Number of Unique XOR Triplets I

1. Handle the base cases for small n.
   • If n = 1, the only possible triplet is (1,1,1) giving 1^1^1=1. The result is 1.
   • If n = 2, the unique values are {1, 2}. The result is 2.
2. For n >= 3, observe the pattern of the unique values.
   • n=3: {1,2,3}. 1^2^3=0. Unique values are {0,1,2,3}. Size 4.
   • n=4: {1,2,3,4}. Triplets generate {0,5,6,7}. Unique values are {0,1,2,3,4,5,6,7}. Size 8.
3. Generalize the pattern: For n >= 3, the set of unique XOR values is {0, 1, ..., p-1}, where p is the smallest power of 2 that is strictly greater than n.
4. The size of this set is p. This can be calculated efficiently using bit manipulation.
5. Implement the logic: check for n=1 and n=2, otherwise calculate and return p.

The key insight is that since nums is a permutation of {1, ..., n}, the problem is independent of the order of elements in nums and only depends on n. The set of unique values is {1, ..., n} igcup {a^b^c | a,b,c ext{ are distinct in } {1,...,n}}.

By analyzing small cases:

• n=1: nums=[1]. Values: {1}. Size: 1.
• n=2: nums is perm of {1,2}. Values: {1,2}. Size: 2.
• n=3: nums is perm of {1,2,3}. Values: {1,2,3} and 1^2^3=0. Total: {0,1,2,3}. Size: 4.
• n=4: nums is perm of {1,2,3,4}. Values: {1,2,3,4} and {1^2^3=0, 1^2^4=7, 1^3^4=6, 2^3^4=5}. Total: {0,1,2,3,4,5,6,7}. Size: 8.

For n >= 3, it can be shown that the set of all possible XOR triplet values forms a complete range of numbers from 0 to p-1, where p is the smallest power of 2 strictly greater than n. The size of this set is p.

This value p can be found efficiently. For a given n, p is 2^k where k is the smallest integer such that 2^k > n. This is equivalent to finding the next power of two after n.

For example, if n=5, the next power of two is 8. If n=8, the next power of two is 16.

This leads to a very simple and fast implementation.

1class Solution {2    public int countUniqueXorTriplets(int[] nums) {3        int n = nums.length;4 5        if (n == 1) {6            return 1;7        }8        if (n == 2) {9            return 2;10        }11 12        // For n >= 3, the result is the smallest power of 2 strictly greater than n.13        // This can be found using bit manipulation.14        // Integer.highestOneBit(n) finds the largest power of 2 less than or equal to n.15        // If n is a power of 2, say n = 8, highestOneBit(8) is 8. We need 16. So 8 << 1.16        // If n is not a power of 2, say n = 7, highestOneBit(7) is 4. We need 8. So 4 << 1.17        // This logic holds for n >= 1, but we've handled n=1,2 separately for clarity.18        // Let's re-check the formula for n=3: highestOneBit(3) is 2. 2 << 1 = 4. Correct.19        return Integer.highestOneBit(n) << 1;20    }21}

1. Initialize an empty HashSet called uniqueValues to store the unique XOR triplet results.
2. Get the length of the input array nums, let it be n.
3. Use three nested loops to iterate through all possible triplets of indices (i, j, k) such that 0 <= i <= j <= k < n.
4. In the innermost loop, calculate the XOR sum: value = nums[i] ^ nums[j] ^ nums[k].
5. Add the calculated value to the uniqueValues set. The set automatically handles duplicates.
6. After the loops complete, the number of unique XOR triplets is the size of the uniqueValues set. Return this size.

This method involves a brute-force enumeration of all valid triplets. We use three nested loops. The outer loop runs from i = 0 to n-1, the middle loop from j = i to n-1, and the inner loop from k = j to n-1. This structure ensures that we only consider triplets (i, j, k) where i <= j <= k.

For each triplet, we compute the XOR sum of the corresponding elements from the nums array. To count the number of unique results, we use a HashSet. Each computed XOR value is added to the set. If the value is already present, the set remains unchanged. Finally, the total number of unique values is simply the size of the hash set.

1import java.util.HashSet;2import java.util.Set;3 4class Solution {5    public int countUniqueXorTriplets(int[] nums) {6        int n = nums.length;7        Set<Integer> uniqueValues = new HashSet<>();8 9        for (int i = 0; i < n; i++) {10            for (int j = i; j < n; j++) {11                for (int k = j; k < n; k++) {12                    uniqueValues.add(nums[i] ^ nums[j] ^ nums[k]);13                }14            }15        }16 17        return uniqueValues.size();18    }19}

1. Identify the set of possible values: Realize that the set of unique XOR values is the union of {1, ..., n} and {a^b^c | a,b,c ext{ are distinct in } {1,...,n}}. This can be shown to be equivalent to the set {p^c | p ext{ in } P_2, c ext{ in } S}, where S = {1,...,n} and P_2 = {a^b | a,b ext{ distinct in } S}.
2. Represent sets as polynomials: Create boolean arrays (polynomials) polyS and polyP2 of size N (smallest power of 2 > 2n) to represent the characteristic functions of the sets S and P_2.
3. Compute polyP2 using FWHT: The set P_2 is (S igoplus S) \setminus \{0\}. Its polynomial can be computed via XOR convolution of polyS with itself: poly(S igoplus S) = IFWHT(FWHT(polyS) igodot FWHT(polyS)). Set the coefficient for 0 to be false to get polyP2.
4. Compute the final set polynomial: The final set of values is P_2 igoplus S. Its polynomial is found by another XOR convolution: poly(Result) = IFWHT(FWHT(polyP2) igodot FWHT(polyS)).
5. Count unique values: The number of unique values is the number of non-zero coefficients in the final polynomial poly(Result).

This approach transforms the combinatorial problem into an algebraic one using polynomial convolution.

1. Problem Reformulation: The set of all possible XOR values is S igcup P_3, where S = {1, ..., n} and P_3 = {a^b^c | a,b,c ext{ distinct in } S}. This can be shown to be equivalent to P_2 igoplus S = {p^c | p ext{ in } P_2, c ext{ in } S}, where P_2 = {a^b | a,b ext{ distinct in } S}.

2. XOR Convolution: We can find the elements of a sumset A igoplus B by computing the XOR convolution of their characteristic polynomials. The FWHT algorithm computes this convolution in O(N log N) time, where N is the size of the polynomial domain (a power of two large enough to hold all values).

3. Algorithm Steps: a. Choose N to be the smallest power of 2 greater than 2n. b. Create a polynomial polyS for S = {1, ..., n}. c. Use FWHT to compute polyP2 from polyS by first computing the convolution polyS * polyS. d. Use FWHT again to compute the convolution polyP2 * polyS to get the final result polynomial. e. Count the number of non-zero entries in the result polynomial.

1import java.util.Arrays;2 3class Solution {4    // This is a conceptual implementation. For competitive programming,5    // a more optimized FWHT implementation would be used.6    public int countUniqueXorTriplets(int[] nums) {7        int n = nums.length;8        if (n == 0) return 0;9 10        int N = 1;11        while (N <= 2 * n) {12            N <<= 1;13        }14 15        long[] polyS = new long[N];16        for (int i = 1; i <= n; i++) {17            polyS[i] = 1;18        }19 20        long[] h_S = Arrays.copyOf(polyS, N);21        fwht(h_S, false);22 23        long[] h_S_conv_S = new long[N];24        for (int i = 0; i < N; i++) {25            h_S_conv_S[i] = h_S[i] * h_S[i];26        }27        fwht(h_S_conv_S, true);28 29        long[] polyP2 = new long[N];30        for (int i = 1; i < N; i++) {31            if (h_S_conv_S[i] > 0) {32                polyP2[i] = 1;33            }34        }35 36        long[] h_P2 = Arrays.copyOf(polyP2, N);37        fwht(h_P2, false);38 39        long[] h_Q = new long[N];40        for (int i = 0; i < N; i++) {41            h_Q[i] = h_P2[i] * h_S[i];42        }43        fwht(h_Q, true);44 45        int count = 0;46        for (int i = 0; i < N; i++) {47            if (h_Q[i] > 0) {48                count++;49            }50        }51        return count;52    }53 54    private void fwht(long[] a, boolean inverse) {55        int n = a.length;56        for (int len = 1; len < n; len <<= 1) {57            for (int i = 0; i < n; i += 2 * len) {58                for (int j = 0; j < len; j++) {59                    long u = a[i + j];60                    long v = a[i + len + j];61                    a[i + j] = u + v;62                    a[i + len + j] = u - v;63                }64            }65        }66 67        if (inverse) {68            for (int i = 0; i < n; i++) {69                a[i] /= n;70            }71        }72    }73}