1442. Count Triplets That Can Form Two Arrays of Equal XOR
Explanation
To solve this problem, we can iterate through all possible triplets (i, j, k) and calculate the values of a and b. We then check if a == b for each triplet and count the valid triplets. The key observation here is that a == b is equivalent to arr[i] ^ ... ^ arr[j] == arr[j] ^ ... ^ arr[k].
We can precalculate the XOR prefix sums of the given array to efficiently calculate the XOR of any subarray. Using this, we can iterate through all possible pairs of indices (i, j) and use a hashmap to store the count of XOR values encountered so far. Then, for each valid k, we check if arr[i] ^ ... ^ arr[j] == arr[j] ^ ... ^ arr[k] and increment the count accordingly.
Time Complexity: O(n) where n is the length of the input array arr. Space Complexity: O(n) for the hashmap to store XOR values.
class Solution {
public int countTriplets(int[] arr) {
int n = arr.length;
int[] xorPrefix = new int[n + 1];
for (int i = 0; i < n; i++) {
xorPrefix[i + 1] = xorPrefix[i] ^ arr[i];
}
int count = 0;
Map<Integer, Integer> xorCount = new HashMap<>();
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int a = xorPrefix[i] ^ xorPrefix[j];
if (xorCount.containsKey(a)) {
count += xorCount.get(a);
}
if (xorPrefix[j] == xorPrefix[i]) {
count++;
}
xorCount.put(xorPrefix[j], xorCount.getOrDefault(xorPrefix[j], 0) + 1);
}
}
return count;
}
}Code Editor (Testing phase)
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