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1879. Minimum XOR Sum of Two Arrays
Description
You are given two integer arrays nums1 and nums2 of length n.
The XOR sum of the two integer arrays is (nums1[0] XOR nums2[0]) + (nums1[1] XOR nums2[1]) + ... + (nums1[n - 1] XOR nums2[n - 1]) (0-indexed).
- For example, the XOR sum of
[1,2,3]and[3,2,1]is equal to(1 XOR 3) + (2 XOR 2) + (3 XOR 1) = 2 + 0 + 2 = 4.
Rearrange the elements of nums2 such that the resulting XOR sum is minimized.
Return the XOR sum after the rearrangement.
Example 1:
Input: nums1 = [1,2], nums2 = [2,3] Output: 2 Explanation: Rearrangenums2so that it becomes[3,2]. The XOR sum is (1 XOR 3) + (2 XOR 2) = 2 + 0 = 2.
Example 2:
Input: nums1 = [1,0,3], nums2 = [5,3,4] Output: 8 Explanation: Rearrangenums2so that it becomes[5,4,3]. The XOR sum is (1 XOR 5) + (0 XOR 4) + (3 XOR 3) = 4 + 4 + 0 = 8.
Constraints:
n == nums1.lengthn == nums2.length1 <= n <= 140 <= nums1[i], nums2[i] <= 107
Solutions
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class Solution { public int minimumXORSum(int[] nums1, int[] nums2) { int n = nums1.length; int[][] f = new int[n + 1][1 << n]; for (var g : f) { Arrays.fill(g, 1 << 30); } f[0][0] = 0; for (int i = 1; i <= n; ++i) { for (int j = 0; j < 1 << n; ++j) { for (int k = 0; k < n; ++k) { if ((j >> k & 1) == 1) { f[i][j] = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + (nums1[i - 1] ^ nums2[k])); } } } } return f[n][(1 << n) - 1]; } } -
class Solution { public: int minimumXORSum(vector<int>& nums1, vector<int>& nums2) { int n = nums1.size(); int f[n + 1][1 << n]; memset(f, 0x3f, sizeof(f)); f[0][0] = 0; for (int i = 1; i <= n; ++i) { for (int j = 0; j < 1 << n; ++j) { for (int k = 0; k < n; ++k) { if (j >> k & 1) { f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + (nums1[i - 1] ^ nums2[k])); } } } } return f[n][(1 << n) - 1]; } }; -
class Solution: def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int: n = len(nums2) f = [[inf] * (1 << n) for _ in range(n + 1)] f[0][0] = 0 for i, x in enumerate(nums1, 1): for j in range(1 << n): for k in range(n): if j >> k & 1: f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + (x ^ nums2[k])) return f[-1][-1] -
func minimumXORSum(nums1 []int, nums2 []int) int { n := len(nums1) f := make([][]int, n+1) for i := range f { f[i] = make([]int, 1<<n) for j := range f[i] { f[i][j] = 1 << 30 } } f[0][0] = 0 for i := 1; i <= n; i++ { for j := 0; j < 1<<n; j++ { for k := 0; k < n; k++ { if j>>k&1 == 1 { f[i][j] = min(f[i][j], f[i-1][j^(1<<k)]+(nums1[i-1]^nums2[k])) } } } } return f[n][(1<<n)-1] } -
function minimumXORSum(nums1: number[], nums2: number[]): number { const n = nums1.length; const f: number[][] = Array(n + 1) .fill(0) .map(() => Array(1 << n).fill(1 << 30)); f[0][0] = 0; for (let i = 1; i <= n; ++i) { for (let j = 0; j < 1 << n; ++j) { for (let k = 0; k < n; ++k) { if (((j >> k) & 1) === 1) { f[i][j] = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + (nums1[i - 1] ^ nums2[k])); } } } } return f[n][(1 << n) - 1]; }