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Given a Large Binary Number as Input. Now, write a Kotlin Program to Convert Large Binary Number to decimal number.

Input: 110000000000000111111 Output: Equivalent decimal: 1572927

## 1. Program to Convert Binary to Decimal

###### Pseudo Algorithm

- Key Idea is to use string for input instead of int type, and Long for Output just to increase the range of output.
- Initialise a variable
**decimalVal**, that stores equivalent decimal value, with 0 - Extract each character from the string.
- While extracting, check if character is 1. If yes, multiply the extracted character with proper base (power of 2).
- For example, if binary number is 110, decimalVal = 1 * (2^2) + 1 * (2^1) = 6

Sourcecode –

fun main() { println("Enter n:") val binaryN = readLine() val countDigit = binaryN!!.length -1 var decimalVal: Long = 0 var base = 1 for(pos in countDigit downTo 0) { val curChar = binaryN[pos] if(curChar == '1') { decimalVal += base } base *= 2 if(!(curChar == '1' || curChar == '0')) { decimalVal = -1 println("$binaryN is not binary number.") break } } if(decimalVal != (-1).toLong()) { println("Equivalent decimal: $decimalVal") } }

When you run the program, output will be –

Enter n: 110000000000000000000000 Equivalent decimal: 12582912

###### Explanation:

- Run a loop from end to first character of the string.
- At each position, check if character at that position is 1. If yes, add base in
**decimalVal**. - Note that base value is calculated for each position. If you observe carefully, base is actually power of 2 for each position in the string.
- Finally, at postion = 0, we get final decimal value.
- Note that we also check, at every position, if character is 0 or 1. It is neither 0 nor 1, it means given string is not binary. So, print error message.

Thus, we went through Kotlin Program to Convert Large Binary Number to Decimal Number

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