if (process.env.NODE_ENV === "production") {
	app.use(express.static("client/build"));

	app.get("*", (req, res) => {
	res.sendFile(path.resolve(__dirname, "client", "build", "index.html"));
	});
}

const quickSort = arr => {
  const a = [...arr];
  if (a.length < 2) return a;
  const pivotIndex = Math.floor(arr.length / 2);
  const pivot = a[pivotIndex];
  const [lo, hi] = a.reduce(
    (acc, val, i) => {
      if (val < pivot || (val === pivot && i != pivotIndex)) {
        acc[0].push(val);
      } else if (val > pivot) {
        acc[1].push(val);
      }
      return acc;
    },
    [[], []]
  );
  return [...quickSort(lo), pivot, ...quickSort(hi)];
};

quickSort([1, 6, 1, 5, 3, 2, 1, 4]); // [1, 1, 1, 2, 3, 4, 5, 6]

// 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));

function printHeap(heap, index, level) {
    if (index >= heap.length) {
        return;
    }
    console.log("  ".repeat(level) + heap[index]);
    printHeap(heap, 2 * index + 1, level + 1);
    printHeap(heap, 2 * index + 2, level + 1);
}

//You can call this function by passing in the heap array and the index of the root node, which is typically 0, and level = 0.

let heap = [3, 8, 7, 15, 17, 30, 35, 2, 4, 5, 9];
printHeap(heap,0,0)

console.log("xa")

const longestPalindrome = (s) => {
    if(s.length === 1) return s
    
    const odd = longestOddPalindrome(s)
    const even = longestEvenPalindrome(s)
    
    
    if(odd.length >= even.length) return odd
    if(even.length > odd.length) return even
};

const longestOddPalindrome = (s) => {    
    let longest = 0
    let longestSubStr = ""
    
    for(let i = 1; i < s.length; i++) {
        
        let currentLongest = 1
        let currentLongestSubStr = ""
        let left = i - 1
        let right = i + 1
        
        while(left >= 0 && right < s.length) {
            const leftLetter = s[left]
            const rightLetter = s[right]

            left--
            right++

            if(leftLetter === rightLetter) {
                currentLongest += 2
                currentLongestSubStr = s.slice(left+1, right)
            } else break
        }
        
                        
        if(currentLongest > longest) {
            if(currentLongestSubStr) longestSubStr = currentLongestSubStr 
            else longestSubStr = s[i]
            longest = currentLongest
        }
    }
    
    return longestSubStr
}

const longestEvenPalindrome = (s) => {    
    let longest = 0
    let longestSubStr = ""
    
    for(let i = 0; i < s.length - 1; i++) {
        if(s[i] !== s[i+1]) continue;
        
        let currentLongest = 2
        let currentLongestSubStr = ""
        let left = i - 1
        let right = i + 2
        
        while(left >= 0 && right < s.length) {
            const leftLetter = s[left]
            const rightLetter = s[right]

            left--
            right++
            
            if(leftLetter === rightLetter) {
                currentLongest += 2
                currentLongestSubStr = s.slice(left+1, right)
            } else break
        }
                
                
        if(currentLongest > longest) {
            if(currentLongestSubStr) longestSubStr = currentLongestSubStr
            else longestSubStr = s[i] + s[i+1]
    
            longest = currentLongest
        }
    }
    
    return longestSubStr
}
console.log(longestPalindrome("babad"))