time complexity of O(n) to handle large arrays efficiently by avoiding nested loops or unnecessary sorting.
Array with negative numbers	The algorithm should correctly handle negative numbers by taking the absolute difference when calculating the cost of decrementing elements.
Integer overflow when calculating costs	Use appropriate data types (e.g., long) or modulo operations if the problem specifies a maximum cost to prevent integer overflow.