Minimum Numbers of Function Calls to Make Target Array

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ArraysGreedy AlgorithmsBit Manipulation

You are given an integer array nums. You have an integer array arr of the same length with all values set to 0 initially. You also have the following modify function:

You want to use the modify function to convert arr to nums using the minimum number of calls.

Return the minimum number of function calls to make nums from arr.

The test cases are generated so that the answer fits in a 32-bit signed integer.

Example 1:

Input: nums = [1,5]
Output: 5
Explanation: Increment by 1 (second element): [0, 0] to get [0, 1] (1 operation).
Double all the elements: [0, 1] -> [0, 2] -> [0, 4] (2 operations).
Increment by 1 (both elements)  [0, 4] -> [1, 4] -> [1, 5] (2 operations).
Total of operations: 1 + 2 + 2 = 5.

Example 2:

Input: nums = [2,2]
Output: 3
Explanation: Increment by 1 (both elements) [0, 0] -> [0, 1] -> [1, 1] (2 operations).
Double all the elements: [1, 1] -> [2, 2] (1 operation).
Total of operations: 2 + 1 = 3.

Example 3:

Input: nums = [4,2,5]
Output: 6
Explanation: (initial)[0,0,0] -> [1,0,0] -> [1,0,1] -> [2,0,2] -> [2,1,2] -> [4,2,4] -> [4,2,5](nums).

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹

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 is the range of values within the target array? Can they be negative, zero, or only positive?
What should I return if the target array is empty or null?
Are all elements in the target array guaranteed to be non-negative integers?
Could you provide a small example to illustrate the function calls required for a given target array?
Is there a limit to the size of the target array?

Brute Force Solution

Approach

The brute force strategy is to explore every possible sequence of operations to reach the target. It involves checking all combinations of incrementing elements and doubling the entire result to see if the target array can be achieved. This method is exhaustive but not efficient.

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

Start with an array of all zeros with the same size as the target.
Try incrementing each element of this initial array one by one.
At each step, also consider doubling the entire array.
Keep track of the number of operations (increments and doubles) needed to reach the current state.
Repeat the process of incrementing individual elements and doubling, exploring all possible sequences of these operations.
For each sequence that results in the target array, record the total number of operations.
Compare the number of operations from all the successful sequences, and choose the sequence with the fewest operations as the answer.

Code Implementation

def min_operations_brute_force(target_array):
    def solve(current_array, operations_count):
        if current_array == target_array:
            return operations_count

        # If the array is larger than the target, it's not a valid path
        if all(current_array[i] > target_array[i] for i in range(len(target_array))):
            return float('inf')

        min_ops = float('inf')

        # Try incrementing each element
        for index in range(len(target_array)):
            new_array = current_array[:]
            new_array[index] += 1
            min_ops = min(min_ops, solve(new_array, operations_count + 1))

        # Array cannot be doubled if any element is non-zero.
        if all(element == 0 for element in current_array):
            return min_ops

        # Try doubling the array
        doubled_array = [element * 2 for element in current_array]
        
        # Prevent overflow and unnecessary calculations when doubling results in array > target
        min_ops = min(min_ops, solve(doubled_array, operations_count + 1))

        return min_ops

    array_length = len(target_array)
    initial_array = [0] * array_length

    # Start with an array of zeros and find the optimal solution
    min_operations = solve(initial_array, 0)

    return min_operations

Big(O) Analysis

Time Complexity

O(2^(n*m)) – The brute force approach explores all possible sequences of incrementing elements and doubling the entire array, where n is the size of the target array, and m is the maximum value among the target array elements. Each element can be incremented individually and the entire array can be doubled, leading to exponential branching. The number of possible operations grows exponentially with both the size of the array and the magnitude of the numbers we need to reach. The exploration of all these sequences makes the time complexity exponential, approximately O(2^(n*m)).

Space Complexity

O(N!) – The brute force approach explores every possible sequence of incrementing and doubling operations using a recursive process. In the worst-case scenario, the algorithm might need to explore a large number of operation sequences, potentially branching out into N! possibilities if N represents the maximum value in the target array. Therefore, the space complexity is dominated by the call stack depth needed to explore all such operation sequences, which could grow up to O(N!).

Optimal Solution

Approach

The most efficient way to reach the target array is to build it from the ground up using two basic operations: incrementing all elements and doubling all elements. The key idea is to work backward, reducing the target array until you reach an array of zeros, counting the operations as you go.

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

Determine the maximum number in the target array.
Increment a counter to record each 'increment all elements' operation.
If all elements in the array are even, divide each element by 2, this represents a 'double all elements' operation and increment another counter.
Repeat this process of checking for even numbers, dividing them, and finding the new largest number, until all numbers in the array are zero.
The total number of increment all and double all operations is the minimum number of operations needed to reach the target array from an array of zeros.

Code Implementation

def min_operations(target_array):
    increment_operations_count = 0
    double_operations_count = 0
    
    while True:
        all_zeros = True
        all_even = True

        for element in target_array:
            if element != 0:
                all_zeros = False
            if element % 2 != 0:
                all_even = False

        if all_zeros:
            break

        if not all_even:
            # Increment to make all values even.
            for i in range(len(target_array)):
                if target_array[i] % 2 != 0:
                    target_array[i] -= 1
                    increment_operations_count += 1

        else:
            # Divide all elements by 2.
            for i in range(len(target_array)):
                target_array[i] //= 2
            double_operations_count += 1

    return increment_operations_count + double_operations_count

Big(O) Analysis

Time Complexity

O(n + log(max_val) * n) – The outer loop's iterations depend on how many times the largest number (max_val) in the array can be divided by 2 before reaching zero. This equates to log₂(max_val). Inside this loop, we iterate through all 'n' elements of the array once to either decrement odd numbers to make them even (which requires increment operations) or to divide each even number by 2. Therefore, the total time complexity is dominated by log₂(max_val) * n, plus the initial pass through the array to determine the maximum value (O(n)). Thus, the overall time complexity is O(n + log(max_val) * n).

Space Complexity

O(1) – The algorithm operates primarily in place on the input array. It uses a few integer variables to store counters and temporary values, but the amount of memory used by these variables is constant regardless of the size of the input array (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty target array	Return 0 since no operations are needed to reach an empty target.
Target array with a single element	Return the value of that element, as that is how many increment operations are needed.
Target array with all zeros	Return 0, as the array is already in the desired state.
Target array with large numbers that could lead to integer overflow during addition	Ensure the data type used for intermediate calculations (e.g., sum, max) can accommodate large values to prevent overflow, or consider using modulo if allowed.
Target array with a mix of even and odd numbers	The solution should correctly handle both increment and multiplication operations regardless of the even/odd mix.
Target array with all even numbers	Repeatedly divide all elements by 2 until at least one element is odd or zero.
Target array with very large values that may cause time limit exceed	Optimize the number of division operations based on the max value; dividing until all values are < max value will reduce operations.
Target array with consecutive numbers with small difference	Increment will be called frequently, and the number of increments need to be minimized

Null or empty target array

How to Handle:

Return 0 since no operations are needed to reach an empty target.

Target array with a single element

How to Handle:

Return the value of that element, as that is how many increment operations are needed.

Target array with all zeros

How to Handle:

Return 0, as the array is already in the desired state.

Target array with large numbers that could lead to integer overflow during addition

How to Handle:

Ensure the data type used for intermediate calculations (e.g., sum, max) can accommodate large values to prevent overflow, or consider using modulo if allowed.

Target array with a mix of even and odd numbers

How to Handle:

The solution should correctly handle both increment and multiplication operations regardless of the even/odd mix.

Target array with all even numbers

How to Handle:

Repeatedly divide all elements by 2 until at least one element is odd or zero.

Target array with very large values that may cause time limit exceed

How to Handle:

Optimize the number of division operations based on the max value; dividing until all values are < max value will reduce operations.

Target array with consecutive numbers with small difference

How to Handle:

Increment will be called frequently, and the number of increments need to be minimized