Bit Manipulation
Bit Manipulation
- Basic operation
- Basic use case
- Interview Questions
How Integer is stored in Memory
Integer -> 4 bytes -> 32 bits
Long -> 8 bytes -> 64 bits
Boolean/Char -> 1 byte
Two's complement
How double is stored in Memory
Basic Operation
- Bitwise NOT
- ~0111 = 1000
- Bitwise OR
- 0101 | 1010 = 1111
- Bitwise AND
- 0101 & 1100 = 0100
- Bitwise XOR
- 0101 ^ 0011 = 0110
Basic Operation
- Bitwise SHIFT
- 0001 << 3 = 1000
- 1100 >> 1 = 1110 -- arithmetic
- 1100 >>> 1 = 0110 -- logical
Basic use case (Tricks)
- Even & Odd
- n & 1 == 1 ? Odd : Even
- Double
- a / 2 = a >> 1;
- a * (2^m) = a << m;
- Negative number
- a = 101101100
- -a = 010010011 + 1 = 010010100
- Bit NOT on every bit and then plus 1.
Basic use case (Tricks)
- Specific bit operation
- Get kth bit: (a >> k) & 1
- Set kth bit 0: a = a & ~(1 << k)
- Set kth bit 1: a = a | (1 << k)
- Rightmost 1-bit: a & -a
- Absolute value ***
- |a| = a ^ (a >> 31) - (a >> 31)
Basic use case (Tricks)
- XOR
- a ^ a = 0
- a ^ 0 = a
- Swap (int a, int b)
- a ^= b;
- b ^= a; // b = b ^ (a' ^ b) = a'
- a ^= b; // a = (a' ^ b') ^ a' = b'
Interview Questions
- Bit operation
- Single Number
- Number of 1 Bits
- Reverse Bits
- Reduce storage & improve efficiency
- Combination
Single Number
Given an array of integers, every element appears twice except for one. Find that single one.
Single Number
Given an array of integers, every element appears twice except for one. Find that single one.
public int singleNumber(int[] A) {
for (int i = 1; i < A.length; i++) {
A[i] ^= A[i-1];
}
return A[A.length-1];
}Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1' bits it has. (also known as the Hamming weight)
Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1' bits it has. (also known as the Hamming weight)
public int hammingWeight(int n) {
int result = 0;
for (int i = 0; i < 32; i++) {
result += n & 1;
n >>= 1;
}
return result;
}Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1' bits it has. (also known as the Hamming weight)
public int hammingWeight(int n) {
int res = 0;
for (int i = 0; i < 32; i++) {
if (((1 << i) & n) != 0) {
res++;
}
}
return res;
}Reverse Bits
Reverse bits of a given 32 bits unsigned integer.
Reverse Bits
Reverse bits of a given 32 bits unsigned integer.
public int reverseBits(int n) {
int reversed = 0;
for (int i = 0; i < 32; i++) {
reversed <<= 1;
reversed |= n & 1;
n >>= 1;
}
return reversed;
}Single Number II
Given an array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
Instead of using XOR directly, we need to count how many times 1 shows on each bit, it could be 1 or 3.
Single Number II
Single Number II
public int singleNumber(int[] nums) {
int[] digits = new int[32];
int res = 0;
for (int i = 0; i < 32; i++) {
for (int j = 0; j < nums.length; j++) {
digits[i] += ((nums[j] >> i) & 1);
}
res ^= ((digits[i] % 3) << i);
}
return res;
}Single Number II
public int singleNumber(int[] nums) {
int ans = 0;
for(int i = 0; i < 32; i++) {
int sum = 0;
for(int j = 0; j < nums.length; j++) {
if(((nums[j] >> i) & 1) == 1) {
sum++;
sum %= 3;
}
}
if(sum != 0) {
ans |= sum << i;
}
}
return ans;
}Single Number III
Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
For example:
Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].
Single Number III
We can use XOR to get the result, which is the XOR of the two numbers.
Then how we split them into two numbers?
Find the first bit that is different. Then the XOR value will be 1.
How to find the first bit 1?
Single Number III
public int[] singleNumber(int[] nums) {
// Get the XOR of the two numbers we need to find
int diff = 0;
for (int num : nums) {
diff ^= num;
}
// Get its last set bit
diff &= -diff;
int[] rets = new int[2];
for (int num : nums)
{
if ((num & diff) == 0) // the bit is not set
{
rets[0] ^= num;
}
else // the bit is set
{
rets[1] ^= num;
}
}
return rets;
}Combination
Given n different numbers,return all combinations.
- n numbers, 2^n combinations
- 2^n status can be represented by n-bits integer
- {1, 2, 3, 4}, 1100 -> {1, 2}
- {1, 2, 3, 4}, 0000 -> {}
Combination
Given n different numbers,return all combinations.
public static ArrayList<ArrayList<Integer>> combination(int[] nums) {
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
for (int i = 0; i < (1 << nums.length); i++) {
results.add(generate(nums, i));
}
return results;
}
public static ArrayList<Integer> generate(int[] nums, int index) {
ArrayList<Integer> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if ((index & 1) == 1) {
result.add(nums[i]);
}
index >>= 1;
}
return result;
}