<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 {
    int Max(const vector<int>& nums) {
        int max1 = INT_MIN;
        for (int i : nums) {
            max1 = max(max1, i);
        }
        return max1;
    }

public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        if (nums.empty()) return nullptr;

        vector<pair<TreeNode*, vector<int>>> k;
        TreeNode* root = new TreeNode(Max(nums)); // Initial root node
        k.push_back({root, nums}); // Push root and the entire nums array
        bool flag = false; // Tracking if root is processed

        while (!k.empty()) {
            vector<pair<TreeNode*, vector<int>>> k1;
            for (auto& nodePair : k) {
                TreeNode* currentNode = nodePair.first;
                vector<int>& currentNums = nodePair.second;

                int maxIndex = find(currentNums.begin(), currentNums.end(), currentNode->val) - currentNums.begin();

                // Left subtree
                if (maxIndex > 0) {
                    vector<int> leftNums(currentNums.begin(), currentNums.begin() + maxIndex);
                    int leftMax = Max(leftNums);
                    currentNode->left = new TreeNode(leftMax);
                    k1.push_back({currentNode->left, leftNums});
                }

                // Right subtree
                if (maxIndex + 1 < currentNums.size()) {
                    vector<int> rightNums(currentNums.begin() + maxIndex + 1, currentNums.end());
                    int rightMax = Max(rightNums);
                    currentNode->right = new TreeNode(rightMax);
                    k1.push_back({currentNode->right, rightNums});
                }
            }
            k = k1;
        }
        return root;
    }
};

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