I am trying to find an efficient way to get the majority of corresponding bit in a group of same sized integers. For example in the given integers:
12 => 1 1 0 0
9 => 1 0 0 1
9 => 1 0 0 1
3 => 0 0 1 1
the majority of bit (column wise) can be represented as:
1 0 0 1
If there are equal number of 1's and 0's (column wise), it should be 1. I would like to achieve this using integers without converting the integer to binary string if possible.
In C, this could be something like this:
unsigned arr[] = { 12, 9, 9, 3 }; // values to loop through
unsigned len = sizeof(arr) / sizeof(arr[0]); // # of values
unsigned result = 0; // collect result here
unsigned bits = sizeof(arr[0]) * 8; // number of bits to consider
unsigned threshold = len / 2 + ((len & 1) ? 1 : 0); // minimum # of 1 to get a 1
unsigned bit = 1 << (bits - 1); // bit mask; point to most significant bit position
for (unsigned pos = 0; pos < bits; pos++) {
unsigned ones = 0;
for (unsigned i = 0; i < len; i++) {
if ((bit & arr[i]) != 0) {
ones++;
}
}
result = (result << 1) + ((ones >= threshold) ? 1 : 0);
bit = bit >> 1;
}
printf("\n%d\n", result);
The outer loop shifts a bitmask from the most-significant to the least-significant end. The inner loop counts the bits set to 1. If the 1 bits exceed a threshold, a 1 is shifted into the result. Otherwise, a 0 is shifted in. At the end, the result holds the majority bits as requested.
This could be further optimized but was not to keep readability up.
Simplified for special case len==4
unsigned arr[] = { 12, 9, 9, 3 }; // values to loop through
unsigned len = sizeof(arr) / sizeof(arr[0]); // # of values
unsigned result = 0; // collect result here
// for len==4, majority means that 2 or more bits are 1
for (unsigned i = 0; i < len - 1; i++)
for (unsigned j = i + 1; j < len; j++) {
result = result | (arr[i] & arr[j]);
}
printf("\n%d\n", result);
