Decrease Elements To Make Array Zigzag
<p>Given an array <code>nums</code> of integers, a <em>move</em>&nbsp;consists of choosing any element and <strong>decreasing it by 1</strong>.</p>

<p>An array <code>A</code> is a&nbsp;<em>zigzag array</em>&nbsp;if either:</p>

<ul>
	<li>Every even-indexed element is greater than adjacent elements, ie.&nbsp;<code>A[0] &gt; A[1] &lt; A[2] &gt; A[3] &lt; A[4] &gt; ...</code></li>
	<li>OR, every odd-indexed element is greater than adjacent elements, ie.&nbsp;<code>A[0] &lt; A[1] &gt; A[2] &lt; A[3] &gt; A[4] &lt; ...</code></li>
</ul>

<p>Return the minimum number of moves to transform the given array <code>nums</code> into a zigzag array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can decrease 2 to 0 or 3 to 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [9,6,1,6,2]
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
</ul>
Greedy; Array
class Solution {
public:
    int movesToMakeZigzag(vector<int>& nums) {
        vector<int> ans(2);
        int n = nums.size();
        for (int i = 0; i < 2; ++i) {
            for (int j = i; j < n; j += 2) {
                int d = 0;
                if (j) d = max(d, nums[j] - nums[j - 1] + 1);
                if (j < n - 1) d = max(d, nums[j] - nums[j + 1] + 1);
                ans[i] += d;
            }
        }
        return min(ans[0], ans[1]);
    }
};