--- description: 'Author: @wingkwong, @vigneshshiv, @radojicic23 | https://leetcode.com/problems/number-of-1-bits/' --- # 0191 - Number of 1 Bits (Easy) ## Problem Link https://leetcode.com/problems/number-of-1-bits/ ## Problem Statement Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the [Hamming weight](http://en.wikipedia.org/wiki/Hamming_weight)). **Note:** * Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned. * In Java, the compiler represents the signed integers using [2's complement notation](https://en.wikipedia.org/wiki/Two's_complement). Therefore, in **Example 3**, the input represents the signed integer. `-3`. **Example 1:** ``` Input: n = 00000000000000000000000000001011 Output: 3 Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits. ``` **Example 2:** ``` Input: n = 00000000000000000000000010000000 Output: 1 Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit. ``` **Example 3:** ``` Input: n = 11111111111111111111111111111101 Output: 31 Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits. ``` **Constraints:** * The input must be a **binary string** of length `32`. **Follow up:** If this function is called many times, how would you optimize it? ## Approach 1: Built-in Function ```cpp class Solution { public: int hammingWeight(uint32_t n) { // or return bitset<32>(n).count(); return __builtin_popcount(n); } }; ``` ```py class Solution: def hammingWeight(self, n: int) -> int: return bin(n).count('1') ``` ```go func hammingWeight(num uint32) int { return bits.OnesCount32(num); } ``` ```rs impl Solution { pub fn hammingWeight (n: u32) -> i32 { n.count_ones() as i32 } } ``` ```java public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { return Integer.bitCount(n); } } ``` ```javascript /** * @param {number} n - a positive integer * @return {number} */ var hammingWeight = function(n) { return n.toString(2).split('1').length - 1 }; ``` ## Approach 2: Bit Manipulation We check each parity of teach bit. Increase $$ans$$ by 1 if the bit is set. ```cpp class Solution { public: int hammingWeight(uint32_t n) { int ans = 0; while (n) { if (n & 1) ans++; n >>= 1; // or n /= 2; } return ans; } }; ``` ```java public class Solution { // you need to treat n as an unsigned value // public int hammingWeight(int n) { int ones = 0; // n > 0, fails to return the correct the answer because of Integer MAX_VALUE. // Integer.MAX_VALUE + 1 is -2147483648, so it's not greater than 0, so n will not enter into loop while (n != 0) { ones += (n & 1); // Why can't we use n >>= 1? // Since n is 32 bit binary number, >> operator does shift by keeping signed bit position same as before // Take a look at SO reference - https://stackoverflow.com/questions/2811319/difference-between-and n >>>= 1; } return ones; } } ``` ```python class Solution: def hammingWeight(self, n: int) -> int: res = 0 while n: if (n & 1): res += 1 # same as --> n //= 2 n >>= 1 return res ``` ```javascript /** * @param {number} n - a positive integer * @return {number} */ var hammingWeight = function(n) { let res = 0; while (n != 0) { if (n & 1) res++; n /= 2 } return res; }; ``` ## Approach 3: n & (n - 1) We can optimise approach 2. Instead of checking all digits, we can use `n & (n - 1)` to remove the rightmost set bit. ``` n n n - 1 n & (n - 1) -- ---- ---- ------- 0 0000 0111 0000 1 0001 0000 0000 2 0010 0001 0000 3 0011 0010 0010 4 0100 0011 0000 5 0101 0100 0100 6 0110 0101 0100 7 0111 0110 0110 8 1000 0111 0000 9 1001 1000 1000 10 1010 1001 1000 11 1011 1010 1010 12 1100 1011 1000 13 1101 1100 1100 14 1110 1101 1100 15 1111 1110 1110 ``` ```cpp class Solution { public: int hammingWeight(uint32_t n) { int ans = 0; for (; n; n = n & (n - 1)) ans++; return ans; } }; ``` ```java public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int ones = 0; // Since n is 32 bit binary number, count 1's till that range for (int i = 0; i < 32; i++) { ones += (n & 1); n >>= 1; } return ones; } } ``` ```python class Solution: def hammingWeight(self, n: int) -> int: res = 0 while n: n = n & (n - 1) res += 1 return res ```