<p>You are given an integer array <code>nums</code> with no duplicates. A <strong>maximum binary tree</strong> can be built recursively from <code>nums</code> using the following algorithm:</p>

<ol>
	<li>Create a root node whose value is the maximum value in <code>nums</code>.</li>
	<li>Recursively build the left subtree on the <strong>subarray prefix</strong> to the <strong>left</strong> of the maximum value.</li>
	<li>Recursively build the right subtree on the <strong>subarray suffix</strong> to the <strong>right</strong> of the maximum value.</li>
</ol>

<p>Return <em>the <strong>maximum binary tree</strong> built from </em><code>nums</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/24/tree1.jpg" style="width: 302px; height: 421px;" />
<pre>
<strong>Input:</strong> nums = [3,2,1,6,0,5]
<strong>Output:</strong> [6,3,5,null,2,0,null,null,1]
<strong>Explanation:</strong> The recursive calls are as follow:
- The largest value in [3,2,1,6,0,5] is 6. Left prefix is [3,2,1] and right suffix is [0,5].
    - The largest value in [3,2,1] is 3. Left prefix is [] and right suffix is [2,1].
        - Empty array, so no child.
        - The largest value in [2,1] is 2. Left prefix is [] and right suffix is [1].
            - Empty array, so no child.
            - Only one element, so child is a node with value 1.
    - The largest value in [0,5] is 5. Left prefix is [0] and right suffix is [].
        - Only one element, so child is a node with value 0.
        - Empty array, so no child.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/24/tree2.jpg" style="width: 182px; height: 301px;" />
<pre>
<strong>Input:</strong> nums = [3,2,1]
<strong>Output:</strong> [3,null,2,null,1]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
	<li>All integers in <code>nums</code> are <strong>unique</strong>.</li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:

    #define pii pair<int, int>

    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        
        int n = nums.size();

        // max heap
        set<pii, greater<pii>> maxHeap;
        for (int i = 0; i < n; i++) {
            maxHeap.insert(make_pair(nums[i], i));
        }

        // hash map to maintain idx
        unordered_map<TreeNode*, int> nodeToIdx;

        // construct the tree

        // root
        auto maxItr = maxHeap.begin();
        auto [val, idx] = *maxItr;
        maxHeap.erase(maxItr);
        TreeNode* root = new TreeNode(val);
        nodeToIdx[root] = idx;

        // subsequent nodes of the tree
        while (maxHeap.empty() == false) {
            // create the node
            auto maxItr = maxHeap.begin();
            auto [val, idx] = *maxItr;
            maxHeap.erase(maxItr);
            TreeNode* node = new TreeNode(val);
            nodeToIdx[node] = idx;

            // connect the node
            TreeNode* parent = root;
            while (true) {
                if (idx < nodeToIdx[parent] && parent->left != nullptr) parent = parent->left;
                else if (idx > nodeToIdx[parent] && parent->right != nullptr) parent = parent->right;
                else { // found the parent node
                    if (idx < nodeToIdx[parent]) parent->left = node;
                    else parent->right = node;
                    break;
                }
            }
        }
        
        // return root
        return root;
    }
};
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