Author: robert

Developers flex on each with the strangest of ways. How green is your github commit history. Have you submitted a PR to a open source project. How well can you solve the silliest questions on leetcode / geeksforgeeks / hackerrank?
While normies use money and status as a measure of success. Coders have incelled their way into a paradise where the glow of the monitor is all you need.
Pray for death all you want, it’s not coming. Man up and start grinding.

You are given a 0-indexed string pattern of length n consisting of the characters 'I' meaning increasing and 'D' meaning decreasing.

A 0-indexed string num of length n + 1 is created using the following conditions:

• num consists of the digits '1' to '9', where each digit is used at most once.
• If pattern[i] == 'I', then num[i] < num[i + 1].
• If pattern[i] == 'D', then num[i] > num[i + 1].

Return the lexicographically smallest possible string num that meets the conditions.

Example 1:

Input: pattern = "IIIDIDDD"
Output: "123549876"
Explanation:
At indices 0, 1, 2, and 4 we must have that num[i] < num[i+1].
At indices 3, 5, 6, and 7 we must have that num[i] > num[i+1].
Some possible values of num are "245639871", "135749862", and "123849765".
It can be proven that "123549876" is the smallest possible num that meets the conditions.
Note that "123414321" is not possible because the digit '1' is used more than once.

Example 2:

Input: pattern = "DDD"
Output: "4321"
Explanation:
Some possible values of num are "9876", "7321", and "8742".
It can be proven that "4321" is the smallest possible num that meets the conditions.

Constraints:

• 1 <= pattern.length <= 8
• pattern consists of only the letters 'I' and 'D'.

class Solution {
public:
    string smallestNumber(string pattern) {
        string result;
        stack<int> st;
        for( int i = 0; i<= pattern.size(); i++ ){
            st.push(i+1);
             if ( i == pattern.length() || pattern[i] == 'I' )
                while(!st.empty()){
                    result+=to_string( st.top() );
                    st.pop();        
                }
                
        }
        return result;
    }
};

1863. Sum of All Subset XOR Totals
This is a good problem to grasp basics of backtracking

The XOR total of an array is defined as the bitwise XOR of all its elements, or 0 if the array is empty.

• For example, the XOR total of the array [2,5,6] is 2 XOR 5 XOR 6 = 1.

Given an array nums, return the sum of all XOR totals for every subset of nums. 

Note: Subsets with the same elements should be counted multiple times.

An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b.

Example 1:

Input: nums = [1,3]
Output: 6
Explanation: The 4 subsets of [1,3] are:
- The empty subset has an XOR total of 0.
- [1] has an XOR total of 1.
- [3] has an XOR total of 3.
- [1,3] has an XOR total of 1 XOR 3 = 2.
0 + 1 + 3 + 2 = 6

Example 2:

Input: nums = [5,1,6]
Output: 28
Explanation: The 8 subsets of [5,1,6] are:
- The empty subset has an XOR total of 0.
- [5] has an XOR total of 5.
- [1] has an XOR total of 1.
- [6] has an XOR total of 6.
- [5,1] has an XOR total of 5 XOR 1 = 4.
- [5,6] has an XOR total of 5 XOR 6 = 3.
- [1,6] has an XOR total of 1 XOR 6 = 7.
- [5,1,6] has an XOR total of 5 XOR 1 XOR 6 = 2.
0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28

Example 3:

Input: nums = [3,4,5,6,7,8]
Output: 480
Explanation: The sum of all XOR totals for every subset is 480.

class Solution {
    vector<vector<int>> subset;
    void backtrack(vector<int> &nums,vector<int> &current,int index){
        subset.push_back(current);
        for(int i = index; i < nums.size(); i++ ){
            current.push_back(nums[i]);
            backtrack( nums,  current, i+1);
            current.pop_back();
        }
    }
public:
    
    int subsetXORSum(vector<int>& nums) {
        vector<int> current;
        backtrack(nums,current,0);
        
        int total_sum = 0;
        for (const auto& sub : subset) {
            int xor_sum = 0;
            for (int num : sub) {
                //std::cout<<num<<" ";
                xor_sum ^= num;
            }
            //std::cout<<std::endl;
            total_sum += xor_sum;
        }
        
        return total_sum;
    }
};