# Daily Coding Challenge #131

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.

/*
Find the Difference

Given two strings s and t which consist of only lowercase letters.

String t is generated by random shuffling string s and then add one more letter at a random position.

Find the letter that was added in t.

Example:

Input:
s = "abcd"
t = "abcde"

Output:
e

Explanation:
'e' is the letter that was added.
*/

class Solution {
public:
char findTheDifference(string s, string t) {
// similar to 136. Single Number - https://leetcode.com/problems/single-number/
char ans='\0';
// use XOR to find the difference
for(char c:s+t) ans ^= c;
return ans;
}
};

class Solution2 {
public:
char findTheDifference(string s, string t) {
unordered_map<char, int> m;
char ans;
for(int i=0;i<t.size();i++) m[t[i]-'a']++;
for(int i=0;i<s.size();i++) m[s[i]-'a']--;
for(int i=0;i<26;i++) {
if(m[i]==1){
ans=i+'a';
break;
}
}
return ans;
}
};

static const auto io_sync_off = []() {std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();


/*
Gas Station

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1.

Note:

If there exists a solution, it is guaranteed to be unique.
Both input arrays are non-empty and have the same length.
Each element in the input arrays is a non-negative integer.
Example 1:

Input:
gas  = [1,2,3,4,5]
cost = [3,4,5,1,2]

Output: 3

Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.
Example 2:

Input:
gas  = [2,3,4]
cost = [3,4,3]

Output: -1

Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.

*/

class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int tank=0, total=0, start=0;
for(int i=0;i<gas.size();i++){
tank += gas[i]-cost[i];
total += gas[i]-cost[i];
if(tank < 0){
start = i+1;
tank = 0;
}
}
}
};


There are other programming solutions in the following repositories below. Star and watch for timely updates!

## wingkwong / competitive-programming

### Discussion   