Added leetcode problem:414 and 628 #999

dsa-solutions/lc-solutions/0400-0499/0414-third-maximum-number.md

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+---
+id: third-maximum-number
+title: Third Maximum Number
+sidebar_label: 0414 Third Maximum Number
+tags:
+  - Math
+  - Vector
+  - Set
+  - LeetCode
+  - C++
+description: "This is a solution to the Third Maximum Number problem on LeetCode."
+---
+
+## Problem Description
+
+Given an integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.
+
+### Examples
+
+**Example 1:**
+
+```
+
+Input:  nums = [1,2]
+Output: 2
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,2,3,1]
+Output: 1
+```
+
+**Example 2:**
+
+```
+Input: root = [0]
+Output: [0]
+```
+
+### Constraints
+
+- $1 \leq \text{nums.length} \leq 10^4$.
+- $-2^(31) \leq \text{Node.val} \leq 2^(31)-1$.
+
+### Approach 
+
+To solve this problem(third maximum element) first we will store the number from the array/vector in set to get all the unique number from the given array then if size of the set is less than 3 we will just return maximum element from the array otherwise we will return the third element from the set.
+
+#### Code in C++
+
+```cpp
+class Solution {
+public:
+    int thirdMax(vector<int>& nums) {
+        set<int>b;
+        for(int i=0;i<nums.size();i++){
+            b.insert(nums.at(i));
+        }
+        int n=b.size();
+        if(n<3){
+            return *max_element(nums.begin() ,nums.end());
+        }
+        vector<int>p;
+      for(auto x :b){
+          p.push_back(x);
+      }
+
+      return p.at(n-3);
+    }
+};
+```
+
+

dsa-solutions/lc-solutions/600-699/628-maximum-product-of-three-numbers.md

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+---
+id: maximum-product-of-three-numbers
+title: Maximum Product of Three Numbers
+sidebar_label: 628 Maximum Product of Three Numbers
+tags:
+  - Math
+  - Vector
+  - LeetCode
+  - C++
+description: "This is a solution to the Maximum Product of Three Numbers problem on LeetCode."
+---
+
+## Problem Description
+
+Given an integer array nums, find three numbers whose product is maximum and return the maximum product.
+
+### Examples
+
+**Example 1:**
+
+```
+
+Input: nums = [1,2,3]
+Output: 6
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4]
+Output: 24
+```
+
+**Example 2:**
+
+```
+Input: nums = [-1,-2,-3]
+Output: -6
+```
+
+### Constraints
+
+- $3 \leq \text{nums.length} \leq 10^4$.
+- $-1000 \leq \text{Node.val} \leq 1000$.
+
+### Approach 
+
+To solve this problem(third maximum element) we will sort the given array and then return the product of last three numbers but there are chances that product of two smallest negative number(starting first two numbers generally) with largest positive number will be greater than last three numbers product so we will check that also.
+
+#### Code in C++
+
+```cpp
+class Solution {
+public:
+    int maximumProduct(vector<int>& nums) {
+        sort(nums.begin(),nums.end());
+        int n=nums.size();
+        int l=nums[n-3]*nums[n-2]*nums[n-1]; // prouduct of maximum numbers
+        int p=nums[0]*nums[1]*nums[n-1];// here we are checking if (num[0] and num[1] is negative so may this product will be greater than the product(l) which is product of last three maximum numbers)
+        if(p>l){
+            return p;
+        }
+        return l;
+    }
+};
+```
+
+#### code in Python
+
+```python
+class Solution(object):
+    def maximumProduct(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        nums.sort()
+        p=nums[0]*nums[1]*nums[-1] #Calculate the product of the two smallest numbers (negative numbers) and the largest number in the sorted list.
+        l=nums[-1]*nums[-2]*nums[-3] #Calculate the product of the three largest numbers in the sorted list.
+        return max(l,p)
+
+```
+