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Posts tagged as “xor”

花花酱 LeetCode 2317. Maximum XOR After Operations

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++

花花酱 LeetCode 2220. Minimum Bit Flips to Convert Number

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++

花花酱 LeetCode 136. Single Number

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

Input: nums = [2,2,1]
Output: 1

Example 2:

Input: nums = [4,1,2,1,2]
Output: 4

Example 3:

Input: nums = [1]
Output: 1

Constraints:

  • 1 <= nums.length <= 3 * 104
  • -3 * 104 <= nums[i] <= 3 * 104
  • Each element in the array appears twice except for one element which appears only once.

Solution: XOR

single_number ^ a ^ b ^ c ^ … ^ a ^ b ^ c … = single_number

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

C++

Related Problems

花花酱 LeetCode 1863. Sum of All Subset XOR Totals

The XOR total of an array is defined as the bitwise XOR of all its elements, or 0 if the array is empty.

  • For example, the XOR total of the array [2,5,6] is 2 XOR 5 XOR 6 = 1.

Given an array nums, return the sum of all XOR totals for every subset of nums

Note: Subsets with the same elements should be counted multiple times.

An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b.

Example 1:

Input: nums = [1,3]
Output: 6
Explanation: The 4 subsets of [1,3] are:
- The empty subset has an XOR total of 0.
- [1] has an XOR total of 1.
- [3] has an XOR total of 3.
- [1,3] has an XOR total of 1 XOR 3 = 2.
0 + 1 + 3 + 2 = 6

Example 2:

Input: nums = [5,1,6]
Output: 28
Explanation: The 8 subsets of [5,1,6] are:
- The empty subset has an XOR total of 0.
- [5] has an XOR total of 5.
- [1] has an XOR total of 1.
- [6] has an XOR total of 6.
- [5,1] has an XOR total of 5 XOR 1 = 4.
- [5,6] has an XOR total of 5 XOR 6 = 3.
- [1,6] has an XOR total of 1 XOR 6 = 7.
- [5,1,6] has an XOR total of 5 XOR 1 XOR 6 = 2.
0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28

Example 3:

Input: nums = [3,4,5,6,7,8]
Output: 480
Explanation: The sum of all XOR totals for every subset is 480.

Constraints:

  • 1 <= nums.length <= 12
  • 1 <= nums[i] <= 20

Solution 1: Brute Force

Use an array A to store all the xor subsets, for a given number x
A = A + [x ^ a for a in A]

Time complexity: O(2n)
Space complexity: O(2n)

Python3

花花酱 LeetCode 1835. Find XOR Sum of All Pairs Bitwise AND

The XOR sum of a list is the bitwise XOR of all its elements. If the list only contains one element, then its XOR sum will be equal to this element.

  • For example, the XOR sum of [1,2,3,4] is equal to 1 XOR 2 XOR 3 XOR 4 = 4, and the XOR sum of [3] is equal to 3.

You are given two 0-indexed arrays arr1 and arr2 that consist only of non-negative integers.

Consider the list containing the result of arr1[i] AND arr2[j] (bitwise AND) for every (i, j) pair where 0 <= i < arr1.length and 0 <= j < arr2.length.

Return the XOR sum of the aforementioned list.

Example 1:

Input: arr1 = [1,2,3], arr2 = [6,5]
Output: 0
Explanation: The list = [1 AND 6, 1 AND 5, 2 AND 6, 2 AND 5, 3 AND 6, 3 AND 5] = [0,1,2,0,2,1].
The XOR sum = 0 XOR 1 XOR 2 XOR 0 XOR 2 XOR 1 = 0.

Example 2:

Input: arr1 = [12], arr2 = [4]
Output: 4
Explanation: The list = [12 AND 4] = [4]. The XOR sum = 4.

Constraints:

  • 1 <= arr1.length, arr2.length <= 105
  • 0 <= arr1[i], arr2[j] <= 109

Solution: Bit

(a[0] & b[i]) ^ (a[1] & b[i]) ^ … ^ (a[n-1] & b[i]) = (a[0] ^ a[1] ^ … ^ a[n-1]) & b[i]

We can pre compute that xor sum of array A.

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

C++