LeetCode 2939: Maximum Xor Product Solution
Master LeetCode problem 2939 (Maximum Xor Product), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2939. Maximum Xor Product
Problem Explanation
Explanation:
To solve this problem, we need to find the maximum value of the expression (a XOR x) * (b XOR x) for all possible values of x from 0 to 2^n. We can observe that the value of XOR operation between a and b will be maximized when the corresponding bits are different. Therefore, we aim to find the maximum value of XOR for each position of x.
We can approach this problem by iterating through the bits of x from the most significant bit to the least significant bit. For each bit position, we check if setting the bit to 1 results in a higher XOR value. If it does, we set the bit to 1; otherwise, we set it to 0.
The time complexity of this algorithm is O(n) where n is the number of bits in the input numbers a, b, and n. The space complexity is O(1) as we are not using any extra data structures.
Solution Code
class Solution {
public int maximizeXor(int a, int b, int n) {
int res = 0;
for (int i = n; i >= 0; i--) {
int mask = 1 << i;
int candidate = res | (1 << i);
if ((a & mask) == 0 && (b & mask) == 0 && candidate <= (1 << (n + 1)) - 1) {
res = candidate;
} else if ((a & mask) != 0 && (b & mask) != 0) {
res = candidate - 1;
}
}
return res;
}
}Try It Yourself
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