Reduction Operations to Make the Array Elements Equal

#236 Most AskedMedium

Reduction Operations to Make the Array Elements Equal #236 Most Asked

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ArraysGreedy Algorithms

Given an integer array nums, your goal is to make all elements in nums equal. To complete one operation, follow these steps:

Find the largest value in nums. Let its index be i (0-indexed) and its value be largest. If there are multiple elements with the largest value, pick the smallest i.
Find the next largest value in nums strictly smaller than largest. Let its value be nextLargest.
Reduce nums[i] to nextLargest.

Return the number of operations to make all elements in nums equal.

Example 1:

Input: nums = [5,1,3]
Output: 3
Explanation: It takes 3 operations to make all elements in nums equal:
1. largest = 5 at index 0. nextLargest = 3. Reduce nums[0] to 3. nums = [3,1,3].
2. largest = 3 at index 0. nextLargest = 1. Reduce nums[0] to 1. nums = [1,1,3].
3. largest = 3 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1].

Example 2:

Input: nums = [1,1,1]
Output: 0
Explanation: All elements in nums are already equal.

Example 3:

Input: nums = [1,1,2,2,3]
Output: 4
Explanation: It takes 4 operations to make all elements in nums equal:
1. largest = 3 at index 4. nextLargest = 2. Reduce nums[4] to 2. nums = [1,1,2,2,2].
2. largest = 2 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1,2,2].
3. largest = 2 at index 3. nextLargest = 1. Reduce nums[3] to 1. nums = [1,1,1,1,2].
4. largest = 2 at index 4. nextLargest = 1. Reduce nums[4] to 1. nums = [1,1,1,1,1].

Constraints:

1 <= nums.length <= 5 * 10⁴
1 <= nums[i] <= 5 * 10⁴

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 for the integers in the input array?
Can the input array be empty or null?
Are duplicate values allowed in the input array, and if so, how should they be handled?
Can you provide a concrete example of an input array and the expected output to illustrate the desired behavior?
Is the input array guaranteed to contain at least two distinct elements?

Brute Force Solution

Approach

The brute force method involves trying every possible combination of operations to make all numbers in the group equal. We essentially explore every conceivable path until we find one that leads to the desired outcome: a group of identical numbers.

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

Find the largest number in the group.
Consider this largest number as the target value that all other numbers must eventually equal.
For each number in the group that is not equal to the largest number, simulate the process of repeatedly reducing the largest number until it matches the current number.
Count how many reduction operations are needed for each such number.
Sum up all those counts to get the total number of operations needed to make all numbers equal to the current largest number.
Repeat this entire process, but consider the second largest number, then the third largest number, and so on as our target.
For each target, remember the total number of reduction operations needed.
Finally, pick the target that required the smallest number of reduction operations to make every number equal to it. That's your answer.

Code Implementation

def reduction_operations_brute_force(numbers):
    unique_numbers = sorted(list(set(numbers)))
    minimum_operations = float('inf')

    # Iterate through each unique number as a potential target
    for target_value in unique_numbers:
        total_operations = 0

        # Calculate operations needed to reach the current target
        for number in numbers:
            operations_needed = 0
            current_number = number

            # Simulate reduction operations
            while current_number != target_value:
                current_number = unique_numbers[unique_numbers.index(current_number) - 1]
                operations_needed += 1

            total_operations += operations_needed

        # Store the minimum number of required operations
        minimum_operations = min(minimum_operations, total_operations)

    return minimum_operations

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each distinct number in the input array of size n as a potential target value. For each target value, it then iterates through the array again to simulate reduction operations on elements that are larger than the current target. In the worst case, each element might require a reduction step down to the target, resulting in approximately n operations for each target. Since we have n potential targets, the overall time complexity becomes O(n*n), which simplifies to O(n²).

Space Complexity

O(1) – The described brute force algorithm primarily uses a fixed number of variables to store the largest number, the target value, loop counters, and the number of reduction operations needed. No auxiliary data structures such as lists, hash maps, or recursion are used. Therefore, the space required remains constant irrespective of the input array's size N, resulting in a space complexity of O(1).

Optimal Solution

Approach

The goal is to make all numbers in a group the same by repeatedly changing the largest number. The most efficient way is to figure out how many unique values there are, and each of these unique values (besides the smallest one) will require a change.

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

First, identify all the different numbers that appear in the group.
Then, count how many unique numbers there are.
Finally, since we don't need to change the smallest number, the number of operations needed is one less than the total number of different numbers.

Code Implementation

def reductionOperations(numbers) -> int:
    unique_numbers = set(numbers)

    # Sorting allows counting operations based on value rank
    sorted_unique_numbers = sorted(list(unique_numbers))

    number_of_unique_numbers = len(sorted_unique_numbers)

    # No operations needed if all elements are the same
    if number_of_unique_numbers <= 1:
        return 0

    # The result will store total operations needed
    total_operations = 0

    # Iterate through array and increment for higher numbers
    for number in numbers:
        for i in range(1, number_of_unique_numbers):
            if number > sorted_unique_numbers[0] and number == sorted_unique_numbers[i]:
                total_operations += number_of_unique_numbers - i

    # This is the alternative approach
    number_of_operations = number_of_unique_numbers - 1

    # This is to count unique values to get operations
    counts = {}
    for number in numbers:
        counts[number] = counts.get(number, 0) + 1
    sorted_counts = sorted(counts.keys())

    # Calculates the total number of operations needed
    operations = 0
    for i in range(len(sorted_counts) - 1, 0, -1):
        operations += counts[sorted_counts[i]]

    # This is the optimal approach
    unique_values = sorted(list(set(numbers)))

    # Each value except min needs reduction operation
    return len(unique_values) - 1

Big(O) Analysis

Time Complexity

O(n log n) – Identifying unique elements in an array of size n typically involves sorting the array first, which has a time complexity of O(n log n). After sorting, finding the number of distinct elements can be done in O(n) time by iterating through the sorted array. Since O(n log n) dominates O(n), the overall time complexity is O(n log n).

Space Complexity

O(N) – The algorithm identifies unique numbers from the input. In the worst case, all N elements of the input array could be unique, requiring a data structure (like a set) to store these unique values. Therefore, the auxiliary space used to store the unique numbers grows linearly with the input size N. The space complexity is proportional to the number of unique elements in the array, which can be at most N.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since no operations are needed on an empty array.
Array with only one element	Return 0 immediately as all elements are already equal.
Array with all identical elements	Return 0 immediately, as no reduction operations are required.
Array with maximum integer values	The algorithm relies on comparisons, and the number of operations. Ensure there isn't overflow during the accumulation of the number of steps/operations, using long if necessary.
Large array size that could cause a time complexity issue when sorting (if sorting is used)	Ensure that the chosen sorting algorithm (if used) is efficient (e.g., O(n log n)), or consider a counting sort approach if the range of numbers is limited.
Negative numbers in the array	The algorithm should work correctly with negative numbers as it relies on comparing the numbers rather than their absolute values.
Array contains duplicate minimum values.	Since we want to reduce everything to the minimum, these should be skipped from the start so algorithm works.
Extremely skewed distribution of values	The algorithm's performance might be impacted if most elements are significantly larger than the smallest, possibly leading to a large number of reduction operations and in this instance no changes are needed to the algorithm itself.

Null or empty input array

How to Handle:

Return 0 immediately since no operations are needed on an empty array.

Array with only one element

How to Handle:

Return 0 immediately as all elements are already equal.

Array with all identical elements

How to Handle:

Return 0 immediately, as no reduction operations are required.

Array with maximum integer values

How to Handle:

The algorithm relies on comparisons, and the number of operations. Ensure there isn't overflow during the accumulation of the number of steps/operations, using long if necessary.

Large array size that could cause a time complexity issue when sorting (if sorting is used)

How to Handle:

Ensure that the chosen sorting algorithm (if used) is efficient (e.g., O(n log n)), or consider a counting sort approach if the range of numbers is limited.

Negative numbers in the array

How to Handle:

The algorithm should work correctly with negative numbers as it relies on comparing the numbers rather than their absolute values.

Array contains duplicate minimum values.

How to Handle:

Since we want to reduce everything to the minimum, these should be skipped from the start so algorithm works.

Extremely skewed distribution of values

How to Handle:

The algorithm's performance might be impacted if most elements are significantly larger than the smallest, possibly leading to a large number of reduction operations and in this instance no changes are needed to the algorithm itself.