Is Minimum Operations to Make Array Non Decreasing easy or hard?

The problem is typically categorized as Medium. The brute force idea is simple, but recognizing that you can directly add the required difference instead of simulating each increment requires a greedy insight. Once that observation is made, the implementation becomes a clean O(n) scan.

Minimum Operations to Make Array Non Decreasing Python/Java solution

The implementation is straightforward in Python, Java, or C++. Maintain a variable storing the maximum value seen so far. When a smaller element appears, add the difference to the answer and treat the element as updated to the maximum. The code uses a single loop and constant extra memory.

How to solve Minimum Operations to Make Array Non Decreasing in O(n)?

Iterate through the array while maintaining a variable like `prev_max`. If the current value is less than `prev_max`, add `prev_max - nums[i]` to the operation count. Otherwise update `prev_max` to the current value. This greedy strategy ensures each element becomes at least as large as the previous element with minimal adjustments.

What is the best approach for Minimum Operations to Make Array Non Decreasing?

The best approach is a greedy single-pass algorithm that tracks the maximum value seen so far. When the current element is smaller than this value, increase it conceptually to match and add the difference to the operation count. This guarantees the minimal adjustment needed while maintaining the non-decreasing property. The algorithm runs in O(n) time and O(1) space.

Is Minimum Operations to Make Array Non Decreasing asked at Google/Amazon/Meta?

Variants of this problem appear in interviews at large tech companies because it tests greedy reasoning and array traversal patterns. Companies like Amazon, Google, and Meta frequently ask questions involving maintaining order constraints, prefix maxima, or minimal adjustments to arrays.

What data structure is used in Minimum Operations to Make Array Non Decreasing?

The solution primarily relies on simple array traversal and a few scalar variables. No complex data structures are required. The algorithm keeps track of a running prefix maximum and a counter for total operations while scanning the array.

What is the time complexity of Minimum Operations to Make Array Non Decreasing?

The optimal solution runs in O(n) time because the array is scanned once from left to right. Each element requires only constant work: compare with the previous maximum and accumulate the difference if needed. Space complexity is O(1) since only a few variables such as the running maximum and operation counter are maintained.

Description Hints 3 Solution Video 14:40

3914. Minimum Operations to Make Array Non Decreasing

Medium54.9% acceptance1 company

You are given an integer array nums of length n.

In one operation, you may choose any subarray nums[l..r] and increase each element in that subarray by x, where x is any positive integer.

Return the minimum possible sum of the values of x across all operations required to make the array non-decreasing.

An array is non-decreasing if nums[i] <= nums[i + 1] for all 0 <= i < n - 1.

Example 1:

Input: nums = [3,3,2,1]

Output: 2

Explanation:

One optimal set of operations:

Choose subarray [2..3] and add x = 1 resulting in [3, 3, 3, 2]
Choose subarray [3..3] and add x = 1 resulting in [3, 3, 3, 3]

The array becomes non-decreasing, and the total sum of chosen x values is 1 + 1 = 2.

Example 2:

Input: nums = [5,1,2,3]

Output: 4

Explanation:

One optimal set of operations:

Choose subarray [1..3] and add x = 4 resulting in [5, 5, 6, 7]

The array becomes non-decreasing, and the total sum of chosen x values is 4.

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Asked at 1 company

Amazon

Topics

Array Greedy

Hints open one at a time

Hint 1

Focus only on positions where the array decreases (nums[i] > nums[i + 1]).

Hint 2

Each such drop must be fixed by increasing a subarray starting at i + 1.

Hint 3

Observe that you only need to pay for the difference at each drop.

14:40

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1class Solution:
2    def minOperations(self, nums: list[int]) -> int:
3        

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Examples

Case 1Case 2

Input

[3,3,2,1]

Input

[5,1,2,3]

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