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 \(num2\).

Python3JavaC++GoTypeScript

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

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;
    }
};

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;
}

Solution 2

Python3JavaC++GoTypeScript

class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:
        cnt1 = num1.bit_count()
        cnt2 = num2.bit_count()
        while cnt1 > cnt2:
            num1 &= num1 - 1
            cnt1 -= 1
        while cnt1 < cnt2:
            num1 |= num1 + 1
            cnt1 += 1
        return num1

class Solution {
    public int minimizeXor(int num1, int num2) {
        int cnt1 = Integer.bitCount(num1);
        int cnt2 = Integer.bitCount(num2);
        for (; cnt1 > cnt2; --cnt1) {
            num1 &= (num1 - 1);
        }
        for (; cnt1 < cnt2; ++cnt1) {
            num1 |= (num1 + 1);
        }
        return num1;
    }
}

class Solution {
public:
    int minimizeXor(int num1, int num2) {
        int cnt1 = __builtin_popcount(num1);
        int cnt2 = __builtin_popcount(num2);
        for (; cnt1 > cnt2; --cnt1) {
            num1 &= (num1 - 1);
        }
        for (; cnt1 < cnt2; ++cnt1) {
            num1 |= (num1 + 1);
        }
        return num1;
    }
};

func minimizeXor(num1 int, num2 int) int {
    cnt1 := bits.OnesCount(uint(num1))
    cnt2 := bits.OnesCount(uint(num2))
    for ; cnt1 > cnt2; cnt1-- {
        num1 &= (num1 - 1)
    }
    for ; cnt1 < cnt2; cnt1++ {
        num1 |= (num1 + 1)
    }
    return num1
}

function minimizeXor(num1: number, num2: number): number {
    let cnt1 = bitCount(num1);
    let cnt2 = bitCount(num2);
    for (; cnt1 > cnt2; --cnt1) {
        num1 &= num1 - 1;
    }
    for (; cnt1 < cnt2; ++cnt1) {
        num1 |= num1 + 1;
    }
    return num1;
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

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