3125. Maximum Number That Makes Result of Bitwise AND Zero 🔒

Problem Description

Given an integer n, your task is to return the maximum integer x such that x <= n, and the bitwise AND of all the numbers within the range [x, n] results in 0.

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Intuition

To solve this problem, we need to identify the highest bit that is set to 1 in the binary representation of n. The goal is to find the maximum integer x such that:

x is less than or equal to n.
The bitwise AND operation for the range [x, n] results in 0.

For the range [x, n] to AND to 0, x must be such that when it is incremented by 1, the result combined using AND with x is 0. This scenario occurs when x is made up of all bits set to 1 up to the position just below the highest bit set in n. Essentially, if the binary length of n is k bits, x is formed by setting all k-1 bits to 1 (i.e., x = 2^{k-1} - 1). This results in x being the largest value less than n that satisfies the bitwise AND condition.

Learn more about Greedy and Sorting patterns.

Solution Approach

The solution leverages bit manipulation to find the desired value of x. Here’s a step-by-step breakdown of the approach:

Identify the Highest Bit: First, determine the position of the highest set bit in the binary representation of n. This can be effectively achieved using the bit_length() method on the integer n, which returns the number of bits required to represent n in binary.
Calculate x: Using the position identified, compute x as 2^{(n.bit_length() - 1)} - 1. This formula works because 2^{k} - 1 gives a binary number with all bits set to 1 up to the k-1 position. Here, k is the number of bits in the binary representation of n.
Return the Result: This computed value of x ensures that it is the maximum number less than or equal to n where the bitwise AND with numbers in the range [x, n] equals 0.

The implementation of this logic in code is concise:

class Solution:
    def maxNumber(self, n: int) -> int:
        return (1 << (n.bit_length() - 1)) - 1

In this implementation, 1 << (n.bit_length() - 1) effectively computes 2^{(n.bit_length() - 1)}, shifting the 1 left to the desired bit position. Subtracting 1 then results in a number with all bits set below this position, giving the required x.

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Example Walkthrough

Let's walk through an example to better understand the solution approach. Suppose we have n = 10. Our goal is to find the maximum integer x such that x <= 10 and the bitwise AND of all numbers in the range [x, 10] is 0.

Identify the Binary Representation: First, let's determine the binary representation of 10, which is 1010.
Find the Highest Set Bit: The highest set bit in 10 is in the 2(^3) position (counting from zero). The binary length of 10 is, therefore, 4.
Calculate x: We use the bit length to compute x as (2^{(4 - 1)} - 1 = 2^3 - 1 = 8 - 1 = 7). In binary, x equals 0111.
Verification: Now, check if this value of x satisfies the condition. Consider the range [7, 10]:
- 7 in binary: 0111
- 8 in binary: 1000
- 9 in binary: 1001
- 10 in binary: 1010
The bitwise AND of all these numbers is 0000, which is indeed 0. Therefore, x = 7 is the maximum integer meeting our conditions.

Thus, applying the bit manipulation technique effectively gives us the correct result: x = 7.

Solution Implementation

1class Solution:
2    def maxNumber(self, n: int) -> int:
3        # Calculate the bit length of the number n
4        bit_length = n.bit_length()
5
6        # Calculate the maximum number with all 1s in binary representation
7        # This is equivalent to (2^(bit_length-1)) - 1
8        max_number_with_ones = (1 << (bit_length - 1)) - 1
9
10        # Return the computed maximum number
11        return max_number_with_ones
12

1class Solution {
2    public long maxNumber(long n) {
3        // Find the position of the highest set bit in 'n'
4        int highestSetBitPosition = 63 - Long.numberOfLeadingZeros(n);
5
6        // Calculate the maximum number with all bits set below the highest set bit position.
7        // This is achieved by left-shifting 1 to create a number with all bits set '0', then subtracting 1.
8        return (1L << highestSetBitPosition) - 1;
9    }
10}
11

1#include <iostream>
2#include <climits>
3
4class Solution {
5public:
6    // Method to find the maximum number with the same number of bits set as the given number
7    long long maxNumber(long long n) {
8        // Calculate the number of leading zero bits in the binary representation of n
9        int leadingZeroBits = __builtin_clzll(n);
10      
11        // Shift left by (63 - leadingZeroBits), which gives us the highest bit position of n
12        // Subtract 1 to fill all lower bits with 1
13        return (1LL << (63 - leadingZeroBits)) - 1;
14    }
15};
16

1// Function to find the maximum number with the same number of bits set as the given number
2function maxNumber(n: bigint): bigint {
3    // Calculate the number of leading zero bits in the binary representation of n
4    let leadingZeroBits: number = countLeadingZeros(n);
5  
6    // Shift left by (63 - leadingZeroBits), which gives us the highest bit position of n
7    // Subtract 1 to fill all lower bits with 1
8    return (1n << BigInt(63 - leadingZeroBits)) - 1n;
9}
10
11// Helper function to count leading zeros in a 64-bit integer
12function countLeadingZeros(n: bigint): number {
13    let count: number = 0;
14    const totalBits: number = 64;
15  
16    // Iterate over the bits from left to right
17    for (let i = totalBits - 1; i >= 0; i--) {
18        if ((n & (1n << BigInt(i))) === 0n) {
19            count++;
20        } else {
21            break;
22        }
23    }
24    return count;
25}
26

Time and Space Complexity

The time complexity of the code is O(log n) because the bit_length() function determines the number of bits required to represent the integer n in binary, which is proportional to the logarithm of n. Therefore, computing bit_length() takes O(log n) time.

The space complexity of the code is O(1) because it uses a constant amount of additional space, independent of the input size n.

Learn more about how to find time and space complexity quickly.

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