Given an fraction decimal number n and integer k, convert decimal number n into equivalent binary number up-to k precision after decimal point.
Input: n = 2.47, k = 5 Output: 10.01111 Input: n = 6.986 k = 8 Output: 110.11111100
A) Convert the integral part of decimal to binary equivalent
- Divide the decimal number by 2 and store remainders in array.
- Divide the quotient by 2.
- Repeat step 2 until we get the quotient equal to zero.
- Equivalent binary number would be reverse of all remainders of step 1.
B) Convert the fractional part of decimal to binary equivalent
- Multiply the fractional decimal number by 2.
- Integral part of resultant decimal number will be first digit of fraction binary number.
- Repeat step 1 using only fractional part of decimal number and then step 2.
C) Combine both integral and fractional part of binary number.
Let's take an example for n = 4.47 k = 3 Step 1: Conversion of 4 to binary 1. 4/2 : Remainder = 0 : Quotient = 2 2. 2/2 : Remainder = 0 : Quotient = 1 3. 1/2 : Remainder = 1 : Quotient = 0 So equivalent binary of integral part of decimal is 100. Step 2: Conversion of .47 to binary 1. 0.47 * 2 = 0.94, Integral part: 0 2. 0.94 * 2 = 1.88, Integral part: 1 3. 0.88 * 2 = 1.76, Integral part: 1 So equivalent bianry of fractional part of decimal is .011 Step 3: Combined the result of step 1 and 2. Final answer can be written as: 100 + .011 = 100.011
C++ Program to demonstrate above steps:
Output: 100.011 110.11111
Time complexity: O(len(n))
Auxiliary space: O(len(n))
where len() is the total digits contain in number n.
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