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2429. Minimize XOR

Description

Given two positive integers num1 and num2, find the positive integer x such that:

  • x has the same number of set bits as num2, and
  • The value x XOR num1 is minimal.

Note that XOR is the bitwise XOR operation.

Return the integer x. The test cases are generated such that x is uniquely determined.

The number of set bits of an integer is the number of 1's in its binary representation.

 

Example 1:

Input: num1 = 3, num2 = 5
Output: 3
Explanation:
The binary representations of num1 and num2 are 0011 and 0101, respectively.
The integer 3 has the same number of set bits as num2, and the value 3 XOR 3 = 0 is minimal.

Example 2:

Input: num1 = 1, num2 = 12
Output: 3
Explanation:
The binary representations of num1 and num2 are 0001 and 1100, respectively.
The integer 3 has the same number of set bits as num2, and the value 3 XOR 1 = 2 is minimal.

 

Constraints:

  • 1 <= num1, num2 <= 109

Solutions

Solution 1: Greedy + Bit Manipulation

According to the problem description, we first calculate the number of set bits $cnt$ in $num2$, then enumerate each bit of $num1$ from high to low. If the bit is $1$, we set the corresponding bit in $x$ to $1$ and decrement $cnt$ by $1$, until $cnt$ is $0$. If $cnt$ is still not $0$ at this point, we start from the low bit and set each bit of $num1$ that is $0$ to $1$, and decrement $cnt$ by $1$, until $cnt$ is $0$.

The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the maximum value of $num1$ and $num2$.

  • class Solution {
    public int minimizeXor(int num1, int num2) {
        int cnt = Integer.bitCount(num2);
        int x = 0;
        for (int i = 30; i >= 0 && cnt > 0; --i) {
            if ((num1 >> i & 1) == 1) {
                x |= 1 << i;
                --cnt;
            }
        }
        for (int i = 0; cnt > 0; ++i) {
            if ((num1 >> i & 1) == 0) {
                x |= 1 << i;
                --cnt;
            }
        }
        return x;
    }
}

  • class Solution {
public:
    int minimizeXor(int num1, int num2) {
        int cnt = __builtin_popcount(num2);
        int x = 0;
        for (int i = 30; ~i && cnt; --i) {
            if (num1 >> i & 1) {
                x |= 1 << i;
                --cnt;
            }
        }
        for (int i = 0; cnt; ++i) {
            if (num1 >> i & 1 ^ 1) {
                x |= 1 << i;
                --cnt;
            }
        }
        return x;
    }
};

  • class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:
        cnt = num2.bit_count()
        x = 0
        for i in range(30, -1, -1):
            if num1 >> i & 1 and cnt:
                x |= 1 << i
                cnt -= 1
        for i in range(30):
            if num1 >> i & 1 ^ 1 and cnt:
                x |= 1 << i
                cnt -= 1
        return x


  • func minimizeXor(num1 int, num2 int) int {
	cnt := bits.OnesCount(uint(num2))
	x := 0
	for i := 30; i >= 0 && cnt > 0; i-- {
		if num1>>i&1 == 1 {
			x |= 1 << i
			cnt--
		}
	}
	for i := 0; cnt > 0; i++ {
		if num1>>i&1 == 0 {
			x |= 1 << i
			cnt--
		}
	}
	return x
}

  • function minimizeXor(num1: number, num2: number): number {
    let cnt = 0;
    while (num2) {
        num2 &= num2 - 1;
        ++cnt;
    }
    let x = 0;
    for (let i = 30; i >= 0 && cnt > 0; --i) {
        if ((num1 >> i) & 1) {
            x |= 1 << i;
            --cnt;
        }
    }
    for (let i = 0; cnt > 0; ++i) {
        if (!((num1 >> i) & 1)) {
            x |= 1 << i;
            --cnt;
        }
    }
    return x;
}

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