Skip to content

201. Bitwise AND of Numbers Range

Description

Given two integers left and right that represent the range [left, right], return the bitwise AND of all numbers in this range, inclusive.

 

Example 1:

Input: left = 5, right = 7
Output: 4

Example 2:

Input: left = 0, right = 0
Output: 0

Example 3:

Input: left = 1, right = 2147483647
Output: 0

 

Constraints:

  • 0 <= left <= right <= 231 - 1

Solutions

Solution 1

Thinking

AND-ing every integer in \([left,right]\) is impossible when the span approaches \(2^{31}\). The bitwise AND of a contiguous range is their common binary prefix: lower bits are cleared by some number in the range.

While \(left < right\), clear the lowest set bit of \(right\) (\(right \mathrel{\&}= right-1\)) until \(right \le left\). The remaining \(right\) is that common prefix.

1
2
3
4
5
class Solution:
    def rangeBitwiseAnd(self, left: int, right: int) -> int:
        while left < right:
            right &= right - 1
        return right

1
2
3
4
5
6
7
8
class Solution {
    public int rangeBitwiseAnd(int left, int right) {
        while (left < right) {
            right &= (right - 1);
        }
        return right;
    }
}

1
2
3
4
5
6
7
8
9
class Solution {
public:
    int rangeBitwiseAnd(int left, int right) {
        while (left < right) {
            right &= (right - 1);
        }
        return right;
    }
};

1
2
3
4
5
6
func rangeBitwiseAnd(left int, right int) int {
    for left < right {
        right &= (right - 1)
    }
    return right
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
/**
 * @param {number} left
 * @param {number} right
 * @return {number}
 */
var rangeBitwiseAnd = function (left, right) {
    while (left < right) {
        right &= right - 1;
    }
    return right;
};

1
2
3
4
5
6
7
8
public class Solution {
    public int RangeBitwiseAnd(int left, int right) {
        while (left < right) {
            right &= (right - 1);
        }
        return right;
    }
}

Comments