Zero Array Transformation I

You are given an integer array nums of length n and a 2D array queries, where queries[i] = [lᵢ, rᵢ]. For each queries[i]:

• Select a subset of indices within the range [lᵢ, rᵢ] 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 <= 10^5
• 0 <= nums[i] <= 10^5
• 1 <= queries.length <= 10^5
• queries[i].length == 2
• 0 <= lᵢ <= rᵢ < nums.length

Let's analyze this problem. A naive approach would be to try all possible subsets for each query. However, this would be highly inefficient and would likely result in a Time Limit Exceeded (TLE) error.

Naive Approach For each query [l, r], iterate through all possible subsets of the range [l, r] and decrement the elements at the selected indices. Check if it's possible to make the array zero after processing all the queries.

This approach would have exponential time complexity, which is not feasible for the given constraints.

Optimal Approach

The crucial idea is to consider the differences between adjacent elements in the nums array. Let diff[i] = nums[i] - nums[i-1] for i > 0, and diff[0] = nums[0]. We can transform the nums array into a zero array if and only if we can satisfy the queries while respecting the differences.

For each query [l, r], we need to determine how to use it to adjust the differences within that range. We can think of this as distributing the values in nums to the right using the queries. It is possible if and only if for each prefix, the sum of nums is non-negative.

Let's walk through an example: nums = [4,3,2,1], queries = [[1,3],[0,2]]

1. First Query [1, 3]: We want to decrease nums[1], nums[2], and nums[3] by some amounts such that we can potentially make the array zero. In other words, we can choose any subset of indices from 1 to 3 and decrement them by 1.
2. Second Query [0, 2]: Now we operate on the range 0 to 2.

Here's the detailed algorithm:

1. Create a difference array: diff[i] = nums[i] - nums[i-1] (where nums[-1] = 0).
2. Iterate through queries:
   • For each query [l, r], calculate the prefix sum of nums and try to make it zero.

def solve():
    nums = [4, 3, 2, 1]
    queries = [[1, 3], [0, 2]]

    def check_possible(arr, queries):
        for l, r in queries:
            possible = False

            # Try all subsets of indices in [l, r]
            for i in range(1 << (r - l + 1)):
                temp_arr = arr[:]
                subset = []
                for j in range(r - l + 1):
                    if (i >> j) & 1:
                        subset.append(l + j)

                # Decrement the values at the selected indices by 1
                for index in subset:
                    temp_arr[index] -= 1

                # Check if the array is zero
                if all(x == 0 for x in temp_arr):
                    possible = True
                    break

            if not possible:
                return False

        return True

    if check_possible(nums, queries):
        print("true")
    else:
        print("false")


solve()

def solve():
    nums = [1, 0, 1]
    queries = [[0, 2]]

    def check_possible(arr, queries):
        for l, r in queries:
            possible = False
            for i in range(1 << (r - l + 1)):
                temp_arr = arr[:]
                subset = []
                for j in range(r - l + 1):
                    if (i >> j) & 1:
                        subset.append(l + j)
                for index in subset:
                    temp_arr[index] -= 1
                if all(x == 0 for x in temp_arr):
                    possible = True
                    break
            if not possible:
                return False
        return True

    if check_possible(nums, queries):
        print("true")
    else:
        print("false")

solve()

def is_possible_to_transform(nums, queries):
    n = len(nums)
    diff = [0] * n
    diff[0] = nums[0]
    for i in range(1, n):
        diff[i] = nums[i] - nums[i - 1]

    for l, r in queries:
        can_transform = False

        # Iterate through all possible subsets of indices in [l, r]
        for i in range(1 << (r - l + 1)):
            temp_diff = diff[:]
            subset = []

            # Build the subset of indices to decrement
            for j in range(r - l + 1):
                if (i >> j) & 1:
                    subset.append(l + j)

            # Create a temporary array to simulate the decrements
            temp_nums = [0] * n
            temp_nums[0] = temp_diff[0]
            for k in range(1, n):
                temp_nums[k] = temp_diff[k] + temp_nums[k - 1]

            # Decrement the selected indices by 1 in the temporary array
            for index in subset:
                temp_nums[index] -= 1

            # Check if it is possible to transform to zero array with these subsets
            temp_diff = [0] * n
            temp_diff[0] = temp_nums[0]
            for k in range(1, n):
                temp_diff[k] = temp_nums[k] - temp_nums[k - 1]

            can_transform = True

            is_all_zero = True
            temp_nums = [0] * n
            temp_nums[0] = temp_diff[0]
            if temp_nums[0] < 0:
                is_all_zero = False
            for k in range(1, n):
                temp_nums[k] = temp_diff[k] + temp_nums[k-1]
                if temp_nums[k] < 0:
                    is_all_zero = False
            if is_all_zero and all([x==0 for x in temp_nums]):
                return True
    return False

Big(O) Run-time Analysis

The time complexity of the given is_possible_to_transform function is quite high due to the nested loops and the subset generation. Let's break it down:

1. Difference Array Creation: Creating the difference array diff takes O(n) time, where n is the length of the nums array. This is because we iterate through the nums array once to calculate the differences.

2. Iterating Through Queries: The outer loop iterates through each query in the queries list. Let q be the number of queries. So, this loop contributes O(q) to the overall time complexity.

3. Subset Generation: Inside the outer loop, for each query [l, r], we generate all possible subsets of the indices within the range [l, r]. The number of subsets of a set with k elements is 2^k. In this case, k = r - l + 1. Therefore, the time complexity of generating all subsets is O(2^(r-l+1)).

4. Simulating Decrements and Checking for Zero Array: For each subset, we create a temporary array temp_nums and simulate the decrements. Then, we check if it’s a zero array. These operations take O(n) time because we iterate through the entire array to perform decrements and check for zero.

Combining these components:

• O(n) for creating the initial difference array.
• O(q) for iterating through queries.
• For each query: O(2^(r-l+1)) for generating subsets.
• For each subset: O(n) for simulating decrements and checking for zero.

Thus, the overall time complexity is: O(n + q * 2^(r-l+1) * n)

In the worst case, r - l + 1 can be close to n, so the time complexity can approach: O(n + q * 2^n * n) which simplifies to O(q * n * 2^n)

This is an exponential time complexity, making it highly inefficient for larger input sizes.

Big(O) Space Usage Analysis

1. Difference Array: The diff array requires O(n) space, where n is the length of the nums array.

2. Temporary Arrays: Inside the main loop, we create temporary arrays temp_nums and temp_diff, each of size n. Thus, this contributes O(n) space.

3. Subset Storage: The subset list stores the indices of the current subset, which in the worst case can be of size n. So it contributes O(n) space.

Therefore, the overall space complexity is: O(n + n + n) = O(n)

The space complexity is linear with respect to the size of the input array nums.

Edge Cases

1. Empty nums array: If nums is empty, return true since an empty array is considered a zero array.
2. Empty queries array: If queries is empty, return true if nums is already a zero array, otherwise false.
3. Large ranges in queries: If r - l is large for some queries, the subset generation will take exponential time. The algorithm will become inefficient. There are no specific code changes to handle this; rather, this limitation should be considered when using this function.
4. Negative numbers in nums: The algorithm assumes non-negative numbers. If negative numbers are present and the problem requires transforming to exactly zero, the approach may need adjustments.
5. Large values in nums: If nums contains very large values, the intermediate computations might cause overflow issues. In this case, you may consider using appropriate data types or scaling down the input, if possible.