Quotient and remainder dividing by 2^k (a power of 2)
You are given an positive integer n as dividend and another number m (form of 2^k), you have to find quotient and remainder without performing actual division.
Input : n = 43, m = 8 Output : Quotient = 5, Remainder = 3 Input : n = 58, m = 16 Output : Quotient = 3, Remainder = 10
In this we are using bitwise representation of a number for understanding the role of division of any number by divisor of form 2^k. All numbers which are power of two includes only 1 set bits in their representation and we will use this property.
For finding remainder we will take logical AND of the dividend (n) and divisor minus 1 (m-1), this will give only the set bits of dividend right to the set bit of divisor which is our actual remainder in that case.
Further, the left part of the dividend (from the position of set bit in divisor) would be considered for quotient. So, from dividend (n) removing all bits right from the position of set bit of divisor will result into quotient, and right shifting the dividend log2(m) times will do this job for finding the quotient.
- Remainder = n & (m-1)
- Quotient = (n >> log2(m) )
Note : Log2(m) will give the number of bits present in the binary representation of m.
Remainder = 3 Quotient = 5
Time Complexity: O(1)
Auxiliary Space: O(1)
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