Convert a number into 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.
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.
Above explained approach is implemented in below code,
Time Complexity: O(N)
Auxiliary Space: O(1)
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