Math

isPrime

public static boolean isPrime(int number) {
    if(number == 1 ){
        return false;
    }
    for (int i = 2; i < number; i++) {
        if (number % i == 0) {
            return false;
        }
    }
    return true;
}

Digital Root

https://en.wikipedia.org/wiki/Digital_root#Congruence_formula

Fibonacci

>>> def fibonacci():
...         a, b = 0, 1
...         while True:
...             yield b
...             a, b = b, a + b
...
>>> c = fibonacci()
>>> c.next()
1
>>> c.next()
1
>>> c.next()
2
>>> c.next()
3
>>> c.next()
5

class Solution {
    public int fibonacci(int n) {
        int a = 0, b = 1;
        while(--n > 0) {
            int c = a + b;
            a = b;
            b = c;
        }
        return a;
    }
}

Sum of Array

JavaScript

Use .reduce():

> a = [1, 2, 3, 4]
[ 1, 2, 3, 4 ]
> a.reduce((x, y) => x + y)
10

The full signature is:

array.reduce(function(total, currentValue, currentIndex, arr), initialValue)

However only total and currentValue are required.

Cumulative Sum

JavaScript

> const cumSum = [];
> [1, 2, 3, 4].reduce((a, b, i) => (cumSum[i] = a + b), 0);
> cumSum
[1, 3, 6, 10]

Max / Min / Mean

Go

Go has math.Max to compare 2 numbers, but only for float:

func Max(x, y float64) float64

Round

Java

Integer.valueOf(Math.round(value));

Log

Java

These 2 are equivalent:

Math.log1p(x)
Math.log(x+1)

Fibonacci

Python

def fibonacci():
    a, b = 0, 1
    while True:
        yield b
        a, b = b, a + b

Bit Manipulation

To convert integers to Binary or Hex String:

System.out.println(Integer.toBinaryString(96));
System.out.println(Integer.toHexString(96));

To convert Binary or Hex string to integer:

Integer.parseInt("10010101010101010101010", 2);
Integer.parseInt("ffff", 16);

Basic operations:

// ffffffff
int max = ~0;

// 111111111111111
int ones = (1<<15) - 1;

// 11111111111111111000000000000000
int leftOnes = max - ones;

// 11111
int rightOnes = (1<<5) - 1;

// 11111111111111111000000000011111
int mask = leftOnes | rightOnes;

// 10010101000000000001010
int masked = Integer.parseInt("10010101010101010101010", 2) & mask;

Swap Two Integers

a = a ^ b;
b = a ^ b;
a = a ^ b;

Check If a Number is power of 2

n > 0 && ((n & (n-1)) == 0)

e.g.

• k = 256
  • k = 1000000(binary)
  • k - 1 = 0111111
  • k & (k - 1) = 0

Count 1 in an Integer

int count(int n)
{
    int cnt = 0;
    while(n) {
        cnt += n & 1;
        n >>= 1;
    }
    return cnt;
}

or

int count(int n)
{
    int cnt = 0;
    while(n) {
        n &= (n-1);
        cnt++;
    }
    return cnt;
}

Count Trailing 0s in N!

int count(int n)
{
    int cnt = 0;
    while(n) {
        cnt += n / 5;
        n /= 5;
    }
    return cnt;
}

Find the Position of the Last 1 in Binary Result of N!

int count(int n)
{
    int cnt = 0;
    while(n) {
        n >>= 1;
        cnt += n;
    }
    return cnt;
}

Plus with Binary Operations

Add two integers without +, -, *, /

int bplus(int a, int b)
{
    if (b == 0)
        return a;
    int sum = a ^ b;
    int carry = a & b;
    bplus(sum, carry<<1);
}

Get Max without if-else

int getMax(int a, int b) {
    int c = a - b;
    int k = (c >> 31) & 0x1;
    int max = a - k * c;
    return max;
}