Zero Array Transformation III

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ArraysGreedy Algorithms

You are given an integer array nums of length n and a 2D array queries where queries[i] = [l_i, r_i].

Each queries[i] represents the following action on nums:

Decrement the value at each index in the range [l_i, r_i] in nums by at most 1.
The amount by which the value is decremented can be chosen independently for each index.

A Zero Array is an array with all its elements equal to 0.

Return the maximum number of elements that can be removed from queries, such that nums can still be converted to a zero array using the remaining queries. If it is not possible to convert nums to a zero array, return -1.

Example 1:

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

Output: 1

Explanation:

After removing queries[2], nums can still be converted to a zero array.

Using queries[0], decrement nums[0] and nums[2] by 1 and nums[1] by 0.
Using queries[1], decrement nums[0] and nums[2] by 1 and nums[1] by 0.

Example 2:

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

Output: 2

Explanation:

We can remove queries[2] and queries[3].

Example 3:

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

Output: -1

Explanation:

nums cannot be converted to a zero array even after using all the queries.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵
1 <= queries.length <= 10⁵
queries[i].length == 2
0 <= l_i <= r_i < nums.length

Try coding it in LeetCode

Zero Array Transformation III

Medium

Asked by:

17 views

Topics:

ArraysGreedy Algorithms

You are given an integer array nums of length n and a 2D array queries where queries[i] = [l_i, r_i].

Each queries[i] represents the following action on nums:

Decrement the value at each index in the range [l_i, r_i] in nums by at most 1.
The amount by which the value is decremented can be chosen independently for each index.

A Zero Array is an array with all its elements equal to 0.

Return the maximum number of elements that can be removed from queries, such that nums can still be converted to a zero array using the remaining queries. If it is not possible to convert nums to a zero array, return -1.

Example 1:

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

Output: 1

Explanation:

After removing queries[2], nums can still be converted to a zero array.

Using queries[0], decrement nums[0] and nums[2] by 1 and nums[1] by 0.
Using queries[1], decrement nums[0] and nums[2] by 1 and nums[1] by 0.

Example 2:

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

Output: 2

Explanation:

We can remove queries[2] and queries[3].

Example 3:

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

Output: -1

Explanation:

nums cannot be converted to a zero array even after using all the queries.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵
1 <= queries.length <= 10⁵
queries[i].length == 2
0 <= l_i <= r_i < nums.length

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 integer ranges within the input array? Can I expect negative numbers or zeros?
What should be returned if the array is already all zeros, or if no valid transformation sequence exists?
Can the input array be empty or null? If so, what is the expected behavior?
If there are multiple valid transformation sequences, is any valid sequence acceptable, or is there a specific criteria for choosing one?
Is the input array guaranteed to only contain integers, or could it contain other data types?

Brute Force Solution

Approach

The brute force method aims to discover if one set of numbers can be transformed into another by repeatedly choosing a pair of numbers and making them both zero. It operates by exhaustively testing every possible pairing until a solution is found or all combinations are exhausted.

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

Consider every possible pair of numbers in the initial set.
For each pair, imagine setting both numbers to zero.
Check if setting this specific pair to zero allows you to eventually transform the initial set into the desired final set (all zeros). You can determine this by repeating the pairing process on the remaining numbers.
If the transformation works for a specific pair, you've found a valid move.
If it doesn't work, try a different pair from the beginning.
Continue this process, testing all possible pairings and subsequent transformations until either a solution is found (the initial set can be transformed into the final set of zeros) or all possible combinations are exhausted (meaning no such transformation is possible).

Code Implementation

def can_transform_to_zero_array_brute_force(number_list):
    def solve(current_list):
        if all(number == 0 for number in current_list):
            return True

        list_length = len(current_list)
        for first_index in range(list_length):
            for second_index in range(first_index + 1, list_length):
                # Create a copy to avoid modifying the original list during recursion
                next_list = current_list[:]
                # Simulate setting the pair to zero.

                next_list[first_index] = 0
                next_list[second_index] = 0

                # Recursively check if the transformation leads to all zeros.

                if solve(next_list):
                    return True
        return False

    # Check if the array can be transformed to zero array
    return solve(number_list)

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach considers all possible pairs in the array of size n. There are n choose 2, or n(n-1)/2 initial pairs to consider. For each of these pairs, the algorithm recursively tries to transform the remaining elements to zero, potentially leading to a combinatorial explosion of possibilities. In the worst-case, the algorithm might explore all possible combinations of pairs, resulting in factorial time complexity due to the branching recursion. Therefore, the time complexity is O(n!).

Space Complexity

O(N) – The brute force method, as described, implicitly uses recursion to explore possible pairings. Each recursive call creates a new stack frame. In the worst case, every pairing decision leads to another recursive call until all numbers are potentially paired off. Therefore, the maximum depth of the recursion, and hence the size of the call stack, can be proportional to N, where N is the number of elements in the initial set. This implies that the auxiliary space required is O(N) due to the recursion stack.

Optimal Solution

Approach

The optimal approach transforms the input array into a zero array by strategically canceling out elements. It identifies patterns and combines elements to minimize the number of operations. This avoids brute force which tries all combinations.

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

First, focus on the differences between adjacent numbers. Think of these differences as the key to making the numbers equal.
Next, look for places where numbers are the same. If you find these, you can start to cancel out the differing values around them.
Now, consider the effect of changing one number on its neighbors. Aim to make the numbers more alike, eventually getting everything to zero.
Continue changing the numbers locally, spreading the effect of those changes to adjacent elements.
Repeat this process until all the numbers become zero. Each step reduces the overall difference in the array, leading to the final zero array.

Code Implementation

def zero_array_transformation(input_array):
    array_length = len(input_array)
    operations_count = 0

    while True:
        all_zeros = True
        for element in input_array:
            if element != 0:
                all_zeros = False
                break
        if all_zeros:
            break

        changed = False
        for index in range(array_length):
            if input_array[index] != 0:
                # Find a neighbor to cancel out with
                for neighbor_index in range(max(0, index - 1), min(array_length, index + 2)):
                    if index != neighbor_index and input_array[index] == -input_array[neighbor_index]:
                        input_array[index] = 0
                        input_array[neighbor_index] = 0
                        operations_count += 1
                        changed = True
                        break

                if changed:
                    break

        if not changed:
            # If no direct cancellations possible, try to reduce differences.
            for index in range(array_length):
                if input_array[index] != 0:
                    for neighbor_index in range(max(0, index - 1), min(array_length, index + 2)):
                        if index != neighbor_index:
                            #Reduce absolute difference
                            if abs(input_array[index]) > 0 and abs(input_array[neighbor_index]) >= 0:

                                # Propagate value to neighbor
                                input_array[neighbor_index] -= input_array[index]
                                input_array[index] = 0
                                operations_count += 1
                                changed = True
                                break

                if changed:
                    break

            if not changed:
                # If still stuck, try shifting values to make progress
                for index in range(array_length):
                    if input_array[index] != 0:
                        #Distribute non-zero value
                        value_to_distribute = input_array[index] // 2
                        input_array[index] -= value_to_distribute
                        if index > 0:
                            input_array[index-1] += value_to_distribute
                        if index < array_length - 1:
                            input_array[index+1] += value_to_distribute

                        operations_count += 1
                        changed = True
                        break

                if not changed:
                    return -1 # Indicates failure

    return operations_count

Big(O) Analysis

Time Complexity

O(n²) – The provided approach iteratively attempts to equalize adjacent elements to transform the array to all zeros. In the worst-case scenario, each element might need to be adjusted based on its neighbors, and this process could propagate across the array multiple times. Correcting a single element might require iterating through the array to propagate the changes. Therefore, in the worst case, for each of the n elements, the process might require adjustments that potentially iterate over the entire array, resulting in approximately n iterations for each element. This results in a time complexity of O(n²).

Space Complexity

O(1) – The described approach primarily focuses on in-place transformations by manipulating adjacent numbers and canceling out differences. It doesn't mention the creation of any auxiliary data structures like lists, hash maps, or stacks to store intermediate results or visited locations. The algorithm's space complexity is thus dominated by a few variables used for tracking indices or performing calculations, resulting in constant extra space regardless of the input array size N. Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list or null indicating no transformation possible.
Array with only one element	Return an empty list because a pair is needed.
Array with all elements being zero	Check if an array of all zeros should return empty or all pairs (0,0) depending on if transformation is possible.
Array with a very large number of elements leading to potential memory issues	Consider space complexity when implementing with large inputs and choose an appropriate data structure
Array contains duplicate numbers, where those duplicate numbers form multiple solutions	Handle correctly through index tracking in the hash map to consider all possible pairs.
No solution exists (i.e., no valid sequence of operations can transform the array)	Return an empty list or specific error code to indicate failure to transform.
Input array contains negative numbers	The algorithm needs to correctly handle negative numbers and their interactions during subtraction.
Input array contains extremely large positive or negative numbers that might cause integer overflow	Use appropriate data types (e.g., long) to prevent integer overflows and handle potential overflow exceptions.