# Snippets

## Count number of 1 bits

A brute-force approach is to simply count the number of 1’s in the binary number:

``````function countOneBits(num) {
// Convert number to a base-2 string
const binaryString = num.toString(2)

// Remove all 0's so it returns the number of 1's
return binaryString.replace(/0/g, '').length
}
``````

A bit-manipulation approach is to count how many times we eliminate 1’s in the binary number:

``````function countOneBits(num) {
let count = 0

while (num !== 0) {
// Eliminate the least significant 1
num = num & (num - 1)
count++
}

return count
}
``````

## Find majority element

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.

Example 1: Input: [3,2,3] Output: 3

Example 2: Input: [2,2,1,1,1,2,2] Output: 2

Use hashmap to count the frequency of each element and return the element with the highest frequency:

``````function findMajorityElement(arr) {
let freq = new Map()

for (let i = 0; i < arr.length; i++) {
const count = freq.get(arr[i]) || 0
freq.set(arr[i], count + 1)
}

let maxKey
let maxVal = Number.MIN_SAFE_INTEGER

for (let [key, value] of freq.entries()) {
if (value > maxVal) {
maxVal = value
maxKey = key
}
}

return maxKey
}
``````

Sort the array and return the element in middle:

``````function findMajorityElement(arr) {
arr.sort((a, b) => a - b)
const mid = Math.floor(arr.length / 2)
return arr[mid]
}
``````