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191. Number of 1 Bits

Description

Given a positive integer n, write a function that returns the number of set bits in its binary representation (also known as the Hamming weight).

 

Example 1:

Input: n = 11

Output: 3

Explanation:

The input binary string 1011 has a total of three set bits.

Example 2:

Input: n = 128

Output: 1

Explanation:

The input binary string 10000000 has a total of one set bit.

Example 3:

Input: n = 2147483645

Output: 30

Explanation:

The input binary string 1111111111111111111111111111101 has a total of thirty set bits.

 

Constraints:

  • 1 <= n <= 231 - 1

 

Follow up: If this function is called many times, how would you optimize it?

Solutions

Solution 1

Thinking

Count set bits. Checking every bit always looks at \(32\) positions. \(n\mathbin{\&}(n-1)\) clears the lowest \(1\), so the loop runs once per set bit.

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class Solution:
    def hammingWeight(self, n: int) -> int:
        ans = 0
        while n:
            n &= n - 1
            ans += 1
        return ans

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public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ans = 0;
        while (n != 0) {
            n &= n - 1;
            ++ans;
        }
        return ans;
    }
}

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class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        while (n) {
            n &= n - 1;
            ++ans;
        }
        return ans;
    }
};

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func hammingWeight(num uint32) int {
    ans := 0
    for num != 0 {
        num &= num - 1
        ans++
    }
    return ans
}

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function hammingWeight(n: number): number {
    let ans: number = 0;
    while (n !== 0) {
        ans++;
        n &= n - 1;
    }
    return ans;
}

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impl Solution {
    pub fn hammingWeight(n: u32) -> i32 {
        n.count_ones() as i32
    }
}

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/**
 * @param {number} n - a positive integer
 * @return {number}
 */
var hammingWeight = function (n) {
    let ans = 0;
    while (n) {
        n &= n - 1;
        ++ans;
    }
    return ans;
};

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int hammingWeight(uint32_t n) {
    int ans = 0;
    while (n) {
        n &= n - 1;
        ans++;
    }
    return ans;
}

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class Solution {
    fun hammingWeight(n: Int): Int {
        var count = 0
        var num = n
        while (num != 0) {
            num = num and (num - 1)
            count++
        }
        return count
    }
}

Solution 2

Thinking

Solution 1 uses \(n\mathbin{\&}(n-1)\). \(\textit{lowbit}=n\mathbin{\&}-n\) isolates the lowest \(1\); subtract and repeat. Same idea, the form used in Fenwick trees.

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class Solution:
    def hammingWeight(self, n: int) -> int:
        ans = 0
        while n:
            n -= n & -n
            ans += 1
        return ans

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public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ans = 0;
        while (n != 0) {
            n -= (n & -n);
            ++ans;
        }
        return ans;
    }
}

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class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        while (n) {
            n -= (n & -n);
            ++ans;
        }
        return ans;
    }
};

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func hammingWeight(num uint32) int {
    ans := 0
    for num != 0 {
        num -= (num & -num)
        ans++
    }
    return ans
}

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function hammingWeight(n: number): number {
    let count = 0;
    while (n) {
        n -= n & -n;
        count++;
    }
    return count;
}

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impl Solution {
    pub fn hammingWeight(mut n: u32) -> i32 {
        let mut res = 0;
        while n != 0 {
            n &= n - 1;
            res += 1;
        }
        res
    }
}

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/**
 * @param {number} n
 * @return {number}
 */
var hammingWeight = function (n) {
    let count = 0;
    while (n) {
        n -= n & -n;
        count++;
    }
    return count;
};

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class Solution {
    fun hammingWeight(n: Int): Int {
        var count = 0
        var num = n
        while (num != 0) {
            num -= num and (-num)
            count++
        }
        return count
    }
}

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