Algorithm

1. Move all the 0s in the array to the end.

function moveZeros(arr) {
    let zeros = 0
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] === 0) {
            zeros += 1
        } else {
            arr[i - zeros] = arr[i]
        }
    }

    for (let i = arr.length - zeros; i < arr.length; i++) {
        arr[i] = 0
    }
}

let arr = [1, 3, 4, 0, 8, 9, 0, 7, 20, 3, 0, 6, 11]
moveZeros(arr)
console.log(arr)

2. Given two strings s and t , write a function to determine if t is an anagram of s.

function lowerDictionary() {
    let obj = {}

    for (let i = 97; i < 123; i++) {
        obj[String.fromCharCode(i)] = 0
    }

    return obj
}

function isAnagram(s1, s2) {
    let d1 = lowerDictionary()
    let d2 = lowerDictionary()

    for (let item of s1) {
        d1[item] += 1
    }
    for (let item of s2) {
        d2[item] += 1
    }

    return JSON.stringify(d1) === JSON.stringify(d2)
}

console.log(isAnagram("ten", "net"))

3. Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
   You may assume that the array is non-empty and the majority element always exist in the array.

function majorityElement(nums) {
    let stack = []

    nums.forEach(element => {
        if (stack.length === 0) {
            stack.push(element)
        } else {
            if (element === stack[stack.length - 1]) {
                stack.push(element)
            } else {
                stack.pop()
            }
        }
    })

    return stack.pop()
}

4. A frog can jump one or two steps at a time. How many jumping methods are there when the frog jumps up an n-step? (Different order corresponds to different results)

function jumpFloor(number) {
    if (number <= 0) {
        return 0
    } else if (number === 1) {
        return 1
    } else if (number === 2) {
        return 2
    } else {
        return jumpFloor(number - 1) + jumpFloor(number - 2)
    }
}

5. A frog can jump up a step or two at a time... It can also jump to level n at a time. Find out how many jumping methods the frog can use to jump up an n-step.

function jumpFloorII(number) {
    if (number <= 0) {
        return 0
    } else if (number === 1) {
        return 1
    } else {
        return 2 * jumpFloorII(number - 1)
    }
}