Fraction to Recurring Decimal
Problem Statement :
Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. If multiple answers are possible, return any of them. It is guaranteed that the length of the answer string is less than 104 for all the given inputs. Example 1: Input: numerator = 1, denominator = 2 Output: "0.5" Example 2: Input: numerator = 2, denominator = 1 Output: "2" Example 3: Input: numerator = 4, denominator = 333 Output: "0.(012)" Constraints: -231 <= numerator, denominator <= 231 - 1 denominator != 0
Solution :
Solution in C :
long abs1(long a){
return a < 0 ? -a : a;
}
long repIndex(long* arr, long len, long num){
for(int i = 0; i < len; i++){
if(arr[i] == num)
return i;
}
return -1;
}
int length(int num){
int len = num == 0 ? 1 : 0;
while(num != 0){
num /= 10;
len ++;
}
return len;
}
char * fractionToDecimal(int numerator, int denominator){ //Assume: 1 6 Result: 0.1(6)
long num = numerator, den = denominator; //cast to long
long integer = num / den; //0
int sign = 0;
int iLen = length(integer); //1
// Set positive for all
if((long)num * den < 0){
sign = 1;
integer = -integer;
num = abs1(num);
den = abs1(den);
}
//=========================================================================================
bool repeat = 0;
int fLen = 0, index = -1; //fLen: Fraction length, index: repeat at index (0.16666 -> index = 1)
if(num % den){ //mod is not 0
fLen ++; //if mod is not 0, it has dot.
int maxUnrepeated = length(den) + 1 /*2*/, realUnrepeated = 0;
long *mods = malloc(maxUnrepeated * sizeof(long)); //Store unrepeated mods * 10. In the case of 1/6, it would be 1 * 10 and 4 * 10.
long n = num % den * 10; //0.16666 -> n = 10
do{
if(realUnrepeated < maxUnrepeated){
mods[realUnrepeated] = n;
realUnrepeated ++;
}
fLen ++;
n = n % den * 10;
index = repIndex(mods, realUnrepeated, n); //0.16666 -> index = 1
if(index != -1){ //if index != -1, n is found in mods[]
repeat = 1;
fLen += 2; //()
break;
}
}while(n);
}
//=========================================================================================
int len = sign + iLen + fLen; //0.1(6) -> 0 + 1 + 5
char* r = malloc((len + 1) * sizeof(char)), *p = r;
//copy sign
if(sign) *p++ = '-';
//copy integer
for(int i = iLen - 1; i >= 0; i--){
p[i] = (integer % 10) + '0';
integer /= 10;
}
if(fLen){
p += iLen; //move the pointer to the fractional part
*p++ = '.';
}
if(repeat){
p[index] = '(';
r[len - 1] = ')';
fLen -= 3;
}
//copy fraction
long n = num, mod;
for(int i = 0; i < fLen; i++){
mod = n % den; //4
n = mod * 10;
if(p[i]=='(')p++;
p[i] = n / den + '0';
}
r[len] = '\0';
return r;
}
Solution in C++ :
class Solution {
public String fractionToDecimal(int numerator, int denominator) {
if(numerator == 0){
return "0";
}
StringBuilder sb = new StringBuilder("");
if(numerator<0 && denominator>0 || numerator>0 && denominator<0){
sb.append("-");
}
long divisor = Math.abs((long)numerator);
long dividend = Math.abs((long)denominator);
long remainder = divisor % dividend;
sb.append(divisor / dividend);
if(remainder == 0){
return sb.toString();
}
sb.append(".");
HashMap<Long, Integer> map = new HashMap<Long, Integer>();
while(remainder!=0){
if(map.containsKey(remainder)){
sb.insert(map.get(remainder), "(");
sb.append(")");
break;
}
map.put(remainder, sb.length());
remainder*= 10;
sb.append(remainder/dividend);
remainder%= dividend;
}
return sb.toString();
}
}
Solution in Java :
class Solution {
StringBuilder sb;
public String fractionToDecimal(int numerator, int denominator) {
if(numerator == 0){
return "0";
}
sb = new StringBuilder();
addSign(numerator,denominator);
divideNumbers(numerator,denominator);
return sb.toString();
}
private void addSign(int a , int b){
if((a > 0 && b > 0) || (a < 0 && b < 0)){
return;
}
sb.append("-");
}
private void divideNumbers(int a,int b){
long num = Math.abs((long)a);
long den = Math.abs((long)b);
sb.append(num/den);
num %= den;
if(num == 0){
return;
}
appendDecimalPart(num,den);
}
private void appendDecimalPart(long num, long den){
HashMap<Long,Integer> map = new HashMap<>();
sb.append(".");
while(num !=0){
num *= 10;
sb.append(num/den);
num %= den;
if(map.containsKey(num)){
sb.insert(map.get(num),"(");
sb.append(")");
return;
}
else{
map.put(num,sb.length());
}
}
}
}
Solution in Python :
class Solution:
def fractionToDecimal(self, numerator: int, denominator: int) -> str:
# Handle edge cases
if numerator == 0:
return "0"
if denominator == 0:
return ""
# Initialize result and check for negative sign
result = ""
if (numerator < 0) ^ (denominator < 0):
result += "-"
numerator, denominator = abs(numerator), abs(denominator)
# Integer part of the result
result += str(numerator // denominator)
# Check if there is a fractional part
if numerator % denominator == 0:
return result
result += "."
# Use a dictionary to store the position of each remainder
remainder_dict = {}
remainder = numerator % denominator
# Keep adding the remainder to the result until it repeats or the remainder becomes 0
while remainder != 0 and remainder not in remainder_dict:
remainder_dict[remainder] = len(result)
remainder *= 10
result += str(remainder // denominator)
remainder %= denominator
# Check if there is a repeating part
if remainder in remainder_dict:
result = result[:remainder_dict[remainder]] + "(" + result[remainder_dict[remainder]:] + ")"
return result
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