1144. Decrease Elements To Make Array Zigzag

Problem

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 (Java)

class Solution {
    public int movesToMakeZigzag(int[] nums) {
        int ans;
        int n = nums.length;
        int cur = 0;
        if (n == 1) {
            return 0;
        }
        int[] clone = nums.clone();
        for (int i = 0; i < n; i += 2) {
            if (i == 0) {
                if (nums[i] <= nums[i + 1]) {
                    cur += (nums[i + 1] - nums[i] + 1);
                    nums[i + 1] = nums[i] - 1;
                }
            } else if (i == n - 1) {
                if (nums[i] <= nums[i - 1]) {
                    cur += (nums[i - 1] - nums[i] + 1);
                }
            } else {
                if (nums[i] <= nums[i + 1]) {
                    cur += (nums[i + 1] - nums[i] + 1);
                    nums[i + 1] = nums[i] - 1;
                }
                if (nums[i] <= nums[i - 1]) {
                    cur += (nums[i - 1] - nums[i] + 1);
                }
            }
        }
        ans = cur;
        cur = 0;
        nums = clone;
        for (int i = 1; i < n; i += 2) {
            if (i != n - 1 && nums[i] <= nums[i + 1]) {
                cur += (nums[i + 1] - nums[i] + 1);
                nums[i + 1] = nums[i] - 1;
            }

            if (nums[i] <= nums[i - 1]) {
                cur += (nums[i - 1] - nums[i] + 1);
            }
        }
        ans = Math.min(ans, cur);
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).