Today I have done Generate Random Point in a Circle and leetcode's March LeetCoding Challenge with cpp.

Generate Random Point in a Circle

Description

Given the radius and x-y positions of the center of a circle, write a function randPoint which generates a uniform random point in the circle.

Note:

1. input and output values are in floating-point.
2. radius and x-y position of the center of the circle is passed into the class constructor.
3. a point on the circumference of the circle is considered to be in the circle.
4. randPoint returns a size 2 array containing x-position and y-position of the random point, in that order.

Example 1:

Input: 
["Solution","randPoint","randPoint","randPoint"]
[[1,0,0],[],[],[]]
Output: [null,[-0.72939,-0.65505],[-0.78502,-0.28626],[-0.83119,-0.19803]]

Example 2:

Input: 
["Solution","randPoint","randPoint","randPoint"]
[[10,5,-7.5],[],[],[]]
Output: [null,[11.52438,-8.33273],[2.46992,-16.21705],[11.13430,-12.42337]]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution's constructor has three arguments, the radius, x-position of the center, and y-position of the center of the circle. randPoint has no arguments. Arguments are always wrapped with a list, even if there aren't any.

Solution

constexpr double pi = 3.14159265358979732384626433832795;

class Solution {
  mt19937 gen;
  double x;
  double y;
  double r;
  uniform_real_distribution<double> distr;
  
public:
  Solution(double radius, double x_center, double y_center): r(radius), x(x_center), y(y_center), distr(0.0, 1.0){}
    
  vector<double> randPoint() {
    double angle = 2 * pi * distr(gen);
    double length = r * sqrt(distr(gen));
    return vector<double>{x + length * cos(angle), y + length * sin(angle)};
  }
};

March LeetCoding Challenge 11

Description

Coin Change

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Example 4:

Input: coins = [1], amount = 1
Output: 1

Example 5:

Input: coins = [1], amount = 2
Output: 2

Constraints:

• $1 \le coins.length \le 12$
• $1 \le coins[i] \le 2^{31} - 1$
• $0 \le amount \le 10^4$

Solution

class Solution {
public:
  int coinChange(vector<int>& coins, int amount) {
    vector<int> dp(amount + 1, amount+1);
    dp[0] = 0;
    for(auto coin : coins){
      for(int i = coin; i <= amount; ++i) {
        dp[i] = min(dp[i], dp[i - coin] + 1);
      }
    }
    return dp.back() > amount ? -1 : dp.back();
  }
};