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Apply Operations to Make Sum of Array Greater Than or Equal to k #1 Most Asked

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Zscaler

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Greedy Algorithms

You are given a positive integer k. Initially, you have an array nums = [1].

You can perform any of the following operations on the array any number of times (possibly zero):

Choose any element in the array and increase its value by 1.
Duplicate any element in the array and add it to the end of the array.

Return the minimum number of operations required to make the sum of elements of the final array greater than or equal to k.

Example 1:

Input: k = 11

Output: 5

Explanation:

We can do the following operations on the array nums = [1]:

Increase the element by 1 three times. The resulting array is nums = [4].
Duplicate the element two times. The resulting array is nums = [4,4,4].

The sum of the final array is 4 + 4 + 4 = 12 which is greater than or equal to k = 11.
The total number of operations performed is 3 + 2 = 5.

Example 2:

Input: k = 1

Output: 0

Explanation:

The sum of the original array is already greater than or equal to 1, so no operations are needed.

Constraints:

1 <= k <= 10⁵

Try coding it in LeetCode

Apply Operations to Make Sum of Array Greater Than or Equal to k #1 Most Asked

Medium

Zscaler

2 views

Topics:

Greedy Algorithms

You are given a positive integer k. Initially, you have an array nums = [1].

You can perform any of the following operations on the array any number of times (possibly zero):

Choose any element in the array and increase its value by 1.
Duplicate any element in the array and add it to the end of the array.

Return the minimum number of operations required to make the sum of elements of the final array greater than or equal to k.

Example 1:

Input: k = 11

Output: 5

Explanation:

We can do the following operations on the array nums = [1]:

Increase the element by 1 three times. The resulting array is nums = [4].
Duplicate the element two times. The resulting array is nums = [4,4,4].

The sum of the final array is 4 + 4 + 4 = 12 which is greater than or equal to k = 11.
The total number of operations performed is 3 + 2 = 5.

Example 2:

Input: k = 1

Output: 0

Explanation:

The sum of the original array is already greater than or equal to 1, so no operations are needed.

Constraints:

1 <= k <= 10⁵

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 constraints on the size of the array and the value of 'k'? Can 'k' be negative?
Can the elements in the array be negative, zero, or floating-point numbers?
If it's impossible to make the sum of the array greater than or equal to 'k' using the given operations, what should I return?
Are there any specific operations that I should prioritize if multiple operation sequences can achieve the target sum?
What data type should I use to represent the numbers in the array and 'k' (e.g., int, long)? Is overflow a concern?

Brute Force Solution

Approach

We need to make the sum of some numbers greater than or equal to a target value by doubling some of them. The brute force method means we will explore every single possible combination of which numbers to double.

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

Start by considering the original set of numbers without doubling any of them. See if their sum is already big enough.
Then, try doubling only the first number. Calculate the new sum and see if it's big enough.
Next, try doubling only the second number, then only the third number, and so on, checking the sum each time.
After trying each number by itself, start trying all pairs of numbers. Double the first and second, then the first and third, and so on, always checking the sum.
Continue this process by trying all possible groups of three numbers, then four numbers, and so on, until you've considered doubling all the numbers at once.
Each time you double a group of numbers, check if the resulting sum is greater than or equal to the target value.
Keep track of the smallest number of doubling operations it took to reach a sum that is big enough.

Code Implementation

def apply_operations_brute_force(numbers, target_value):
    number_of_elements = len(numbers)
    minimum_operations = float('inf')

    for i in range(1 << number_of_elements):
        current_sum = 0
        operations_count = 0
        modified_numbers = list(numbers)

        # Iterate through the bits to determine which numbers to double
        for j in range(number_of_elements):
            if (i >> j) & 1:
                # Double the number if the j-th bit is set
                modified_numbers[j] *= 2
                operations_count += 1

        current_sum = sum(modified_numbers)

        # Check if the current sum meets the target
        if current_sum >= target_value:
            # If target is met, update the minimum operations
            minimum_operations = min(minimum_operations, operations_count)

    if minimum_operations == float('inf'):
        return -1
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

O(2^n) – The described algorithm explores all possible subsets of the input array to determine which elements to double. For an array of size n, there are 2^n possible subsets (each element can either be included or excluded in a subset). The algorithm iterates through each of these subsets, calculating the sum of the modified array for each subset. Therefore, the time complexity is driven by the number of subsets it has to evaluate which grows exponentially with input size n. Consequently, the overall time complexity is O(2^n).

Space Complexity

O(1) – The described brute-force solution explores combinations of numbers to double, but the plain English explanation doesn't mention the use of any auxiliary data structures like lists, sets, or maps to store these combinations. It only involves calculating sums and comparing them against the target value. Therefore, the space complexity is dominated by a few variables used to track the current sum and the minimum number of doublings, which require constant space, irrespective of the input array size N.

Optimal Solution

Approach

The core idea is to use the limited operation wisely to maximize the increase in elements. We prioritize multiplying the smallest numbers by two because this gives the biggest boost to the sum with the operations we are allowed. This avoids testing every possible combination.

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

First, calculate the current total sum of all the numbers.
If the current sum is already greater than or equal to the target, you don't need to do anything, and you are done.
If the current sum is less than the target, find a way to increase it. For this, keep the numbers in order from smallest to largest.
Then, for each operation available, choose the smallest number and double it. After doubling, re-insert the doubled value into the list in its correct sorted position.
Recalculate the total sum of all the numbers. If it's now greater than or equal to the target, you're done, and you can return the number of doubling operations done.
If the total sum is still less than the target, repeat the process of doubling the smallest number and updating the total sum until you reach the target or you have used all of your available operations.
If you have used all the operations and the total sum is still less than the target, it's impossible to reach the target, return -1.
If you have reached the target before using all the operations, return the number of operations it took.

Code Implementation

def apply_operations(numbers, target, operation_limit):
    current_sum = sum(numbers)
    operations_used = 0

    if current_sum >= target:
        return 0

    numbers.sort()

    while current_sum < target and operations_used < operation_limit:
        smallest_number = numbers[0]

        # Prioritize smallest for max sum increase.
        doubled_number = smallest_number * 2
        numbers.pop(0)

        # Insert the doubled number in sorted order.
        import bisect
        bisect.insort(numbers, doubled_number)

        current_sum += smallest_number
        operations_used += 1

        # Check if target has been reached.
        if current_sum >= target:
            return operations_used

    # Impossible if operations are exhausted.
    if current_sum < target:
        return -1

    return operations_used

Big(O) Analysis

Time Complexity

O(n log n + m*n) – Initially, sorting the array of n elements takes O(n log n) time. Then, we have a loop that iterates at most m times, where m is the number of allowed operations. Inside the loop, we double the smallest number (O(1)), and then re-insert it into the sorted list. Re-inserting into the sorted list takes O(n) time because it may require shifting existing elements in the worst case. Therefore, the loop takes O(m*n) time. Combining these two phases, the overall time complexity is O(n log n + m*n).

Space Complexity

O(N) – The algorithm uses a sorted list (implicitly described as 'keeping the numbers in order from smallest to largest') to store the numbers. In the worst-case scenario, this sorted list will contain all N numbers from the input array. Therefore, the auxiliary space required to store this sorted list is proportional to the input size N. This leads to a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since no operations can be performed, indicating the target cannot be reached.
k is already less than or equal to the sum of the array	Return 0 immediately as no operations are needed.
Array contains only zeros and k > 0	Return -1 since the target k can never be reached with only zero elements.
Array contains large positive numbers that could lead to integer overflow when summed.	Use long data types for intermediate sum calculations to prevent overflow and ensure accurate comparison with k.
Array contains only negative numbers and k > sum(array)	If the operation always increases sum, it will eventually meet k or return -1 if sum cannot be increased towards k with operations allowed.
k is extremely large, requiring many operations which might exceed time limits.	Optimize the operation selection to prioritize larger increases in the sum, potentially using a priority queue or other efficient data structure.
Array contains duplicates such that applying operations on different indexes result in same sum changes.	The optimal choice must be selected to keep the operation count minimum, potentially requiring dynamic programming or greedy approach.
No valid solution exists, i.e., the sum of the array can never be greater than or equal to k with permitted operations.	Return -1 after exploring all possible operation combinations or when it's provable that k cannot be reached within the constraints.

Null or empty input array

How to Handle:

Return 0 immediately since no operations can be performed, indicating the target cannot be reached.

k is already less than or equal to the sum of the array

How to Handle:

Return 0 immediately as no operations are needed.

Array contains only zeros and k > 0

How to Handle:

Return -1 since the target k can never be reached with only zero elements.

Array contains large positive numbers that could lead to integer overflow when summed.

How to Handle:

Use long data types for intermediate sum calculations to prevent overflow and ensure accurate comparison with k.

Array contains only negative numbers and k > sum(array)

How to Handle:

If the operation always increases sum, it will eventually meet k or return -1 if sum cannot be increased towards k with operations allowed.

k is extremely large, requiring many operations which might exceed time limits.

How to Handle:

Optimize the operation selection to prioritize larger increases in the sum, potentially using a priority queue or other efficient data structure.

Array contains duplicates such that applying operations on different indexes result in same sum changes.

How to Handle:

The optimal choice must be selected to keep the operation count minimum, potentially requiring dynamic programming or greedy approach.

No valid solution exists, i.e., the sum of the array can never be greater than or equal to k with permitted operations.

How to Handle:

Return -1 after exploring all possible operation combinations or when it's provable that k cannot be reached within the constraints.

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 constraints on the size of the array and the value of 'k'? Can 'k' be negative?
Can the elements in the array be negative, zero, or floating-point numbers?
If it's impossible to make the sum of the array greater than or equal to 'k' using the given operations, what should I return?
Are there any specific operations that I should prioritize if multiple operation sequences can achieve the target sum?
What data type should I use to represent the numbers in the array and 'k' (e.g., int, long)? Is overflow a concern?

Brute Force Solution

Approach

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

Start by considering the original set of numbers without doubling any of them. See if their sum is already big enough.
Then, try doubling only the first number. Calculate the new sum and see if it's big enough.
Next, try doubling only the second number, then only the third number, and so on, checking the sum each time.
After trying each number by itself, start trying all pairs of numbers. Double the first and second, then the first and third, and so on, always checking the sum.
Continue this process by trying all possible groups of three numbers, then four numbers, and so on, until you've considered doubling all the numbers at once.
Each time you double a group of numbers, check if the resulting sum is greater than or equal to the target value.
Keep track of the smallest number of doubling operations it took to reach a sum that is big enough.

Code Implementation

def apply_operations_brute_force(numbers, target_value):
    number_of_elements = len(numbers)
    minimum_operations = float('inf')

    for i in range(1 << number_of_elements):
        current_sum = 0
        operations_count = 0
        modified_numbers = list(numbers)

        # Iterate through the bits to determine which numbers to double
        for j in range(number_of_elements):
            if (i >> j) & 1:
                # Double the number if the j-th bit is set
                modified_numbers[j] *= 2
                operations_count += 1

        current_sum = sum(modified_numbers)

        # Check if the current sum meets the target
        if current_sum >= target_value:
            # If target is met, update the minimum operations
            minimum_operations = min(minimum_operations, operations_count)

    if minimum_operations == float('inf'):
        return -1
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, calculate the current total sum of all the numbers.
If the current sum is already greater than or equal to the target, you don't need to do anything, and you are done.
If the current sum is less than the target, find a way to increase it. For this, keep the numbers in order from smallest to largest.
Then, for each operation available, choose the smallest number and double it. After doubling, re-insert the doubled value into the list in its correct sorted position.
Recalculate the total sum of all the numbers. If it's now greater than or equal to the target, you're done, and you can return the number of doubling operations done.
If the total sum is still less than the target, repeat the process of doubling the smallest number and updating the total sum until you reach the target or you have used all of your available operations.
If you have used all the operations and the total sum is still less than the target, it's impossible to reach the target, return -1.
If you have reached the target before using all the operations, return the number of operations it took.

Code Implementation

def apply_operations(numbers, target, operation_limit):
    current_sum = sum(numbers)
    operations_used = 0

    if current_sum >= target:
        return 0

    numbers.sort()

    while current_sum < target and operations_used < operation_limit:
        smallest_number = numbers[0]

        # Prioritize smallest for max sum increase.
        doubled_number = smallest_number * 2
        numbers.pop(0)

        # Insert the doubled number in sorted order.
        import bisect
        bisect.insort(numbers, doubled_number)

        current_sum += smallest_number
        operations_used += 1

        # Check if target has been reached.
        if current_sum >= target:
            return operations_used

    # Impossible if operations are exhausted.
    if current_sum < target:
        return -1

    return operations_used

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since no operations can be performed, indicating the target cannot be reached.
k is already less than or equal to the sum of the array	Return 0 immediately as no operations are needed.
Array contains only zeros and k > 0	Return -1 since the target k can never be reached with only zero elements.
Array contains large positive numbers that could lead to integer overflow when summed.	Use long data types for intermediate sum calculations to prevent overflow and ensure accurate comparison with k.
Array contains only negative numbers and k > sum(array)	If the operation always increases sum, it will eventually meet k or return -1 if sum cannot be increased towards k with operations allowed.
k is extremely large, requiring many operations which might exceed time limits.	Optimize the operation selection to prioritize larger increases in the sum, potentially using a priority queue or other efficient data structure.
Array contains duplicates such that applying operations on different indexes result in same sum changes.	The optimal choice must be selected to keep the operation count minimum, potentially requiring dynamic programming or greedy approach.
No valid solution exists, i.e., the sum of the array can never be greater than or equal to k with permitted operations.	Return -1 after exploring all possible operation combinations or when it's provable that k cannot be reached within the constraints.

Null or empty input array

How to Handle:

Return 0 immediately since no operations can be performed, indicating the target cannot be reached.

k is already less than or equal to the sum of the array

How to Handle:

Return 0 immediately as no operations are needed.

Array contains only zeros and k > 0

How to Handle:

Return -1 since the target k can never be reached with only zero elements.

Array contains large positive numbers that could lead to integer overflow when summed.

How to Handle:

Use long data types for intermediate sum calculations to prevent overflow and ensure accurate comparison with k.

Array contains only negative numbers and k > sum(array)

How to Handle:

If the operation always increases sum, it will eventually meet k or return -1 if sum cannot be increased towards k with operations allowed.

k is extremely large, requiring many operations which might exceed time limits.

How to Handle:

Optimize the operation selection to prioritize larger increases in the sum, potentially using a priority queue or other efficient data structure.

Array contains duplicates such that applying operations on different indexes result in same sum changes.

How to Handle:

The optimal choice must be selected to keep the operation count minimum, potentially requiring dynamic programming or greedy approach.

No valid solution exists, i.e., the sum of the array can never be greater than or equal to k with permitted operations.

How to Handle:

Return -1 after exploring all possible operation combinations or when it's provable that k cannot be reached within the constraints.

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