What is the time complexity of Minimum Increment Operations to Make Array Beautiful?

The optimal solution for Minimum Increment Operations to Make Array Beautiful runs in O(n) time and uses O(1) space. This complexity is linear because we only make a single pass through the array, adjusting values as necessary.

What is the brute force solution for Minimum Increment Operations to Make Array Beautiful?

The brute force approach for Minimum Increment Operations to Make Array Beautiful is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray of size 3 or more to ensure that at least one element in each subarray is greater than or equal to k. If not, we increment the necessary elements until the condition is met. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Increment Operations to Make Array Beautiful?

Minimum Increment Operations to Make Array Beautiful uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Sliding Window, Greedy Approach.

What are the interview tips for Minimum Increment Operations to Make Array Beautiful?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as arrays with fewer than 3 elements. Practice explaining your thought process as you code; this helps interviewers understand your approach.

What are common mistakes when solving Minimum Increment Operations to Make Array Beautiful?

Failing to check all elements in the sliding window correctly. Not accounting for overlapping increments when adjusting elements.

#2919

Minimum Increment Operations to Make Array Beautiful

Medium

ArrayDynamic ProgrammingSliding WindowGreedy Approach

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to ensure that every subarray of size 3 has at least one element greater than or equal to k. We only need to check the last element of each window and adjust accordingly, minimizing the number of increments.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to count increments and iterate through the array from index 0 to n-1.
2Step 2: For each index, check if the current element is less than k.
3Step 3: If it is, calculate how much to increment it to reach k and add that to the increments count.
4Step 4: Adjust the next two elements in the window to ensure they are also at least k.

solution.py11 lines

1def minIncrementForBeautiful(nums, k):
2    increments = 0
3    n = len(nums)
4    for i in range(n):
5        if nums[i] < k:
6            increments += k - nums[i]
7            if i + 1 < n:
8                nums[i + 1] = max(nums[i + 1], k)
9            if i + 2 < n:
10                nums[i + 2] = max(nums[i + 2], k)
11    return increments

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Complexity note: This complexity is linear because we only make a single pass through the array, adjusting values as necessary.

1To ensure every subarray of size 3 has a maximum >= k, at least one element in each subarray must be adjusted.
2Increment operations should be minimized by focusing on the last element of each sliding window.

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