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Formatted question description: https://leetcode.ca/all/2429.html

2429. Minimize XOR

  • Difficulty: Medium.
  • Related Topics: .
  • Similar Questions: Maximum XOR of Two Numbers in an Array, Maximum XOR With an Element From Array.

Problem

Given two positive integers num1 and num2, find the 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 <= 10^9

Solution (Java, C++, Python)

  • class Solution {
    public int minimizeXor(int num1, int num2) {
        int cnt = Integer.bitCount(num2);
        int ans = 0;
        for (int i = 30; i >= 0; --i) {
            if (((num1 >> i) & 1) == 1) {
                ans |= 1 << i;
                if (--cnt == 0) {
                    return ans;
                }
            }
        }
        for (int i = 0; i < 31; ++i) {
            if (((num1 >> i) & 1) == 0) {
                ans |= 1 << i;
                if (--cnt == 0) {
                    return ans;
                }
            }
        }
        return 0;
    }
}

  • class Solution {
public:
    int minimizeXor(int num1, int num2) {
        int cnt = __builtin_popcount(num2);
        int ans = 0;
        for (int i = 30; i >= 0; --i) {
            if ((num1 >> i) & 1) {
                ans |= 1 << i;
                if (--cnt == 0) {
                    return ans;
                }
            }
        }
        for (int i = 0; i < 31; ++i) {
            if (((num1 >> i) & 1) == 0) {
                ans |= 1 << i;
                if (--cnt == 0) {
                    return ans;
                }
            }
        }
        return 0;
    }
};

  • class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:
        cnt = num2.bit_count()
        ans = 0
        for i in range(30, -1, -1):
            if (num1 >> i) & 1:
                ans |= 1 << i
                cnt -= 1
                if cnt == 0:
                    return ans
        for i in range(31):
            if (num1 >> i) & 1 == 0:
                ans |= 1 << i
                cnt -= 1
                if cnt == 0:
                    return ans
        return 0


  • func minimizeXor(num1 int, num2 int) int {
	cnt := bits.OnesCount(uint(num2))
	ans := 0
	for i := 30; i >= 0; i-- {
		if num1>>i&1 == 1 {
			ans |= 1 << i
			cnt--
			if cnt == 0 {
				return ans
			}
		}
	}
	for i := 0; i < 31; i++ {
		if num1>>i&1 == 0 {
			ans |= 1 << i
			cnt--
			if cnt == 0 {
				return ans
			}
		}
	}
	return 0
}

  • function minimizeXor(num1: number, num2: number): number {
    let ans = num1;
    let target = num1
        .toString(2)
        .split('')
        .reduce((a, c) => a + Number(c), 0);
    let count = num2
        .toString(2)
        .split('')
        .reduce((a, c) => a + Number(c), 0);
    while (count > target) {
        ans |= ans + 1;
        count -= 1;
    }
    while (count < target) {
        ans &= ans - 1;
        count += 1;
    }
    return ans;
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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