Approach: Common binary prefix

def rangeBitwiseAnd(a, b):
    shift = 0
    while a < b:
        a >>= 1
        b >>= 1
        shift += 1
    return a << shift

Complexity: O(log n) time, O(1) space.

Full Solution — all languages

class Solution:
    def rangeBitwiseAnd(self, a, b):
        shift = 0
        while a < b:
            a >>= 1
            b >>= 1
            shift += 1
        return a << shift

import java.util.*;
class Solution {
    public int rangeBitwiseAnd(int a, int b) {
        int shift = 0;
        while (a < b) {
            a >>= 1;
            b >>= 1;
            shift++;
        }
        return a << shift;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int rangeBitwiseAnd(int a, int b) {
        int shift = 0;
        while (a < b) {
            a >>= 1;
            b >>= 1;
            shift++;
        }
        return a << shift;
    }
};

/**
 * @param {number} a
 * @param {number} b
 * @return {number}
 */ var rangeBitwiseAnd = function(a, b) {
    let shift = 0;
    while (a < b) {
        a >>= 1;
        b >>= 1;
        shift++;
    }
    return a << shift;
};

Bitwise AND of Numbers Range

Example 1

Example 2

Example 3

Constraints