I'm trying to write the Power method and now the question is how to write it in a better way. Here is my code, I've verified it works well for all the Power conditions, however I can't think on a better solution.
The real issue is how I express the power with a Negative exponent in O(log)n condition
public static double raiseToPowerIterative(double x, int n) {
double sol = 1;
if (n == 0) {
return 1;
}
if (n > 0) {
for (int i = 1; i <= n;i++) {
sol *= x;
}
return sol;
} else if (n < 0){
for (int i = -1; i >= n; i--) {
sol /= x;
}
}
return sol;
}
You need some additional math rules, you are using:
xn = x.x. ... .x (n times)
With O(n)
However
x2k = xk . xk
x2k+1 = x.x2k
Which can deliver O(²log n), as you are doing exponents n → n/2 → n/2/2 → ... → 1.
Where you only need to calculate xk once.
A recursive solution might be easiest; calling oneself for xk.
Your solution:
That does not work, I think. As probably past limit date of the work.
Wrong:
double power(double x, int n) {
if (n == 0) {
return 1;
}
boolean negative = ( n < 0 );
if (negative) {
n = -n;
}
double sol = 1;
while (n > 0) {
if (n%2 == 1) {
sol *= x;
}
x *= x;
n /= 2;
}
if (negative) {
return 1 / sol;
}
return sol;
}
Correct recursive:
double power(double x, int n) {
if (n == 0) {
return 1;
} else if (n < 0) {
return 1 / power(x, -n);
}
double sqrPower = power(x, n/2);
int p = sqrPower * sqrPower;
if (n % 1 == 1) {
p *= x;
}
return p;
}
That is, one has to do something with the result of the recursive call.
For an iterative loop one would need a stack to reconstruct the result.
Correct iterative:
double power(double x, int n) {
if (n == 0) {
return 1;
}
boolean invert = n < 0;
if (invert) {
n = -n;
}
Stack<Double> factors = new Stack<>();
while (n > 0) {
factors.push(n % 1 == 1 ? x : 1.0);
n /= 2;
}
double sol = 1;
while (!factors.empty()) {
double f = factors.pop();
sol = sol * sol * f;
}
return invert ? 1/sol : sol;
}
The problem is:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).
The solution from the website I search is:
public class Solution {
public static int reverse(int x) {
int ret = 0;
boolean zero = false;
while (!zero) {
ret = ret * 10 + (x % 10);
x /= 10;
if(x == 0){
zero = true;
}
}
return ret;
}
public static void main(String[] args) {
int s = 1000000003;
System.out.println(reverse(s));
}
}
However when s = 1000000003, the console prints -1294967295 instead of 3000000001. So this solution still does not solve the overflow problem if we cannot use exception. Any help here?(Although there is a hint: add an extra parameter, I still cannot figure out what parameter I should add)
There's no need for any data type other than int.
Just make sure when there's an operation that increases a number, reversing the operation should give you the previous number. Otherwise, there's overflow.
public int reverse(int x) {
int y = 0;
while(x != 0) {
int yy = y*10 + x%10;
if ((yy - x%10)/10 != y) return 0;
else y = yy;
x = x/10;
}
return y;
}
Above most of the answers having a trivial problem is that the int variable possibly might overflow. You can try this : x = -2147483648 as parameter.
There has an easy way to solve the problem. Convert x to long, and check if the result >= Integer.MAX_VALUE, otherwise return 0.
The solution passed all test cases on https://leetcode.com/problems/reverse-integer/
This is a java version.
public int reverse(int x) {
long k = x;
boolean isNegtive = false;
if(k < 0){
k = 0 - k;
isNegtive = true;
}
long result = 0;
while(k != 0){
result *= 10;
result += k % 10;
k /= 10;
}
if(result > Integer.MAX_VALUE) return 0;
return isNegtive ? 0 - ((int)result) : (int)result;
}
C# version
public int Reverse(int x)
{
long value = 0;
bool negative = x < 0;
long y = x;
y = Math.Abs(y);
while (y > 0)
{
value *= 10;
value += y % 10;
y /= 10;
}
if(value > int.MaxValue)
{
return int.MaxValue;
}
int ret = (int)value;
if (negative)
{
return 0 - ret;
}
else
{
return ret;
}
}
Python version
def reverse(self, x):
isNegative = x < 0
ret = 0
x = abs(x)
while x > 0:
ret *= 10
ret += x % 10
x /= 10
if ret > 1<<31:
return 0
if isNegative:
return 0 - ret
else:
return ret
This java code handles the overflow condition:
public int reverse(int x) {
long reverse = 0;
while( x != 0 ) {
reverse = reverse * 10 + x % 10;
x = x/10;
}
if(reverse > Integer.MAX_VALUE || reverse < Integer.MIN_VALUE) {
return 0;
} else {
return (int) reverse;
}
}
This is an old question, but anyway let me have a go at it too! I just solved it on leetcode. With this check, you never hit the overflow/ underflow in either direction, and I think the code is more concise than all the listed codes. It passes all test cases.
public int reverse(int x) {
int y = 0;
while(x != 0) {
if(y > Integer.MAX_VALUE/10 || y < Integer.MIN_VALUE/10) return 0;
y *= 10;
y += x % 10;
x /= 10;
}
return y;
}
you can try this code using strings in java
class Solution {
public int reverse(int x) {
int n = Math.abs(x);
String num = Integer.toString(n);
StringBuilder sb = new StringBuilder(num);
sb.reverse();
String sb1;
sb1 = sb.toString();
int foo;
try {
foo = Integer.parseInt(sb1);
}
catch (NumberFormatException e){
foo = 0;
}
if(x < 0){
foo *= -1;
}
return foo;
}
}
My soluton for this problem is to convert integer inputed to c-string, then everthing will be easy.
class Solution {
public:
int reverse(int x) {
char str[11];
bool isNegative = false;
int i;
int ret = 0;
if ( x < 0 ) {
isNegative = true;
x = -x;
}
i = 0;
while ( x != 0 ) {
str[i++] = x % 10 + '0';
x = x / 10;
}
str[i] = '\0';
if ( (isNegative && strlen(str) == 10 && strcmp(str, "2147483648") > 0) || (!isNegative && strlen(str) == 10 && strcmp(str, "2147483647") > 0) ) {
cout << "Out of range!" << endl;
throw new exception();
}
i = 0;
int strLen = (int)strlen(str);
while ( str[i] != '\0' ) {
ret += ((str[i] - '0') * pow(10.0, strLen - 1 - i));
i++;
}
return (isNegative ? -ret : ret);
}
};
This works:
public class Solution {
public int reverse(int x) {
long tmp = Math.abs((long)x);
long res = 0;
while(tmp >= 10){
res += tmp%10;
res*=10;
tmp=tmp/10;
}
res+=tmp;
if(x<0){
res = -res;
}
return (res>Integer.MAX_VALUE||res<Integer.MIN_VALUE)? 0: (int)res;
}
}
I tried to improve the performance a bit but all I could come up with was this:
public class Solution {
public int reverse(int x) {
long tmp = x;
long res = 0;
if(x>0){
while(tmp >= 10){
res += tmp%10;
res*=10;
tmp=tmp/10;
}
}
else{
while(tmp <= -10){
res += tmp%10;
res*=10;
tmp=tmp/10;
}
}
res+=tmp;
return (res>Integer.MAX_VALUE||res<Integer.MIN_VALUE)? 0: (int)res;
}
}
Its C# equivalent runs 5% faster than the 1st version on my machine, but their server says it is slower, which can't be - I got rid of extra function call here, otherwise it is essentially the same. It places me between 60-30% depending on the language (C# or Java). Maybe their benchmarking code is not very good - if you submit several times - resulting times vary a lot.
Solution In Swift 4.0 (in reference to problem from https://leetcode.com/problems/reverse-integer/description/)
func reverse(_ x : Int) -> Int {
var stringConversion = String(x)
var negativeCharacter = false
var finalreversedString = String()
let signedInt = 2147483647 //Max for Int 32
let unSignedInt = -2147483647 // Min for Int 32
if stringConversion.contains("-"){
stringConversion.removeFirst()
negativeCharacter = true
}
var reversedString = String(stringConversion.reversed())
if reversedString.first == "0" {
reversedString.removeFirst()
}
if negativeCharacter {
finalreversedString = "-\(reversedString)"
} else {
finalreversedString = reversedString
}
return (x == 0 || Int(finalreversedString)! > signedInt || Int(finalreversedString)! < unSignedInt) ? 0 : Int(finalreversedString)!
}
Last night, i have tried this same problem and i have found a simple solution in python, which is given below, here after checking the number type positive or negative, though i have tried in different section for both of them, i have convert the negative number into positive and before returning the reverse number, i had converted the number into negative.
For handling overflow, i have just simply checked with the upper limit of our 32-bit signed number and lower limit of the number, and it accepted my answer, thank you.
class Solution:
def reverse(self, x: int):
reverse = 0
if x > 0:
while x != 0:
remainder = x % 10
if reverse > (2147483647/10):
return 0
reverse = reverse * 10 + remainder
x = int(x / 10)
return reverse
elif x < 0:
x = x * (-1)
while x != 0:
remainder = x % 10
if reverse > ((2147483648)/10):
return 0
reverse = reverse * 10 + remainder
x = int(x / 10)
reverse = reverse * (-1)
return reverse
else:
return 0
public static int reverse(int x) {
boolean pos = x >= +0;
int y = (pos) ? x : -x;
StringBuilder sb = new StringBuilder(
String.valueOf(y));
sb.reverse();
int z = Integer.parseInt(sb.toString());
return pos ? z : -z;
}
public static void main(String[] args) {
for (int i = -10; i < 11; i++) {
System.out.printf("%d r= '%d'\n", i, reverse(i));
}
}
Outputs
-10 r= '-1'
-9 r= '-9'
-8 r= '-8'
-7 r= '-7'
-6 r= '-6'
-5 r= '-5'
-4 r= '-4'
-3 r= '-3'
-2 r= '-2'
-1 r= '-1'
0 r= '0'
1 r= '1'
2 r= '2'
3 r= '3'
4 r= '4'
5 r= '5'
6 r= '6'
7 r= '7'
8 r= '8'
9 r= '9'
10 r= '1'
Did you notice the reverse of 10 and -10? Or 20? You could just return a String, for example
public static String reverse(int x) {
boolean pos = x >= +0;
int y = (pos) ? x : -x;
StringBuilder sb = new StringBuilder(
String.valueOf(y));
sb.reverse();
if (!pos) {
sb.insert(0, '-');
}
return sb.toString();
}
public static void main(String[] args) {
for (int i = -10; i < 11; i++) {
System.out.printf("%d r= '%s'\n", i, reverse(i));
}
}
Works as I would expect.
If you are required to return a 32 bit int, and still need to know if there was an overflow perhaps you could use a flag as an extra parameter. If you were using c or c++ you could use pointers to set the flag, or in Java you can use an array (since Java objects pass by value).
Java example:
public class Solution {
public static int reverse(int x, Boolean[] overflowed) {
int ret = 0;
boolean zero = false;
boolean inputIsNegative = x < 0;
while (!zero) {
ret = ret * 10 + (x % 10);
x /= 10;
if(x == 0){
zero = true;
}
}
//Set the flag
if ( (inputIsNegative && (ret > 0)) || ((!inputIsNegative) && (ret < 0)))
overflowed[0] = new Boolean(true);
else
overflowed[0] = new Boolean(false);
return ret;
}
public static void main(String[] args) {
int s = 1000000004;
Boolean[] flag = {null};
System.out.println(s);
int n = reverse(s,flag); //reverse() will set the flag.
System.out.println(flag[0].booleanValue() ? "Error: Overflow": n );
}
}
Notice if the reversed number is too large for a 32 bit integer the flag will be set.
Hope this helps.
Use string to store the reverse and then print or use long or BigInt
public class Solution {
/**
* OVERFLOW
* #param x
* #return
*/
public int reverse(int x) {
int sign = x>0? 1: -1;
x *= sign;
int ret = 0;
while(x>0) {
ret *= 10;
if(ret<0 || x>10&&ret*10/10!=ret) // overflow
return 0;
ret += x%10;
x /= 10;
}
return ret*sign;
}
public static void main(String[] args) {
assert new Solution().reverse(-2147483412)==-2147483412;
}
}
public class Solution {
public int Reverse(int x) {
var sign = x < 0 ? -1 : 1;
var reverse = 0;
if (x == int.MinValue)
{
return 0;
}
x = Math.Abs(x);
while(x > 0)
{
var remainder = x % 10;
if (reverse > ((int.MaxValue - remainder)/10))
{
return 0;
}
reverse = (reverse*10) + remainder;
x = x/10;
}
return sign * Convert.ToInt32(reverse);
}
}
Here we will use long to handle the the over flow:
public class Solution {
public int reverse(int A) {
// use long to monitor Overflow
long result = 0;
while (A != 0) {
result = result * 10 + (A % 10);
A = A / 10;
}
if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) {
return 0;
} else {
return (int) result;
}
}
}
Well This Suitable Code in Java Can be:-
public class Solution {
public int reverse(int x) {
int r;
long s = 0;
while(x != 0)
{
r = x % 10;
s = (s * 10) + r;
x = x/10;
}
if(s >= Integer.MAX_VALUE || s <= Integer.MIN_VALUE) return 0;
else
return (int)s;
}
}
My solution without using long:
public class ReverseInteger {
public static void main(String[] args) {
int input = Integer.MAX_VALUE;
int output = reverse(input);
System.out.println(output);
}
public static int reverse(int x) {
int remainder = 0;
int result = 0;
if (x < 10 && x > -10) {
return x;
}
while (x != 0) {
remainder = x % 10;
int absResult = Math.abs(result);
int maxResultMultipliedBy10 = Integer.MAX_VALUE / 10;
if (absResult > maxResultMultipliedBy10) {
return 0;
}
int resultMultipliedBy10 = absResult * 10;
int maxRemainder = Integer.MAX_VALUE - resultMultipliedBy10;
if (remainder > maxRemainder) {
return 0;
}
result = result * 10 + remainder;
x = x / 10;
}
return result;
}
}
here is the JavaScript solution.
/**
* #param {number} x
* #return {number}
*/
var reverse = function(x) {
var stop = false;
var res = 0;
while(!stop){
res = res *10 + (x % 10);
x = parseInt(x/10);
if(x==0){
stop = true;
}
}
return (res <= 0x7fffffff && res >= -0x80000000) ? res : 0
};
Taking care if the input is negative
public int reverse(int x)
{
long result = 0;
int res;
int num = Math.abs(x);
while(num!=0)
{
int rem = num%10;
result = result *10 + rem;
num = num / 10;
}
if(result > Integer.MAX_VALUE || result < Integer.MIN_VALUE)
{
return 0;
}
else
{
res = (int)result;
return x < 0 ? -res : res;
}
}
This solution in Java will work:
class Solution {
public int reverse(int x) {
long rev = 0, remainder = 0;
long number = x;
while (number != 0) {
remainder = number % 10;
rev = rev * 10 + remainder;
number = number / 10;
}
if (rev >= Integer.MAX_VALUE || rev <= Integer.MIN_VALUE || x >= Integer.MAX_VALUE || x <= Integer.MIN_VALUE)
return 0;
else
return (int) rev;
}
}
Much simpler solution. Ensure that intermittent result does not exceed INT_MAX or get below INT_MIN
int reverse(int x) {
int y = 0;
while(x != 0) {
if ( (long)y*10 + x%10 > INT_MAX || (long)y*10 + x%10 < INT_MIN) {
std::cout << "overflow occurred" << '\n'
return 0;
}
y = y*10 + x%10;
x = x/10;
}
return y;
}
Here is the solution coded in JS(Javascript, it has passed all the 1032 test cases successfully in Leetcode for the problem (https://leetcode.com/problems/reverse-integer), also as asked in the question about the same.
/**
* #param {number} x
* #return {number}
*/
var reverse = function(x) {
let oldNum = x, newNum = 0, digits = 0, negativeNum = false;
if(oldNum < 0){
negativeNum = true;
}
let absVal = Math.abs(x);
while(absVal != 0){
let r = Math.trunc(absVal % 10);
newNum = (newNum*10) + r; digits++;
absVal = Math.floor(absVal/10);
}
if( !(newNum < Number.MAX_VALUE && newNum >= -2147483648 && newNum <= 2147483647)){
return 0;
}
return negativeNum ? -newNum :newNum;
};
Here is the solution coded in JS(Javascript, it has passed all the 1032 test cases successfully in Leetcode for the problem (https://leetcode.com/problems/reverse-integer), also as asked in the question about the same.
/**
* #param {number} x
* #return {number}
*/
var reverse = function(x) {
let oldNum = x, newNum = 0, digits = 0, negativeNum = false;
if(oldNum < 0){
negativeNum = true;
}
let absVal = Math.abs(x);
while(absVal != 0){
let r = Math.trunc(absVal % 10);
newNum = (newNum*10) + r; digits++;
absVal = Math.floor(absVal/10);
}
if( !(newNum < Number.MAX_VALUE && newNum >= -2147483648 && newNum <= 2147483647)){
return 0;
}
return negativeNum ? -newNum :newNum;
};
The earlier answer was posted by the same user (unregistered). Consider this one.
There are several good solutions posted. Here is my JS solution:
const reverse = function (x) {
const strReversed = x.toString().split("").reverse().join("");
rv =
parseInt(strReversed) > Math.pow(2, 31)
? 0
: Math.sign(x) * parseInt(strReversed);
return rv;
};
I got all 1032 cases to work in python, I don't know how to remove multiple 0's such as 100, 1000, 10000 etc thus I used my if statement multiple times lol.
class Solution:
def reverse(self, x: int) -> int:
string = ""
y = str(x)
ab = list(reversed(y))
if len(ab) > 1 and ab[0] == "0":
ab.remove("0")
if len(ab) > 1 and ab[0] == "0":
ab.remove("0")
if len(ab) > 1 and ab[0] == "0":
ab.remove("0")
if len(ab) > 1 and ab[0] == "0":
ab.remove("0")
if len(ab) > 1 and ab[0] == "0":
ab.remove("0")
if ab[-1] == "-":
ab.remove("-")
ab.insert(0, "-")
for i in ab:
string += i
if int(string) > 2**31 - 1 or int(string) < -2**31:
return 0
return string
public static int reverse(int x) {
if (x == 0) return 0;
int sum = 0;
int y = 0;
while (x != 0) {
int value = (x % 10);
x = x - value;
y = sum;
sum = (sum * 10) + value;
if(sum / 10 != y) return 0;
x = x / 10;
}
return sum;
}
Extracting the first digit and dividing x to ten until x will be equal to 0. Therefore integer will be tokenized its digits.
Every extracted value will be adding the sum value after multiplying the sum by 10. Because adding a new digit means that adding a new 10th to the sum value. Also added if block to check any corruption of data because after 9th digit data will be corrupted.
1032 / 1032 test cases passed.
Status: Accepted
Runtime: 3 ms
Memory Usage: 38 MB
Public int reverse(int A) {
int N, sum = 0;
int rem = 0;
boolean flag = false;
int max = Integer.MAX_VALUE;
int min = Integer.MIN_VALUE;
if (A < 0) {
flag = true;
A = A * -1;} // 123 // 10 1
while (A > 0) {
rem = A % 10;
if (flag == true) {
if ((min + rem) / 10 > -sum) {
return 0;}}else{
if ((max - rem) / 10 < sum) {
return 0;}}
sum = (sum * 10) + rem;
A = A / 10;}
return (flag == true) ? —sum : sum;}}
#java #Algo
def reverse(self, x: int) -> int:
if x<=-2**31 or x>=2**31-1:
return 0
else:
result = 0
number = x
number = abs(number)
while (number) > 0:
newNumber = number % 10
result = result * 10 + newNumber
number = (number // 10)
if x<0:
result = "-"+str(result)
if int(result)<=-2**31:
return 0
return result
else:
if result>=2**31-1:
return 0
return result
if __name__ == '__main__':
obj = Solution()
print(obj.reverse(1534236469))
Note that there are previous solutions that do not work for input: 1000000045
try this:
public int reverse(int A) {
int reverse=0;
int num=A;
boolean flag=false;
if(A<0)
{
num=(-1)*A;
flag=true;
}
int prevnum=0;
while(num>0)
{
int currDigit=num%10;
reverse=reverse*10+currDigit;
if((reverse-currDigit)/10!=prevnum)
return 0;
num=num/10;
prevnum=reverse;
}
if(flag==true)
reverse= reverse*-1;
return reverse;
}
I tried Binary multiplication technique on decimal numbers.
Algorithm:
To multiply two decimal numbers x and y, write them next to each
other, as in the example below. Then repeat the following: divide the first number by 2,
rounding down the result (that is, dropping the :5 if the number was odd), and double the
second number. Keep going till the first number gets down to 1. Then strike out all the rows
in which the first number is even, and add up whatever remains in the second column.
11 13
5 26
2 52
1 104
........
143 (answer)
Code:
class Multiply
{
static int temp;
static int sum;
public static void main(String[] args)
{
int x = Integer.parseInt(args[0]);
int y = Integer.parseInt(args[1]);
int ans = multiply(x , y);
System.out.println(ans);
}
public static int multiply(int x, int y)
{
if(x==1)
{
System.out.println(x+" : "+y);
return y;
}
temp = multiply(x/2, y*2);
if(x%2==0)
{
System.out.println(x+" : "+y);
return temp;
}
else
{
System.out.println(x+" : "+y);
sum = sum+temp;
return sum;
}
}
}
Something is wrong with the recursion i think but i couldn't find what it is!!
When having recursion do not use variables outside the recursive method. It is too confusing. I mean that the recursive method should be self-contained. Here is a working version of your program:
public class Main {
public static void main(String[] args) {
int x = 11;
int y = 13;
int ans = multiply(x, y);
System.out.println(ans);
}
public static int multiply(int x, int y) {
if (x == 1) {
return y;
}
int temp = multiply(x / 2, y * 2);
if (x % 2 != 0) {
temp += y;
}
return temp;
}
}
Your recursion should be like this -
public class Multiply {
static int temp = 0;
static int sum = 0;
public static void main(String[] args) {
int x = Integer.parseInt("11");
int y = Integer.parseInt("9");
int ans = multiply(x, y);
System.out.println(ans);
}
public static int multiply(int x, int y) {
if (x == 1) {
System.out.println(x + " : " + y);
return sum + y;
}
if (x % 2 == 0) {
System.out.println(x + " : " + y);
} else {
System.out.println(x + " : " + y);
sum = sum + y;
}
return multiply(x / 2, y * 2);
}
}
I couldn't resist to post it in a single line
public static int multiply(int x, int y) {
return ((x & 1) > 0 ? y : 0) + ((x & ~1) > 0 ? multiply(x >> 1, y << 1) : 0);
}
I couldn't resist to add an iterative solution: fast, simple and valid
also for negative arguments:
int product(int x, int y) {
boolean positive = x >= 0;
int p = 0;
while (x != 0) {
if (x % 2 != 0) p += y;
x /= 2;
y *= 2;
}
return positive ? p : -p;
}
Here is simple Java implementation which does multiplication without multiplication operator.
public static int multiply(int a, int b) {
int p = 0;
// If a is odd number.
if ((a & 1) > 0) {
p = b;
} //else use the default value in the p.
// If 'a' contains any number larger than one
// than continue recursion.
if (a > 1)
p = p + multiply(a >> 1, b << 1);
return p;
}
public static int multiply(int x, int y) {
if (y == 0 || x == 0) {
return 0;
}
if (x == 1) {
return y;
} else {
return multiply(x >> 1, y << 1);
}
}