Solution(){

srand(time(0));

}

//random invoker

int singleNumber(int A[], int n) {

if (rand()%2){

return singleNumber_1(A, n);

}

return singleNumber_2(A, n);

}

/*

int singleNumber_1(int A[], int n) {

int count[32] = {0};

int result = 0;

for (int i = 0; i < 32; i++) {

for (int j = 0; j < n; j++) {

if ((A[j] >> i) & 1) {

count[i]++;

}

}

result |= ((count[i] % 3) << i);

}

return result;

}

/*

int singleNumber_2(int A[], int n) {

int ones = 0, twos = 0, threes = 0;

for (int i = 0; i < n; i++) {

// `ones & A[i]` the result is the bitmask which the bits appeared twice

twos |= ones & A[i];

// XOR means remove the bit which appeared twice int `ones`

ones ^= A[i];

// count the `three`

threes = ones & twos;

// clear the `ones` and `twos` if the i-th bit had appeared three times.

ones &= ~threes;

twos &= ~threes;

}

return ones;

}