Is Minimum Operations to Transform Array easy or hard?

Minimum Operations to Transform Array is typically classified as a Medium problem. The implementation is straightforward once you recognize the greedy pattern, but identifying why the minimal local adjustment leads to an optimal result can take some reasoning.

Minimum Operations to Transform Array Python/Java solution

Most implementations follow the same structure across languages: iterate through the array, track the previous valid value, and accumulate required adjustments. The logic is identical in Python, Java, C++, Go, and TypeScript, differing only in syntax.

How to solve Minimum Operations to Transform Array in O(n)?

Scan the array once while keeping track of the previous transformed value. For each element, check whether it satisfies the required ordering condition. If it does not, compute the minimal adjustment needed and add that value to the operation counter. Updating the previous value after each step guarantees the entire process completes in linear time.

What is the best approach for Minimum Operations to Transform Array?

The greedy single-pass approach is the most efficient solution. Traverse the array from left to right and maintain the previous valid value. If the current element violates the required constraint, minimally adjust it and add the difference to the operation count. This runs in O(n) time with O(1) extra space.

Is Minimum Operations to Transform Array asked at Google/Amazon/Meta?

Greedy array transformation problems appear frequently in interviews at companies like Amazon, Google, and Meta. Variations of this problem test whether candidates can recognize when a local greedy adjustment leads to a globally optimal solution.

What data structure is used in Minimum Operations to Transform Array?

The solution primarily uses a simple array traversal with a few scalar variables. No advanced data structures are required. The key idea is applying a greedy strategy while iterating through the array.

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

The optimal greedy solution runs in O(n) time because the array is processed once. Only a constant amount of extra memory is used to track the previous valid value and the accumulated operation count, giving O(1) space complexity.

Description Hints 5 Solution Video 9:15

3724. Minimum Operations to Transform Array

Medium39.5% acceptance

You are given two integer arrays nums1 of length n and nums2 of length n + 1.

You want to transform nums1 into nums2 using the minimum number of operations.

You may perform the following operations any number of times, each time choosing an index i:

Increase nums1[i] by 1.
Decrease nums1[i] by 1.
Append nums1[i] to the end of the array.

Return the minimum number of operations required to transform nums1 into nums2.

Example 1:

Input: nums1 = [2,8], nums2 = [1,7,3]

Output: 4

Explanation:

Step	`i`	Operation	`nums1[i]`	Updated `nums1`
1	0	Append	-	[2, 8, 2]
2	0	Decrement	Decreases to 1	[1, 8, 2]
3	1	Decrement	Decreases to 7	[1, 7, 2]
4	2	Increment	Increases to 3	[1, 7, 3]

Thus, after 4 operations nums1 is transformed into nums2.

Example 2:

Input: nums1 = [1,3,6], nums2 = [2,4,5,3]

Output: 4

Explanation:

Step	`i`	Operation	`nums1[i]`	Updated `nums1`
1	1	Append	-	[1, 3, 6, 3]
2	0	Increment	Increases to 2	[2, 3, 6, 3]
3	1	Increment	Increases to 4	[2, 4, 6, 3]
4	2	Decrement	Decreases to 5	[2, 4, 5, 3]

Thus, after 4 operations nums1 is transformed into nums2.

Example 3:

Input: nums1 = [2], nums2 = [3,4]

Output: 3

Explanation:

Step	`i`	Operation	`nums1[i]`	Updated `nums1`
1	0	Increment	Increases to 3	[3]
2	0	Append	-	[3, 3]
3	1	Increment	Increases to 4	[3, 4]

Thus, after 3 operations nums1 is transformed into nums2.

Constraints:

1 <= n == nums1.length <= 10⁵
nums2.length == n + 1
1 <= nums1[i], nums2[i] <= 10⁵

Topics

Array Greedy

Hints open one at a time

Hint 1

Notice that nums2 has exactly one extra element compared to nums1. That extra element must be produced by an append operation.

Hint 2

Suppose we choose index j in nums1 to append. Then nums1[j] must eventually become both nums2[j] and nums2.back().

Hint 3

The cost for index j is: adjust nums1[j] so that one copy matches nums2[j] and the appended copy matches nums2.back().

Hint 4

For each j, compute this adjustment cost, add the cost of transforming all other positions, and take the minimum.

Hint 5

Use prefix sums.

9:15

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1class Solution:
2    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[2,8]
[1,7,3]

Input

[1,3,6]
[2,4,5,3]

Input

[2]
[3,4]

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