Number of 1 Bits (Hamming weight).

Problem. Return the number of 1 bits (the Hamming weight) of a 32-bit integer.

Naive — check every bit, O(32)

int hammingWeight(int n) {
    int count = 0;
    for (int i = 0; i < 32; i++)
        count += (n >>> i) & 1;     // >>> so the sign bit doesn't smear ones
    return count;
}

Use >>> (unsigned shift): with >>, a negative n sign-extends and the loop would over-count. Always 32 iterations regardless of input.

Optimal — Brian Kernighan's, O(set bits)

n & (n - 1) clears the lowest set bit, so the loop runs once per set bit — cheaper for sparse numbers:

int hammingWeight(int n) {
    int count = 0;
    while (n != 0) {
        n &= (n - 1);   // remove the lowest set bit
        count++;
    }
    return count;
}

This works correctly for negative inputs without >>> because we never shift — we only mask. n != 0 terminates after exactly popcount(n) iterations.

Production answer

int hammingWeight(int n) {
    return Integer.bitCount(n);   // typically one CPU POPCNT instruction
}

In real code this is the right answer; in an interview, mention it but be ready to show Kernighan's loop, since they're testing whether you know the trick.