← Bitkernel Interview Insights
I stared at this longer than I'd like to admit.
First, recall that in two's complement, the most negative value has the sign bit (MSB) set to 1. To minimize the value, we want the remaining 1-bits to be in the highest possible positions (closest to the MSB) to maximize the negative contribution. Then, construct the 8-bit pattern and compute its decimal value.
Pro tip: Mention that the smallest value is not simply the minimum possible integer with three 1-bits; you must consider the two's complement weighting where the MSB has negative weight. Also, verify by checking that the bit pattern indeed represents a negative number and that no other arrangement yields a smaller value.
Recall that an 8-bit signed integer uses the most significant bit (bit 7) as the sign bit with weight -2^7, and the remaining bits have positive weights 2^6 down to 2^0.
The number must have exactly three 1-bits and five 0-bits. The sign bit must be 1 to make the number negative (since we want the smallest, most negative value).
To minimize the value, set the sign bit to 1 and place the other two 1-bits in the highest possible positions (bits 6 and 5) to maximize the positive contribution, which actually makes the number less negative? Wait: In two's complement, the value is -128 + sum of positive weights. To get the most negative, we want the sum of positive weights to be as small as possible, so we should place the remaining 1-bits in the lowest possible positions (bits 0 and 1).
Set bit 7 = 1, bit 1 = 1, bit 0 = 1, and all other bits 0. The binary pattern is 10000011. Compute its decimal value: -128 + 2 + 1 = -125.
Check that any other placement of the two remaining 1-bits would yield a larger (less negative) value. For example, placing them at bits 6 and 5 gives -128 + 64 + 32 = -32, which is greater than -125.
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