Today I have done Find Peak Element on leetcode and leetcode's December LeetCoding Challenge with cpp.

Find Peak Element

Description

A peak element is an element that is strictly greater than its neighbors.

Given an integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.

You may imagine that nums[-1] = nums[n] = -∞.

Example 1:

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 5
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

Constraints:

• 1 <= nums.length <= 1000
• -231 <= nums[i] <= 231 - 1
• nums[i] != nums[i + 1] for all valid i.

Follow up: Could you implement a solution with logarithmic complexity?

Solution

logarithmic complexity solution, just use binary search

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        if(nums.size() == 1) return 0;
        int len = nums.size();
        if(nums[0] > nums[1]) return 0;
        if(nums[len-1] > nums[len-2]) return len-1;
        int begin = 0, end = len-1;
        while(begin < end) {
            int mid = (begin + end) / 2;
            if(nums[mid] > nums[mid+1] && nums[mid] > nums[mid-1]) return mid;
            if(nums[mid] < nums[mid-1]) {
                end = mid;
            } else {
                begin = mid;
            }
        }
        return -1;
    }
};

December LeetCoding Challenge 16

Description

Validate Binary Search Tree

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

Example 1:

Input: root = [2,1,3]
Output: true

Example 2:

Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

Solution

inorder traversal

class Solution {
    int prev;
    bool hasPrev = false;
    
public:
    bool isValidBST(TreeNode* root) {
        if(!root) return true;
        if(!isValidBST(root->left)) return false;
        if(!hasPrev) {
            prev = root->val;
            hasPrev = true;
        } else {
            if(root->val <= prev) return false;
            prev = root->val;
        }
        return isValidBST(root->right);
    }
};