How do you solve Decrease Elements To Make Array Zigzag on LeetCode?

Decrease Elements To Make Array Zigzag (LeetCode #1144) 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: Understanding the zigzag pattern is crucial for solving the problem efficiently.

What is the time complexity of Decrease Elements To Make Array Zigzag?

The optimal solution for Decrease Elements To Make Array Zigzag runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only make a single pass through the array, checking conditions for both zigzag patterns simultaneously.

What is the brute force solution for Decrease Elements To Make Array Zigzag?

The brute force approach for Decrease Elements To Make Array Zigzag is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every element in the array and adjusts it to meet the zigzag condition. This method is straightforward but inefficient, as it requires multiple passes through the array for each element. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Decrease Elements To Make Array Zigzag?

Decrease Elements To Make Array Zigzag uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Decrease Elements To Make Array Zigzag?

Always clarify the problem statement and constraints before diving into coding. Discuss your thought process and potential edge cases with the interviewer. Consider both brute-force and optimal solutions, explaining why you choose one over the other.

What are common mistakes when solving Decrease Elements To Make Array Zigzag?

Not considering edge cases, such as arrays of length 2. Failing to account for both even and odd indexed conditions.

#1144

Decrease Elements To Make Array Zigzag

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)

The optimal solution leverages the fact that we can calculate the moves required for both zigzag patterns (even index greater and odd index greater) in a single pass, significantly reducing the time complexity.

⚙️

Algorithm

4 steps

1Step 1: Initialize two move counters for both zigzag patterns.
2Step 2: Iterate through the array, checking conditions for both even and odd indexed elements.
3Step 3: For each element, calculate the required moves and update the respective counter.
4Step 4: Return the minimum of the two move counters.

solution.py16 lines

1def movesToMakeZigzag(nums):
2    even_moves = 0
3    odd_moves = 0
4    n = len(nums)
5    for i in range(n):
6        if i % 2 == 0:  # Even index
7            if i > 0:
8                even_moves += max(0, nums[i] - nums[i - 1] + 1)
9            if i < n - 1:
10                even_moves += max(0, nums[i] - nums[i + 1] + 1)
11        else:  # Odd index
12            if i > 0:
13                odd_moves += max(0, nums[i] - nums[i - 1] + 1)
14            if i < n - 1:
15                odd_moves += max(0, nums[i] - nums[i + 1] + 1)
16    return min(even_moves, odd_moves)

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, checking conditions for both zigzag patterns simultaneously.

1Understanding the zigzag pattern is crucial for solving the problem efficiently.
2Calculating moves for both patterns simultaneously can save time.

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