LeetCode 1017 - Convert to Base -2

https://leetcode.com/problems/convert-to-base-2/

Given a number N, return a string consisting of "0"s and "1"s that represents its value in base -2 (negative two).

The returned string must have no leading zeroes, unless the string is "0".

Example 1:

Input: 2
Output: "110"
Explantion: (-2) ^ 2 + (-2) ^ 1 = 2

Example 2:

Input: 3
Output: "111"
Explantion: (-2) ^ 2 + (-2) ^ 1 + (-2) ^ 0 = 3

Example 3:

Input: 4
Output: "100"
Explantion: (-2) ^ 2 = 4

Note:

1. 0 <= N <= 10^9

https://leetcode.com/problems/convert-to-base-2/discuss/265507/JavaC%2B%2BPython-2-lines-Exactly-Same-as-Base-2

Intuition:

1. Maybe write a base2 function first?
2. How about add minus -?
3. Done.

Explanation:

base2 function is quite basis of basis.
check last digit, shift to right.
base-2 is totally no difference, just add a sign -.

Time Complexity:

O(logN) Time, O(logN) space.
Note that I didn't deal with string concatenation,
and just take this part as O(1).

Instead of create a new string each time,
we can improve this process using some operations join/reverse or data structure list/vector .
Like Java we may need StringBuilder,
but those are not what I want to discuss deeply here.

    public String base2(int N) {
        String res = "";
        while (N != 0) {
            res = Integer.toString(N & 1) + res;
            N = N >> 1;
        }
        return res == ""  ? "0" : res;
    }

    public String baseNeg2(int N) {
        String res = "";
        while (N != 0) {
            res = Integer.toString(N & 1) + res;
            N = -(N >> 1);
        }
        return res == ""  ? "0" : res;
    }

base2 won't work for negative number, instead of >> should be >>>

public String base2(int N) {
String res = "";
while (N != 0) {
res = Integer.toString(N & 1) + res;
N = N >> 1;
}
return res == "" ? "0" : res;
}

https://leetcode.com/problems/convert-to-base-2/discuss/266760/Java-simple-3-lines

    public String baseNeg2(int N) {
        StringBuilder ans = new StringBuilder( N==0 ? "0" : "" );
        for( ; N!=0; N=-(N>>1) ) ans.append( (N&1)==0 ? '0' : '1' );
        return ans.reverse().toString();
    }

https://leetcode.com/problems/convert-to-base-2/discuss/265688/4-line-python-clear-solution-with-explanation

for each number, we first consider its binary mode, then we check 10,1000,100000,... one by one
for example, 6 = '110', so for the second '1', in base -2, it makes the initial number decrease 4(+2 -> -2)
so we jusr make the initial number add 4, then 6 -> 10 = '1010', later we consider the first '1', it makes the initial number decrease 16(+8 -> -8), then we add 16, 10 -> 26 = '11010', now we get the answer.

class Solution:
    def baseNeg2(self, N: int) -> str:
        neg = [1 << i for i in range(1, 33, 2)]
        for mask in neg:
            if N & mask: N += mask*2
        return bin(N)[2:]

https://leetcode.com/problems/convert-to-base-2/discuss/265544/C%2B%2B-Geeks4Geeks

See this Geeks4Geeks article: Convert a number into negative base representation.

Below is the specialized solution for the -2 base.

string baseNeg2(int N, string res = "") {
  while (N != 0) {
    int rem = N % -2;
    N /= -2;
    if (rem < 0) rem += 2, N += 1;
    res = to_string(rem) + res;
  }
  return max(string("0"), res);
}

https://www.geeksforgeeks.org/convert-number-negative-base-representation/

A number n and a negative base negBase is given to us, we need to represent n in that negative base. Negative base works similar to positive base. For example in base 2 we multiply bits to 1, 2, 4, 8 and so on to get actual number in decimal. In case of base -2 we need to multiply bits with 1, -2, 4, -8 and so on to get number in decimal.

Examples:

Input  : n = 13, negBase = -2
Output : 11101
1*(16) + 1*(-8) + 1*(4) + 0*(-2) + 1*(1)  = 13

It is possible to represent a number into any negative base with same procedure (Refer Wiki for details). For simplicity (to get rid of A, B etc characters in output), we are allowing our base to be in between -2 and -10 only.

We can solve this problem similar to solving problem with positive bases but one important thing to remember is, remainder will always be positive whether we work with positive base or negative base but in most compilers, the result of dividing a negative number by a negative number is rounded towards 0, usually leaving a negative remainder.
So whenever we get a negative remainder, we can convert it to positive as below,

Let 
n = (?negBase) * quotient + remainder 
  = (?negBase) * quotient + negBase ? negBase + negBase 
  = (?negBase) * (quotient + 1) + (remainder + negBase). 

So if after doing "remainder = n % negBase" and 
"n = n/negBase", we get negative remainder, we do 
following.
remainder = remainder + (-negBase)
n = n + 1

Example : n = -4, negBase = -3
In C++, we get
    remainder = n % negBase = -4/-3 = -1
    n = n/negBase [Next step for base conversion]
      = -4/-3 
      = 1
To avoid negative remainder, we do,
    remainder = -1 + (-negBase) = -1 - (-3) = 2
    n = n + 1 = 1  + 1 = 2.

So when we will get negative remainder, we will make it positive by adding absolute value of base to it and adding 1 to our quotient.

string toNegativeBase(int n, int negBase)

{

    //  If n is zero then in any base it will be 0 only

    if (n == 0)

        return "0";

  

    string converted = "";

    while (n != 0)

    {

        // Get remainder by negative base, it can be

        // negative also

        int remainder = n % negBase;

        n /= negBase;

  

        // if remainder is negative, add abs(base) to

        // it and add 1 to n

        if (remainder < 0)

        {

            remainder += (-negBase);

            n += 1;

        }

  

        // convert remainder to string add into the result

        converted = toString(remainder) + converted;

    }

  

    return converted;

}

https://www.geeksforgeeks.org/convert-base-decimal-vice-versa/

Given a number and its base, convert it to decimal. The base of number can be anything such that all digits can be represented using 0 to 9 and A to Z. Value of A is 10, value of B is 11 and so on. Write a function to convert the number to decimal.

static int toDeci(String str, 

                  int base)

{

    int len = str.length();

    int power = 1; // Initialize 

                   // power of base

    int num = 0; // Initialize result

    int i;

  

    // Decimal equivalent is 

    // str[len-1]*1 + str[len-1] *

    // base + str[len-1]*(base^2) + ...

    for (i = len - 1; i >= 0; i--)

    {

        // A digit in input number 

        // must be less than 

        // number's base

        if (val(str.charAt(i)) >= base)

        {

        System.out.println("Invalid Number");

        return -1;

        }

  

        num += val(str.charAt(i)) * power;

        power = power * base;

    }

  

    return num;

}




char* fromDeci(char res[], int base, int inputNum)

{

    int index = 0;  // Initialize index of result

  

    // Convert input number is given base by repeatedly

    // dividing it by base and taking remainder

    while (inputNum > 0)

    {

        res[index++] = reVal(inputNum % base);

        inputNum /= base;

    }

    res[index] = '\0';

  

    // Reverse the result

    strev(res);

  

    return res;

}