Transform Array to All Equal Elements

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ArraysGreedy Algorithms

You are given an integer array nums of size n containing only 1 and -1, and an integer k.

You can perform the following operation at most k times:

Choose an index i (0 <= i < n - 1), and multiply both nums[i] and nums[i + 1] by -1.

Note that you can choose the same index i more than once in different operations.

Return true if it is possible to make all elements of the array equal after at most k operations, and false otherwise.

Example 1:

Input: nums = [1,-1,1,-1,1], k = 3

Output: true

Explanation:

We can make all elements in the array equal in 2 operations as follows:

Choose index i = 1, and multiply both nums[1] and nums[2] by -1. Now nums = [1,1,-1,-1,1].
Choose index i = 2, and multiply both nums[2] and nums[3] by -1. Now nums = [1,1,1,1,1].

Example 2:

Input: nums = [-1,-1,-1,1,1,1], k = 5

Output: false

Explanation:

It is not possible to make all array elements equal in at most 5 operations.

Constraints:

1 <= n == nums.length <= 10⁵
nums[i] is either -1 or 1.
1 <= k <= n

Try coding it in LeetCode

Transform Array to All Equal Elements

Medium

Flipkart

2 views

Topics:

ArraysGreedy Algorithms

You are given an integer array nums of size n containing only 1 and -1, and an integer k.

You can perform the following operation at most k times:

Choose an index i (0 <= i < n - 1), and multiply both nums[i] and nums[i + 1] by -1.

Note that you can choose the same index i more than once in different operations.

Return true if it is possible to make all elements of the array equal after at most k operations, and false otherwise.

Example 1:

Input: nums = [1,-1,1,-1,1], k = 3

Output: true

Explanation:

We can make all elements in the array equal in 2 operations as follows:

Choose index i = 1, and multiply both nums[1] and nums[2] by -1. Now nums = [1,1,-1,-1,1].
Choose index i = 2, and multiply both nums[2] and nums[3] by -1. Now nums = [1,1,1,1,1].

Example 2:

Input: nums = [-1,-1,-1,1,1,1], k = 5

Output: false

Explanation:

It is not possible to make all array elements equal in at most 5 operations.

Constraints:

1 <= n == nums.length <= 10⁵
nums[i] is either -1 or 1.
1 <= k <= n

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 data type of the elements in the array, and what is the range of possible values for those elements?
Can the input array be empty or null? If so, what should the function return?
Are we trying to minimize the number of operations, or is any valid solution acceptable?
If the array is already composed of all equal elements, should the function return 0 or take any action?
Can I assume the array is mutable and I can modify it directly, or should I create a copy?

Brute Force Solution

Approach

To make all numbers in the list the same, the most basic approach is to try forcing each number to be the target value one at a time. We'll check how many changes it takes for each potential target to transform all the numbers in the list.

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

First, pick the very first number from the list and imagine that we want every number to be the same as that first number.
Go through the whole list and count how many numbers we'd have to change to match that first number.
Now, repeat this process, but this time, pick the second number from the list and imagine that we want every number to be the same as that second number. Again count how many changes would be required.
Continue this process for every single number in the original list. Each time, keep track of the number of changes it would take.
Finally, compare all the counts. The smallest number of changes we found is the answer. That smallest number represents the fewest changes needed to make all the numbers the same.

Code Implementation

def transform_array_to_all_equal_elements(numbers):
    minimum_changes = float('inf')
    
    for target_value_index in range(len(numbers)):
        target_value = numbers[target_value_index]
        changes_needed = 0
        
        # Iterate through the numbers to count differences.
        for current_number_index in range(len(numbers)):
            if numbers[current_number_index] != target_value:
                changes_needed += 1

        # Update the minimum if we found a smaller change count.
        if changes_needed < minimum_changes:
            minimum_changes = changes_needed

    return minimum_changes

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input array of size n. For each element, it iterates through the entire array again to count the number of changes needed to make all elements equal to the current element. This results in nested loops, where the outer loop iterates n times and the inner loop iterates n times for each iteration of the outer loop. Therefore, the time complexity is proportional to n * n, which simplifies to O(n²).

Space Complexity

O(1) – The algorithm iterates through the input list, comparing each element to every other element. It only uses a few integer variables to store the current element being considered as the target and the count of changes needed. No auxiliary data structures like lists, hash maps, or recursion are used. Therefore, the space complexity is constant and does not depend on the input size N.

Optimal Solution

Approach

The core idea is to figure out which number appears most often in the list. Then, we just need to change all the other numbers to match that most frequent number. This avoids unnecessary changes and leads to the fewest operations.

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

Count how many times each number appears in the list.
Find the number that appears most often.
Calculate how many numbers are *not* the most frequent number.
The number of non-frequent numbers is the answer - the fewest changes needed.

Code Implementation

def transform_array(number_array):
    element_counts = {}
    for element in number_array:
        if element in element_counts:
            element_counts[element] += 1
        else:
            element_counts[element] = 1

    most_frequent_count = 0
    # Find the element with the highest frequency.
    for element_count in element_counts.values():
        if element_count > most_frequent_count:
            most_frequent_count = element_count

    # Calculate the number of elements to transform.
    elements_to_transform = len(number_array) - most_frequent_count

    # This is the minimum number of operations required.
    return elements_to_transform

Big(O) Analysis

Time Complexity

O(n) – The provided solution involves iterating through the input array (of size n) to count the frequency of each number. This initial counting process takes O(n) time. Finding the most frequent number then requires iterating through the frequency counts, which, in the worst case, could involve checking all n unique numbers, still resulting in O(n) time. Finally, calculating the number of non-frequent elements is done by looking again at all n elements. The dominant operation is directly proportional to the number of elements in the array, so the overall time complexity simplifies to O(n).

Space Complexity

O(N) – The algorithm uses a data structure (likely a hash map or dictionary) to count the frequency of each number in the input list. In the worst case, all N numbers in the input array are distinct, requiring the frequency counter to store N key-value pairs. Therefore, the auxiliary space used is proportional to the size of the input array, N. This results in a space complexity of O(N).

Edge Cases

Case	How to Handle
Null input array	Throw IllegalArgumentException or return null to explicitly signify the invalid input
Empty array	Return 0, since no operations are needed to make an empty array equal
Array with one element	Return 0, as a single-element array is already considered equal
Array with all identical elements	Return 0, as no transformations are needed
Array with maximum integer values (Integer.MAX_VALUE)	Handle potential integer overflow when calculating differences and sums during transformations.
Array with minimum integer values (Integer.MIN_VALUE)	Handle potential integer overflow when calculating differences and sums during transformations.
Large array size approaching memory limits	Ensure the algorithm uses memory efficiently, avoiding excessive memory allocation that can cause OutOfMemoryErrors.
Array with a mix of positive, negative, and zero values	The solution must handle the variety of number types correctly without causing unexpected errors