# Daily Coding Challenge #58

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

``````/*
LeetCode - 3Sum

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
*/

class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
// approach:
// sort the array first
// fix one value, then find the another two values
// if the sum is greater than the target, resize the window from the right by 1
// if it is smaller than that, resize the window from the left by 1
// if it is same, add it to a set to de-duplicate the result
int n = nums.size();
vector<vector<int>> ans;
if(n<3) return ans;
set<vector<int>> s;
sort(nums.begin(),nums.end());
for(int i=0;i<n;i++){
int val=nums[i];
int l=i+1;
int r=n-1;
while(l<r){
int sum=val+nums[l]+nums[r];
if(sum==0) s.insert({val,nums[l++],nums[r--]});
else if(sum>0) r--;
else l++;
}
}
for(auto x:s) ans.push_back(x);
return ans;
}
};
``````

``````/*
LeetCode - Maximum Width of Binary Tree
Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.

The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null nodes between the end-nodes are also counted into the length calculation.

Example 1:

Input:

1
/   \
3     2
/ \     \
5   3     9

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
Example 2:

Input:

1
/
3
/ \
5   3

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).
Example 3:

Input:

1
/ \
3   2
/
5

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).
Example 4:

Input:

1
/ \
3   2
/     \
5       9
/         \
6           7
Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

Note: Answer will in the range of 32-bit signed integer.
*/

/**
* 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:
int widthOfBinaryTree(TreeNode* root) {
vector<unsigned long long> s;
// approach: dfs. don't need to care about the value
dfs(root,0,1LL,s);
return ans;
}
private:
long long ans=1;
// use unsigned long long to avoid overflow
void dfs(TreeNode* root, unsigned long long lv, unsigned long long idx, vector<unsigned long long>& s){
if(!root) return;
if(s.size()==lv) s.emplace_back(idx); // storing the leftmost node idx
else if(idx-s[lv]+1>ans) ans=idx-s[lv]+1; //
dfs(root->left,lv+1,idx*2,s);  // the next left node would be idx*2
dfs(root->right,lv+1,idx*2+1,s); // the next left node would be idx*2+1
}
};

//            1
//        /       \
//      2         3
//     /  \        /  \
//   4    5        6   7
``````

The source code is available in corresponding repo below. Star and watch for timely updates!