How do you solve Transform Array to All Equal Elements on LeetCode?

Transform Array to All Equal Elements (LeetCode #3576) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Mismatches can be fixed in pairs.

What is the time complexity of Transform Array to All Equal Elements?

The optimal solution for Transform Array to All Equal Elements runs in O(n) time and uses O(1) space. The linear complexity comes from a single pass through the array to count operations.

What is the brute force solution for Transform Array to All Equal Elements?

The brute force approach for Transform Array to All Equal Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. Try all combinations of operations to see if we can make all elements equal. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Transform Array to All Equal Elements?

Transform Array to All Equal Elements uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Transform Array to All Equal Elements?

Explain your thought process clearly. Discuss edge cases, such as all elements being the same.

What are common mistakes when solving Transform Array to All Equal Elements?

Not accounting for the last element when checking pairs. Overestimating the number of operations needed.

#3576

Transform Array to All Equal Elements

Medium

ArrayGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Use a greedy approach to fix mismatches as we traverse the array. Count the operations needed to make all elements equal.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for operations needed.
2Step 2: Traverse the array and whenever a mismatch occurs (1 followed by -1), increment the operation counter and flip the next element.
3Step 3: After traversal, check if the total operations are within k.

solution.py7 lines

1def canTransform(nums, k):
2    ops = 0
3    for i in range(len(nums) - 1):
4        if nums[i] != nums[i + 1]:
5            ops += 1
6            nums[i + 1] *= -1
7    return ops <= k

ℹ

Complexity note: The linear complexity comes from a single pass through the array to count operations.

1Mismatches can be fixed in pairs.
2The number of operations needed can be counted in a single pass.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.