// There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.

// You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:

// The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B').
// The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9').
// For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.

// Return the number of rods that have all three colors of rings on them.


let rings = "B0B6G0R6R0R6G9";
var countPoints = function(rings) {
    
    let sum = 0;
    
	// Always 10 Rods
    for (let i = 0; i < 10; i++) {
        if (rings.includes(`B${i}`) && rings.includes(`G${i}`) && rings.includes(`R${i}`)) {
            sum+=1;
        }
    }
    
    return sum;
    
};

console.log(countPoints(rings));

console.log("xa")

console.log("hello world")

//JavaScript program to swap two variables

//take input from the users
let a = prompt('Enter the first variable: ');
let b = prompt('Enter the second variable: ');

// XOR operator
a = a ^ b
b = a ^ b
a = a ^ b

console.log(`The value of a after swapping: ${a}`);
console.log(`The value of b after swapping: ${b}`);

class SequentialQueue{	//if you want it to go backwards, too bad
	next(){
		return this.i++;
	}
	constructor(start = 0){
		this.i = start;
	}
}
const que = new SequentialQueue(0);
for(let i = 0; i < 10; i++){
	console.log(que.next());
}

const levenshteinDistance = (s, t) => {
  if (!s.length) return t.length;
  if (!t.length) return s.length;
  const arr = [];
  for (let i = 0; i <= t.length; i++) {
    arr[i] = [i];
    for (let j = 1; j <= s.length; j++) {
      arr[i][j] =
        i === 0
          ? j
          : Math.min(
              arr[i - 1][j] + 1,
              arr[i][j - 1] + 1,
              arr[i - 1][j - 1] + (s[j - 1] === t[i - 1] ? 0 : 1)
            );
    }
  }
  return arr[t.length][s.length];
};

levenshteinDistance('duck', 'dark'); // 2