# Posts tagged as “bit”

You are given an integer array nums of size n.

Consider a non-empty subarray from nums that has the maximum possible bitwise AND.

• In other words, let k be the maximum value of the bitwise AND of any subarray of nums. Then, only subarrays with a bitwise AND equal to k should be considered.

Return the length of the longest such subarray.

The bitwise AND of an array is the bitwise AND of all the numbers in it.

subarray is a contiguous sequence of elements within an array.

Example 1:

Input: nums = [1,2,3,3,2,2]
Output: 2
Explanation:
The maximum possible bitwise AND of a subarray is 3.
The longest subarray with that value is [3,3], so we return 2.


Example 2:

Input: nums = [1,2,3,4]
Output: 1
Explanation:
The maximum possible bitwise AND of a subarray is 4.
The longest subarray with that value is [4], so we return 1.


Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 106

## Solution: Find the largest number

a & b <= a
a & b <= b
if b > a, a & b < b, we choose to start a new sequence of “b” instead of continuing with “ab”

Basically, we find the largest number in the array and count the longest sequence of it. Note, there will be some tricky cases like.
b b b b a b
b a b b b b
We need to return 4 instead of 1.

Time complexity: O(n)
Space complexity: O(1)

## C++

You are given a 0-indexed integer array nums. In one operation, select any non-negative integer x and an index i, then update nums[i] to be equal to nums[i] AND (nums[i] XOR x).

Note that AND is the bitwise AND operation and XOR is the bitwise XOR operation.

Return the maximum possible bitwise XOR of all elements of nums after applying the operation any number of times.

Example 1:

Input: nums = [3,2,4,6]
Output: 7
Explanation: Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
It can be shown that 7 is the maximum possible bitwise XOR.
Note that other operations may be used to achieve a bitwise XOR of 7.

Example 2:

Input: nums = [1,2,3,9,2]
Output: 11
Explanation: Apply the operation zero times.
The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
It can be shown that 11 is the maximum possible bitwise XOR.

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 108

## Solution: Bitwise OR

The maximum possible number MAX = nums[0] | nums[1] | … | nums[n – 1].

We need to prove:
1) MAX is achievable.
2) MAX is the largest number we can get.

nums[i] AND (nums[i] XOR x) means that we can turn any 1 bits to 0 for nums[i].

1) If the i-th bit of MAX is 1, which means there are at least one number with i-th bit equals to 1, however, for XOR, if there are even numbers with i-th bit equal to one, the final results will be 0 for i-th bit, we get a smaller number. By using the operation, we can choose one of them and flip the bit.

**1** XOR **1** XOR **1** XOR **1** = **0** =>
**0** XOR **1** XOR **1** XOR **1** = **1**

2) If the i-th bit of MAX is 0, which means the i-th bit of all the numbers is 0, there is nothing we can do with the operation, and the XOR will be 0 as well.
e.g. **0** XOR **0** XOR **0** XOR **0** = **0**

Time complexity: O(n)
Space complexity: O(1)

## C++

bit flip of a number x is choosing a bit in the binary representation of x and flipping it from either 0 to 1 or 1 to 0.

• For example, for x = 7, the binary representation is 111 and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get 110, flip the second bit from the right to get 101, flip the fifth bit from the right (a leading zero) to get 10111, etc.

Given two integers start and goal, return the minimum number of bit flips to convert start to goal.

Example 1:

Input: start = 10, goal = 7
Output: 3
Explanation: The binary representation of 10 and 7 are 1010 and 0111 respectively. We can convert 10 to 7 in 3 steps:
- Flip the first bit from the right: 1010 -> 1011.
- Flip the third bit from the right: 1011 -> 1111.
- Flip the fourth bit from the right: 1111 -> 0111.
It can be shown we cannot convert 10 to 7 in less than 3 steps. Hence, we return 3.

Example 2:

Input: start = 3, goal = 4
Output: 3
Explanation: The binary representation of 3 and 4 are 011 and 100 respectively. We can convert 3 to 4 in 3 steps:
- Flip the first bit from the right: 011 -> 010.
- Flip the second bit from the right: 010 -> 000.
- Flip the third bit from the right: 000 -> 100.
It can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return 3.


Constraints:

• 0 <= start, goal <= 109

Solution: XOR

start ^ goal will give us the bitwise difference of start and goal in binary format.
ans = # of 1 ones in the xor-ed results.
For C++, we can use __builtin_popcount or bitset<32>::count() to get the number of bits set for a given integer.

Time complexity: O(1)
Space complexity: O(1)

## C++

Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it.

zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).

Example 1:

Input: nums = [0,2,1,5,3,4]
Output: [0,1,2,4,5,3]
Explanation: The array ans is built as follows:
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
= [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]
= [0,1,2,4,5,3]

Example 2:

Input: nums = [5,0,1,2,3,4]
Output: [4,5,0,1,2,3]
Explanation: The array ans is built as follows:
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
= [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]]
= [4,5,0,1,2,3]

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] < nums.length
• The elements in nums are distinct.

Follow-up: Can you solve it without using an extra space (i.e., O(1) memory)?

## Solution 1: Straight forward

Time complexity: O(n)
Space complexity: O(n)

## Solution 2: Follow up: Inplace Encoding

Since nums[i] <= 1000, we can use low 16 bit to store the original value and high 16 bit for new value.

Time complexity: O(n)
Space complexity: O(1)

## C++

Given an integer n, return true if it is a power of two. Otherwise, return false.

An integer n is a power of two, if there exists an integer x such that n == 2x.

Example 1:

Input: n = 1
Output: true
Explanation: 20 = 1


Example 2:

Input: n = 16
Output: true
Explanation: 24 = 16


Example 3:

Input: n = 3
Output: false


Example 4:

Input: n = 4
Output: true


Example 5:

Input: n = 5
Output: false


Constraints:

• -231 <= n <= 231 - 1

## Solution: 1 bit set

Any power of two has only 1 bit set in the binary format. e.g. 1=0b01, 2=0b10, 4=0b100, 8=0b1000

1. use popcount.

## C++

2. Use (n) & (n – 1) to remove the last set bit, so it should be zero.