Bit Manipulation

Author@wkw

Overview

We all know that information is stored in the form of bits in computer memory, which means directly manipulating these bits will yield faster results than performing computation on ordinary data.

Binary uses only $0$ and $1$ to represent a number in a base-2 number system. The series of 0 and 1 are called bits. If the bit is $1$ , then this bit is set. We read binary number from right to left. For example, the binary representation of number $9$ is $1001_2$ which can be calculated by summing up all the set bit: $2^3 + 2^0 = 9_{10}$ . Bit Manipulation utilises different bitwise operations to manipulate bits.

Bitwise Operators

X	Y	X & Y	X \| Y	X ^ Y	~ X
0	0	0	0	0	1
0	1	0	1	1	1
1	0	0	1	1	0
1	1	1	1	0	0

AND (&) Operator

& takes two bit integers to compare. If the bits are both $1$ , then the resulting bit is $1$ , else $0$ .

For example, $0010_2$ & $0011_2 = 0010_2$ because only the second bits from the right are both $1$ .

Usage #1: Check if the rightmost bit is set

Let's say $n$ is $5_{10}$ which is $0101_2$ . If we execute $n$ & $1$ , i.e. $0101_2$ & $0001_2$ , the result is $0001_2$ because only the rightmost bits are both 1 and other bits would return 0.

Usage #2: Check if the i-th bit is set

Let's say $n$ is $5_{10}$ which is $0101_2$ . How to check if the 2-nd, 3-rd, or 4-th bit is set? Using the same idea, the mask would be $0010_2$ , $0100_2$ , $1000_2$ respectively. And you may notice that the mask is always a power of 2. A common way to do it is to use left shift operator (which will be discussed below), i.e. $n$ & $(1 << i)$ where $n$ is the i-th bit to be checked.

Usage #3: Remove the rightmost set bit

We can use $n$ & $(n - 1)$ to remove the rightmost set bit.

n     n     n - 1  n & (n - 1)
--   ----   ----   -------
 0   0000   0111    0000
 1   0001   0000    0000
 2   0010   0001    0000
 3   0011   0010    0010
 4   0100   0011    0000
 5   0101   0100    0100
 6   0110   0101    0100
 7   0111   0110    0110
 8   1000   0111    0000
 9   1001   1000    1000
10   1010   1001    1000
11   1011   1010    1010
12   1100   1011    1000
13   1101   1100    1100
14   1110   1101    1100
15   1111   1110    1110

Example #1: 0231 - Power of Two (Easy)

We know that a power of 2 is a positive number and only has one bit set. We can use $n$ & $(n - 1)$ to see the result is 0 or not to determine if the target value is a power of 2 or not.

In short, $n$ & $(n - 1)$ never have a 1 bit in the same place.

For example, 1000 (8), and 0111(7) - never have 1 bit in the same place for Power of Two

class Solution {
public:
    bool isPowerOfTwo(int n) {
        // 1. check if it is a positive number
        // 2. check the value is 0 after removing the rightmost bit
        return n > 0 && !(n & (n - 1));
    }
};

Example #2: 0191 - Number of 1 Bits

Instead of checking the bits one by one, we can use $n$ & $(n - 1)$ to jump to the next set bit.

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        for (; n; n = n & (n - 1)) ans++;
        return ans;
    }
};

OR (|) Operator

$\vert$ takes two bit integers to compare. If either bits are $1$ , then the resulting bit is $1$ , else $0$ .

For example, $0010_2 | 0011_2 = 0011_2$ because only the first and the second bits from the right are $1$ in either value.

Usage #1: Set the rightmost bit

Let's say $n$ is $4_{10}$ which is $0100_2$ . If we execute $n \vert 1$ , i.e. $0100_2 \vert 0001_2$ , the result is $0101_2$ because $0001_2$ would turn the rightmost bit of $0100_2$ to $1$ since that bit is set.

Usage #2: Set the i-th bit

Let's say $n$ is $4_{10}$ which is $0100_2$ . How to set other bits? Similar to above example, we can use left shift operator (which will be discussed below), i.e. $n \vert (1 << i)$ where $n$ is the i-th bit to be set.

Example: 0421 - Maximum XOR of Two Numbers in an Array

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        int ans = 0, mask = 0;
        for(int i = 31; i >= 0; i--){
           unordered_set<int> s;
           // set i-th in mask
            mask |= (1 << i);
            for (auto x : nums) s.insert(mask & x);
            // set i-th in ans
            int best = ans | (1 << i);
            for(auto pref : s){
                if(s.find(pref ^ best) != s.end()){
                    ans = best;
                    break;
                }
            }

        }
        return ans;
    }
};

XOR (^)

^ takes two bit intergers to compare. If one bit is zero and another bit is one, then the resulting bit is $1$ , else $0$ .

For example, $0010_2$ ^ $0011_2 = 0001_2$ because the first bit got $0$ and $1$ while the second bit got both $1$ .

XOR Properties

X ^ 0 = X
X ^ X = 0
X ^ 1 = ~X (Compliment of that number)
X ^ Y = Y ^ X (Commutativity)
(X ^ Y) ^ Z = X ^ (Y ^ Z) (Associativity)

Example #1: 0268 - Missing Number

Given the fact that we know $n$ distinct numbers in the range $[0, n]$ , we can find the missing number using the above XOR properties.

For example, let's say the input is $[0, 1, 3]$ and we know the the missing number is $2$ . We can compare the index (0, 1, 2) and the value (0, 1, 3) and write $3$ ^ $(0$ ^ $0)$ ^ $(1$ ^ $1)$ ^ $(2$ ^ $3)$ .

Based on property #2, we know $0$ ^ $0$ and $(1$ ^ $1)$ would be $0$ .

Based on property #1, we know that $0$ ^ $1$ would be $1$ . Therefore, we got $3$ ^ $(2$ ^ $3)$ .

Based on property #4, we can rewrite as $2$ ^ $(3$ ^ $3)$ and use property #2 again to get $2$ ^ $0$ .

Based on property #1, we have our final answer which is $2$ .

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        // index = 0 1 2
        // value = 0 1 3
        // n ^ (0 ^ 0) ^ (1 ^ 1) ^ (2 ^ 3)
        // 3 ^ 0 ^ 0 ^ 2 ^ 3
        // 0 ^ 0 ^ 2 ^ 3 ^ 3
        // 2 ^ 0
        // 2
        int n = nums.size(), ans = n;
        for(int i = 0; i < n; i++) ans ^= (i ^ nums[i]);
        return ans;
    }
};

Example #2: 0136 - Single Number

As every element appears twice except for one. We can use property #2 to make all elements which appear twice become $0$ . At the end, there would be $0$ and that element which appears once. Then we use property #1 to get the final answer.

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        // nums: [4,1,2,1,2]
        // 0 ^ 4 ^ 1 ^ 2 ^ 1 ^ 2
        // 0 ^ (1 ^ 1) ^ (2 ^ 2) ^ 4
        // 0 ^ 0 ^ 0 ^ 4
        // 4
        int ans = 0;
        for(auto x : nums) ans ^= x;
        return ans;
    }
};

Example #3: Swap 2 numbers

XOR (^) can be used to swap 2 numbers by changing the bits and reversing it.

Let's say, $a = 4$ $(0100_2)$ , $b = 6$ $(0110_2)$ , we want $a = 6$ and $b = 4$ as our answer.

a = a ^ b => (4 ^ 6) => (0100 ^ 0110) => 2 (0010)
b = b ^ a => (6 ^ 2) => (0110 ^ 0010) => 4 (0100)
a = a ^ b => (2 ^ 4) => (0010 ^ 0100) => 6 (0110)

NOT (~)

~ inverts (flip) all the bits of a bit intergers, which means $1$ would become $0$ and vice versa. If we apply it on a positive integer $x$ , then it is simply $-x-1$ .

For example, if we apply NOT on $~0010_2$ , then the resulting value is $1101_2$ .

The below code snippets are actually equivalent

for (int i = n; i >= 0; i--) {
    // i = n, n - 1, n - 2, ..., 2, 1, 0
}

for (int i = n; ~i; i--) {
    // i = n, n - 1, n - 2, ..., 2, 1, 0
}

Left-Shift

$<<$ shifts the bits to the left. For example, $1 << 1 = 2$ because we shift the $1$ $(0001_2)$ to the left to become $2$ $(0010_2)$ .

Similarily, $1 << 2 = 4$ because we shift the $1$ $(0001_2)$ to the left twice to become $4$ $(0100_2)$ .

And you may find that $1 << n$ is actually $2 ^ n$ . Also $n << m$ means multiplying n by 2 power m. i.e, $n = n * (2^m)$ .

In simple, $n << m$ shifting each bit of n to left m times. Let's say, $n = 8$ and $m = 2$ . $n$ can be represented as $1000_2$ in binary, Therefore, $8 << 2$ = $1000_2 << 2$ = $100000_2 (32)$ , which is same as $(8 * 2^2)$ = $32$

Example #1: 0078 - Subsets

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        int n = nums.size();
        // number of subsets for n elements would be 2 ^ n
        // because for each element, you can choose to take it or not
        // if take = 1, don't take = 0, then we can use bit manipulation
        int p = 1 << n; // 1 * 2 ^ n
        vector<vector<int>> ans;
        for(int i = 0; i < p; i++){
            vector<int> t;
            for(int j = 0; j < n; j++){
               if((1 << j) & i) t.emplace_back(nums[j]);
            }
            ans.emplace_back(t);
        }
        return ans;
    }
};

Right-Shift

$>>$ shifts the bits to the right. For example $3_{10}$ $(0011_2)$ $>> 1$ would become $1$ $(0010_2)$ .

$4 >> 1$ dividing $4$ by $2$ . Also $n >> m$ means dividing $n$ by $2$ power m. i.e, $n = n / (2^m)$

In simple, $n >> m$ shifting each bit of n to right m times. Let's say, $n = 8$ and $m = 2$ . $n$ can be represented as $1000_2$ in binary, Therefore, $8 >> 2$ = $1000_2$ >> $2$$ =$ 0010_2 $(4), which is same as$ (8 / 2^2) $=$ 4$

Example: Check the bits one by one

while (n > 0) {
    int bit = n & 1;
    // do something with bit
    // ...

    // shift bits to the right
    // which is same as n /= 2
    n >>= 1;
}

Two's Compliment and Negative Numbers

Computers typically store integers in two’s complement representation. A positive number is represented as itself while a negative number is represented as the two’s complement of it’s absolute value (with a 1 in its sign bit to indicate that a negative value).

Let’s look at the 4-bit integer $-3$ as an example. If it’s a 4-bit number, we have one bit for the sign and three bits for the value. We want the complement with respect to $2^3$ , which is $8$ .

The complement of $3$ (the absolute value of $-3$ ) with respect to $8$ is $5$ . Binary value of $5$ is $0101_2$ . Therefore, $-3$ in binary as a 4-bit number is $1101_2$ , with the first bit being the sign bit.

In other words, the binary representation of $-K$ (negative K) as a N-bit number is $concat(1, 2^N-1 - K)$ otherwise prepend (prefix) the sign bit $(1)$ with the value for the calculated complement.

Another way to look at this is that we invert the bits in the positive representation and then add $1$ . $3$ is $0011_2$ in binary.

Flip the bits to get 100, add $1$ to get $101_2$ , then prepend the sign bit $1$ to get $1101_2$ .

Positive  Values  Negative  Values
--------  ------  --------  ------
7         0 111   -1        1 111
6         0 110   -2        1 110
5         0 101   -3        1 101
4         0 100   -4        1 100
3         0 011   -5        1 011
2         0 010   -6        1 010
1         0 001   -7        1 001
0         0 000

Common Bit Tasks

Get Bit

It shifts $1$ over by i bits, creating a value that looks like $0100_2$ $(2^i)$ . By performing and AND (&) with num, we clear all bits other than the bit at bit i.

Finally, compare that value to $0$ , if that new value is not zero, then bit i must have a $1$ , otherwise, bit i is a $0$ .

int getBit(int num, int i) {
    return num & (1 << i);
}

Set Bit

It shits $1$ over by i bits, creating a value that looks like $0100_2$ ( $2$ to the power of $i$ ). By performing an $OR$ with num, only the value at bit i will change.

All other bits of the mask are zero and will not affect num.

int setBit(int num, int i) {
    return num | (1 << i);
}

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