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Decrease Elements To Make Array Zigzag

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

Solution Explanation: Decrease Elements To Make Array Zigzag

This problem asks us to find the minimum number of moves needed to transform a given array into a zigzag array. A zigzag array is defined as an array where either every even-indexed element is greater than its adjacent elements, or every odd-indexed element is greater than its adjacent elements. A move consists of decreasing any element by 1.

The optimal approach uses a greedy strategy combined with enumeration. We consider two cases:

  1. Even-indexed elements are greater: We iterate through the even-indexed elements. For each element, we check if it's greater than its neighbors. If not, we calculate how much we need to decrease the current element to satisfy the zigzag condition. We sum up these decreases.

  2. Odd-indexed elements are greater: We do the same as above but for odd-indexed elements.

Finally, we return the minimum of the moves required in both cases.

Detailed Code Explanation (Python):

class Solution:
    def movesToMakeZigzag(self, nums: List[int]) -> int:
        ans = [0, 0]  # Initialize the number of moves for both cases (even and odd)
        n = len(nums)
        for i in range(2):  # Iterate through both cases (i=0 for even, i=1 for odd)
            for j in range(i, n, 2):  # Iterate through even or odd indices depending on i
                d = 0
                if j > 0: #Check left neighbor, avoid index out of bound error
                    d = max(d, nums[j] - nums[j - 1] + 1) #Calculate the required decrease, +1 because we must make it strictly less
                if j < n - 1: #Check right neighbor, avoid index out of bound error
                    d = max(d, nums[j] - nums[j + 1] + 1) #Calculate the required decrease, +1 because we must make it strictly less
                ans[i] += d  # Accumulate the moves for the current case
        return min(ans)  # Return the minimum moves required

Time and Space Complexity Analysis:

  • Time Complexity: O(n), where n is the length of the input array nums. We iterate through the array at most twice (once for even indices and once for odd indices). The operations within each loop (comparisons and max calculations) are constant time.

  • Space Complexity: O(1). The space used is constant regardless of the input size. We only use a few variables to store the results.

Code in Other Languages:

The logic remains the same across all languages; only the syntax changes slightly. The provided code examples in Java, C++, Go, TypeScript, and C# effectively implement this same algorithm. They all achieve a time complexity of O(n) and a space complexity of O(1).