Leetcode 2425: Bitwise XOR of All Pairings
You are given two arrays, nums1 and nums2, each consisting of non-negative integers. You need to calculate the XOR of all possible pairings between the integers in nums1 and nums2. The result should be the XOR of all the numbers from this new array formed by XORing every integer in nums1 with every integer in nums2.
Problem
Approach
Steps
Complexity
Input: You are given two arrays nums1 and nums2. Each integer in nums1 will be paired with every integer in nums2 exactly once.
Example: nums1 = [5, 2], nums2 = [3, 4, 6]
Constraints:
• 1 <= nums1.length, nums2.length <= 10^5
• 0 <= nums1[i], nums2[j] <= 10^9
Output: Return the XOR of all the integers in the array formed by XORing each integer from nums1 with each integer from nums2.
Example: Output: 4
Constraints:
• The output will be an integer, which is the result of XORing all elements from the resulting array.
Goal: To compute the XOR of all pairings of elements from nums1 and nums2 efficiently.
Steps:
• 1. Iterate over all elements in nums1 and nums2.
• 2. For each pair, XOR the elements and accumulate the result.
• 3. After processing all pairs, return the XOR of all the results.
Goal: Make sure to handle the large sizes of the input arrays and optimize the solution to avoid excessive time complexity.
Steps:
• Ensure efficient handling of up to 10^5 elements in both nums1 and nums2.
Assumptions:
• The arrays nums1 and nums2 are non-empty and contain non-negative integers.
• The XOR operation will be applied to all pairings of elements from nums1 and nums2.
• Input: nums1 = [5, 2], nums2 = [3, 4, 6]
• Explanation: We first calculate the pairwise XOR results: 5^3, 5^4, 5^6, 2^3, 2^4, 2^6. These results are then XORed together to get the final result.
Approach: We need to calculate the XOR of all pairings efficiently. Instead of explicitly creating the array, we can optimize the calculation by leveraging properties of XOR.
Observations:
• The XOR operation has the property that it is both commutative and associative, meaning the order of operations does not matter.
• We can avoid explicitly building the nums3 array by directly calculating the XOR as we process the pairings of nums1 and nums2.
Steps:
• 1. XOR all elements in nums1 together to get xr1.
• 2. XOR all elements in nums2 together to get xr2.
• 3. Depending on the parity (even or odd) of the lengths of nums1 and nums2, return the result using the XOR properties.
Empty Inputs:
• Since nums1 and nums2 are non-empty as per the problem constraints, no need to handle empty arrays.
Large Inputs:
• The solution should handle the case where nums1 and nums2 contain up to 10^5 elements.
Special Values:
• The solution must handle large values in nums1 and nums2, which can go up to 10^9.
Constraints:
• Make sure the solution is optimized for time complexity, especially when handling large inputs.
int xorAllNums(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size(), n = nums2.size();
int xr1 = nums1[0], xr2 = nums2[0];
for(int i = 1; i < m; i++)
xr1 ^= nums1[i];
for(int i = 1; i < n; i++)
xr2 ^= nums2[i];
if((n % 2) == 0) {
if((m % 2) == 0) {
return 0;
} else {
return xr2;
}
} else {
if((m % 2) == 0) {
return xr1;
} else {
return xr1 ^ xr2;
}
}
return -1;
}
1 : Code Initialization
int xorAllNums(vector<int>& nums1, vector<int>& nums2) {
Defines the function that will calculate the XOR of two integer vectors.
2 : Variable Declaration
int m = nums1.size(), n = nums2.size();
Declares and initializes variables `m` and `n` to store the sizes of `nums1` and `nums2`.
3 : XOR Initialization
int xr1 = nums1[0], xr2 = nums2[0];
Initializes `xr1` and `xr2` to the first elements of `nums1` and `nums2`, respectively, to begin the XOR operation.
4 : XOR Operation 1
for(int i = 1; i < m; i++)
Loop through the rest of `nums1` starting from index 1.
5 : XOR Operation 2
xr1 ^= nums1[i];
XORs each element of `nums1` with `xr1` to accumulate the result.
6 : XOR Operation 3
for(int i = 1; i < n; i++)
Loop through the rest of `nums2` starting from index 1.
7 : XOR Operation 4
xr2 ^= nums2[i];
XORs each element of `nums2` with `xr2` to accumulate the result.
8 : Condition 1
if((n % 2) == 0) {
Checks if the size of `nums2` is even.
9 : Condition 2
if((m % 2) == 0) {
Checks if the size of `nums1` is even, when `nums2` is even.
10 : Return 0
return 0;
If both `nums1` and `nums2` are of even size, return 0.
11 : Return 1
} else {
If `nums1` is of odd size and `nums2` is of even size.
12 : Return 2
return xr2;
Returns the XOR of `nums2` if `nums1` is of odd size and `nums2` is even.
13 : Else Condition 1
} else {
Else part when `nums2` has odd size.
14 : Condition 4
if((m % 2) == 0) {
Checks if the size of `nums1` is even, when `nums2` is odd.
15 : Return 3
return xr1;
Returns `xr1` if `nums1` is even and `nums2` is odd.
16 : Condition 5
} else {
If both `nums1` and `nums2` are of odd size.
17 : Return 4
return xr1 ^ xr2;
Returns the XOR of both `xr1` and `xr2` if both vectors have odd sizes.
18 : Final Return
return -1;
Return -1 as a fallback if none of the conditions match.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) because we only need to iterate through nums1 and nums2 once.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we only need a few variables to store intermediate results.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus