// Time Complexity : O(N)
// Space Complexity : O(1)
 
var isMonotonic = function(nums) {
    let isMono = null;
    for(let i = 1; i < nums.length; i++) {
        if(isMono === null) {
            if(nums[i - 1] < nums[i]) isMono = 0;
            else if(nums[i - 1] > nums[i]) isMono = 1;
            continue;
        }
        if(nums[i - 1] < nums[i] && isMono !== 0) {
            return false;
        }
        else if(nums[i - 1] > nums[i] && isMono !== 1) {
            return false;
        }
    }
    
    return true;
    
};

let nums1 = [1,2,2,3]
let nums2 = [6,5,4,4]
let nums3 = [1,3,2]

console.log(isMonotonic(nums1));
console.log(isMonotonic(nums2));
console.log(isMonotonic(nums3));

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)

// 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));

const heapsort = arr => {
  const a = [...arr];
  let l = a.length;

  const heapify = (a, i) => {
    const left = 2 * i + 1;
    const right = 2 * i + 2;
    let max = i;
    if (left < l && a[left] > a[max]) max = left;
    if (right < l && a[right] > a[max]) max = right;
    if (max !== i) {
      [a[max], a[i]] = [a[i], a[max]];
      heapify(a, max);
    }
  };

  for (let i = Math.floor(l / 2); i >= 0; i -= 1) heapify(a, i);
  for (i = a.length - 1; i > 0; i--) {
    [a[0], a[i]] = [a[i], a[0]];
    l--;
    heapify(a, 0);
  }
  return a;
};

heapsort([6, 3, 4, 1]); // [1, 3, 4, 6]

const bucketSort = (arr, size = 5) => {
  const min = Math.min(...arr);
  const max = Math.max(...arr);
  const buckets = Array.from(
    { length: Math.floor((max - min) / size) + 1 },
    () => []
  );
  arr.forEach(val => {
    buckets[Math.floor((val - min) / size)].push(val);
  });
  return buckets.reduce((acc, b) => [...acc, ...b.sort((a, b) => a - b)], []);
};

bucketSort([6, 3, 4, 1]); // [1, 3, 4, 6]

const formatDuration = ms => {
  if (ms < 0) ms = -ms;
  const time = {
    day: Math.floor(ms / 86400000),
    hour: Math.floor(ms / 3600000) % 24,
    minute: Math.floor(ms / 60000) % 60,
    second: Math.floor(ms / 1000) % 60,
    millisecond: Math.floor(ms) % 1000
  };
  return Object.entries(time)
    .filter(val => val[1] !== 0)
    .map(([key, val]) => `${val} ${key}${val !== 1 ? 's' : ''}`)
    .join(', ');
};

formatDuration(1001); // '1 second, 1 millisecond'
formatDuration(34325055574); // '397 days, 6 hours, 44 minutes, 15 seconds, 574 milliseconds'