How do you solve Minimum Operations to Transform Array on LeetCode?

Minimum Operations to Transform Array (LeetCode #3724) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The last element of nums2 must be appended.

What is the time complexity of Minimum Operations to Transform Array?

The optimal solution for Minimum Operations to Transform Array runs in O(n) time and uses O(n) space. The complexity is linear since we only traverse the arrays once after appending the last element.

What is the brute force solution for Minimum Operations to Transform Array?

The brute force approach for Minimum Operations to Transform Array is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible combinations of operations on nums1 to match nums2. This involves checking every index for appending and adjusting values, leading to many redundant calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Transform Array?

Minimum Operations to Transform Array uses the following concepts: Array, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Transform Array?

Clarify the operations allowed. Think about edge cases, like empty arrays. Explain your thought process clearly.

What are common mistakes when solving Minimum Operations to Transform Array?

Overlooking the need to append the last element. Not accounting for all elements when calculating operations.

#3724

Minimum Operations to Transform Array

Medium

ArrayGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

We can directly calculate the necessary increments and decrements by focusing on the last element of nums2 and matching it with an element from nums1. This reduces unnecessary operations.

⚙️

Algorithm

3 steps

1Step 1: Append the last element of nums2 to nums1.
2Step 2: For each element in nums2, calculate the difference with the corresponding element in nums1.
3Step 3: Sum the absolute differences to get the total operations needed.

solution.py4 lines

1def minOperations(nums1, nums2):
2    nums1.append(nums2[-1])
3    ops = sum(abs(nums1[i] - nums2[i]) for i in range(len(nums2)))
4    return ops

ℹ

Complexity note: The complexity is linear since we only traverse the arrays once after appending the last element.

1The last element of nums2 must be appended.
2Directly matching elements reduces unnecessary operations.

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