Zero Array Transformation I

Medium

Asked by:

10 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].

For each queries[i]:

Select a subset of indices within the range [l_i, r_i] in nums.
Decrement the values at the selected indices by 1.

A Zero Array is an array where all elements are equal to 0.

Return true if it is possible to transform nums into a Zero Array after processing all the queries sequentially, otherwise return false.

Example 1:

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

Output: true

Explanation:

For i = 0:
- Select the subset of indices as [0, 2] and decrement the values at these indices by 1.
- The array will become [0, 0, 0], which is a Zero Array.

Example 2:

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

Output: false

Explanation:

For i = 0:
- Select the subset of indices as [1, 2, 3] and decrement the values at these indices by 1.
- The array will become [4, 2, 1, 0].
For i = 1:
- Select the subset of indices as [0, 1, 2] and decrement the values at these indices by 1.
- The array will become [3, 1, 0, 0], which is not a Zero Array.

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 I

Medium

Asked by:

10 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].

For each queries[i]:

Select a subset of indices within the range [l_i, r_i] in nums.
Decrement the values at the selected indices by 1.

A Zero Array is an array where all elements are equal to 0.

Return true if it is possible to transform nums into a Zero Array after processing all the queries sequentially, otherwise return false.

Example 1:

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

Output: true

Explanation:

For i = 0:
- Select the subset of indices as [0, 2] and decrement the values at these indices by 1.
- The array will become [0, 0, 0], which is a Zero Array.

Example 2:

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

Output: false

Explanation:

For i = 0:
- Select the subset of indices as [1, 2, 3] and decrement the values at these indices by 1.
- The array will become [4, 2, 1, 0].
For i = 1:
- Select the subset of indices as [0, 1, 2] and decrement the values at these indices by 1.
- The array will become [3, 1, 0, 0], which is not a Zero Array.

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 ranges for the integer values within `nums1` and `nums2`?
What is the maximum length of the input arrays `nums1` and `nums2`?
Are empty arrays or null inputs possible? If so, how should I handle them?
If it's impossible to make `nums1` entirely zeros, should I return `false`, or is there a specific error message or exception I should raise?
Are the arrays guaranteed to be the same length?

Brute Force Solution

Approach

The brute force strategy for transforming an array to all zeros involves trying every possible combination of operations. We essentially explore every potential path to see if it leads to the desired all-zero state. This approach guarantees finding a solution, if one exists, by sheer exhaustive searching.

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

Consider applying an operation to the initial array in every possible way.
For each of those results, consider applying another operation in every possible way.
Continue this process, creating branches for every potential operation at each step.
At each step, check if the current state of the array is all zeros. If it is, you've found a solution.
If you've reached a point where no more operations can be applied and the array isn't all zeros, that path doesn't lead to a solution.
Explore every possible path until a solution is found or all possibilities are exhausted.

Code Implementation

def zero_array_transformation_brute_force(array):
    def apply_operation(current_array, start_index, end_index):
        new_array = current_array[:]
        for index in range(start_index, end_index + 1):
            new_array[index] = 0
        return new_array

    def is_all_zeros(current_array):
        return all(element == 0 for element in current_array)

    def find_solution(current_array, operations):
        if is_all_zeros(current_array):
            return operations

        array_length = len(current_array)
        # Explore possible operations on the array
        for start_index in range(array_length):
            for end_index in range(start_index, array_length):
                new_array = apply_operation(current_array, start_index, end_index)

                # Recursively search for a solution with the updated array

                new_operations = operations + [(start_index, end_index)]
                solution = find_solution(new_array, new_operations)
                if solution:
                    return solution
        return None

    # Start searching for operations from the given input array
    solution = find_solution(array, [])
    return solution

Big(O) Analysis

Time Complexity

O(2^(n*m)) – The brute force approach explores all possible combinations of operations on the array. Assuming we can apply an operation to any subarray, and there are 'm' possible operations that can be applied to each subarray, for an input array of size 'n', we can have approximately n*(n+1)/2 subarrays. Then, consider each of those n*(n+1)/2 subarrays can have m different operations, the algorithm will explore m possible options on each subarray, which grows exponentially with the number of operations and the input size. Therefore, the algorithm checks all combinations which leads to approximately O(2^(n*m)) where n is the array size and m is the number of possible operations.

Space Complexity

O(N!) – The brute force approach explores every possible combination of operations, creating branches for each potential operation at each step. In the worst-case scenario, this could lead to a tree-like structure where each level represents a different operation applied to the array, and the number of branches at each level grows rapidly. The space complexity is driven by the need to store copies of the array at each node of this implicit search tree as well as function call stack frames. Since we are effectively exploring all possible permutations of operations, the auxiliary space used to store these array states can grow up to O(N!), where N is related to the size of the input array as potentially each element or sub-array could be part of an operation.

Optimal Solution

Approach

The goal is to efficiently transform an array into an array of zeros. The most effective strategy involves strategically decreasing elements to indirectly zero out the entire array by targeting differences between adjacent numbers.

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

Examine the initial array and concentrate on pairs of neighboring numbers.
Find the 'bigger' number in each pair.
Reduce the bigger number by the value of its smaller neighbor.
Repeat this process, scanning through all pairs of neighbors from start to finish.
Continue applying this reduction strategy repeatedly. Because you are always making other values smaller, this will eventually reduce every value to zero.

Code Implementation

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

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

        if all_zeros:
            break

        for index in range(array_length - 1):
            # Find the 'bigger' number in each pair
            if input_array[index] < input_array[index + 1]:

                # Reduce the bigger number by its smaller neighbor.
                input_array[index + 1] -= input_array[index]

            elif input_array[index] > input_array[index + 1]:
                input_array[index] -= input_array[index + 1]

    return input_array

Big(O) Analysis

Time Complexity

O(n²) – The outer loop implicitly continues until the array is all zeros. In the worst-case scenario, elements might decrease by only a small amount in each pass. The inner loop iterates through the array of n elements to find adjacent pairs. For each pass of the outer loop, the inner loop takes O(n) time. In the worst case, it might take O(n) passes for all elements to become zero, resulting in a total time complexity of O(n * n) which simplifies to O(n²).

Space Complexity

O(1) – The algorithm operates in-place, modifying the input array directly. It only uses a few constant space variables such as loop counters or temporary variables to hold values during the reduction process. The amount of extra memory used does not depend on the size of the input array, N. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty arrays: nums1 or nums2 is null or has a length of 0.	Return true if both are null or empty, otherwise return false as zeroing is impossible.
Arrays of different lengths: nums1 and nums2 have different lengths.	Return false because the problem states that they must have the same length to perform swaps.
Array with all zeros: nums1 is already all zeros.	Return true immediately since no operations are needed.
Arrays with large values: nums1 and nums2 contain very large numbers (close to Integer.MAX_VALUE).	The algorithm uses boolean logic and thus integer overflow isn't a concern; large numbers themselves don't inherently invalidate the solution.
Arrays with negative numbers: nums1 and nums2 contain negative numbers.	Negative numbers in nums2 don't affect the logic, as we are concerned if those values can be swapped into nums1.
No possible solution: There is no combination of swaps that can make nums1 all zeros.	The algorithm should correctly identify and return false in cases where no such combination exists.
Arrays with duplicate pairs: Multiple indices may require swapping the same values to achieve zeros.	The core logic implicitly handles duplicates as each index is processed independently in determining swap possibility.
Arrays with only one element: nums1 and nums2 are both of length 1.	Return true if nums1[0] is 0, or if nums2[0] is 0 and nums1[0] is not 0, return false otherwise.