Introduction

In this blog post, we will explore how to solve the LeetCode problem "1144" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array nums of integers, a move consists of choosing any element and decreasing it by 1. An array A is a zigzag array if either: Every even-indexed element is greater than adjacent elements, ie. A[0] > A[1] < A[2] > A[3] < A[4] > ... OR, every odd-indexed element is greater than adjacent elements, ie. A[0] < A[1] > A[2] < A[3] > A[4] < ... Return the minimum number of moves to transform the given array nums into a zigzag array. Example 1: Input: nums = [1,2,3] Output: 2 Explanation: We can decrease 2 to 0 or 3 to 1. Example 2: Input: nums = [9,6,1,6,2] Output: 4 Constraints: 1 <= nums.length <= 1000 1 <= nums[i] <= 1000

Explanation

Here's the solution to the zigzag array problem:

• High-Level Approach:

  • Calculate the moves required to transform the array into two different zigzag patterns: one where even indices are greater than their neighbors, and one where odd indices are greater.
  • For each pattern, iterate through the array and, at each index, determine how many moves are needed to satisfy the zigzag condition.
  • Return the minimum of the moves required for the two patterns.

• Complexity:

  • Runtime Complexity: O(n), where n is the length of the input array.
  • Storage Complexity: O(1)

Code

    def movesToMakeZigzag(nums):
    """
    Calculates the minimum number of moves to transform the given array into a zigzag array.

    Args:
        nums: A list of integers.

    Returns:
        The minimum number of moves.
    """

    n = len(nums)
    moves_even = 0
    moves_odd = 0

    nums_even = nums[:]
    nums_odd = nums[:]


    for i in range(n):
        if i % 2 == 0:  # Even index
            # Check left neighbor
            left = float('-inf') if i == 0 else nums_even[i-1]
            # Check right neighbor
            right = float('-inf') if i == n - 1 else nums_even[i+1]

            moves_needed = max(0, nums_even[i] - min(left, right) + 1)
            moves_even += moves_needed

        else: #odd indices are <
            left = float('-inf') if i == 0 else nums_even[i - 1]
            right = float('-inf') if i == n - 1 else nums_even[i + 1]
            moves_needed = max(0, nums_even[i] - min(left, right) + 1)

    for i in range(n):
        if i % 2 != 0:  # Odd index
            # Check left neighbor
            left = float('-inf') if i == 0 else nums_odd[i-1]
            # Check right neighbor
            right = float('-inf') if i == n - 1 else nums_odd[i+1]
            moves_needed = max(0, nums_odd[i] - min(left, right) + 1)
            moves_odd += moves_needed

        else:
            left = float('-inf') if i == 0 else nums_odd[i - 1]
            right = float('-inf') if i == n - 1 else nums_odd[i + 1]
            moves_needed = max(0, nums_odd[i] - min(left, right) + 1)

    return min(moves_even, moves_odd)