https://leetcode.com/problems/balanced-binary-tree/
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary
tree in which the depth of the two subtrees of every node never differ
by more than 1.
public class Solution {
public boolean isBalanced(TreeNode root) {
int ret = getLevel(root);
if(ret < 0)
return false;
return true;
}
public int getLevel(TreeNode node) {
if(node == null)
return 0;
int l = getLevel(node.left);
int r = getLevel(node.right);
if(Math.abs(l - r) > 1)
return -99;
return Math.max(l + 1, r + 1);
}
}
This code is Accepted.
However if I replace -99 with Integer.MIN_VALUE, my code fails. What's the bug?
e.g.
Input: [1,2,null,3,null,4,null,5]
Output: true
Expected: false
Your code is failing because of integer arithmetic which is overflowing. The following code snippet will demonstrate this:
int val = Integer.MIN_VALUE;
System.out.println(val);
val -= 3;
System.out.println(val);
Output:
-2147483648
2147483645
Now consider what is happening in your actual code:
int l = getLevel(node.left);
// l == -2147483648 == Integer.MIN_VALUE assuming your base case is hit
int r = getLevel(node.right);
// assuming positive r, then Math.abs() will return a massively positive number
if (Math.abs(l - r) > 1)
return -99;
In other words, the above if statement will be firing true when it really should have fired false.
Solution:
If you modify the getLevel() method to the following, you should skirt the problems you are having:
public int getLevel(TreeNode node) {
if(node == null)
return 0;
int l = getLevel(node.left);
int r = getLevel(node.right);
if ( (l < 0 ^ r < 0) || Math.abs(l - r) > 1) {
// you can simply return -1 here, since an actual
// level should never have a negative value
return -1;
}
else {
return Math.max(l + 1, r + 1);
}
}
It may fail under some circumstances due to overflow. If l is zero and r is Integer.MIN_VALUE, l-r is actually negative because it overflows. As a result, the condition will fail and the next statement returns max of MIN_VALUE+1 and zero+1.
I dont see how it is not transitive, please someone suggest me the right way.
if both values are null
I return 0, the two other statements are pretty obvious.
Why i got IllegalArgumentException with:
Comparison method violates its general contract
My compare method: (I compare doubles)
#Override
public int compare(HashMap<String, String> lhs, HashMap<String, String> rhs) {
double dist1 = 0;
double dist2 = 0;
int compInt1 = 0;
int compInt2 = 0;
if (lhs.get("dist") != null && rhs.get("dist") != null && !lhs.get("dist").equals("") && !rhs.get("dist").equals("")) {
dist1 = Double.parseDouble(lhs.get("dist").substring(0, lhs.get("dist").length() - 3));
dist2 = Double.parseDouble(rhs.get("dist").substring(0, rhs.get("dist").length() - 3));
dist1 = dist1 * 100;
dist2 = dist2 * 100;
compInt1 = (int) dist1;
compInt2 = (int) dist2;
}
if (compInt1 < compInt2) {
return -1;
} else if (compInt1 >= compInt2) {
return 1;
} else {
return 0;
}
}
Look at this code:
if (compInt1 < compInt2) {
return -1;
} else if (compInt1 >= compInt2) {
return 1;
} else {
return 0;
}
How do you expect that ever to return 0? What values of compInt1 and compInt2 would make both if conditions fail?
This violates symmetry too - it means that compare(x, y) and compare(y, x) can both return 1...
Assuming you're really just trying to compare compInt1 and compInt2 in the obvious way at this point, just replace the whole block with:
// As of Java 7...
return Integer.compare(compInt1, compInt2);
If you're using Java 6 (or earlier) you could use:
return Integer.valueOf(compInt1).compareTo(compInt2);
That's a little inefficient, but it's at least worth using to start with, just to get everything working.
Additionally, I'd strongly recommend extracting the lhs.get("dist") and rhs.get("dist") expressions from your first part - it's horribly repetitive at the moment. And just declare dist1 and dist2 within the block - they're not used elsewhere.
The condition else if (compInt1 >= compInt2) is ambiguous as it would return a value of 1 even if both values are equal. Replace your comparison logic of ints as follows :
return compInt1 - compInt2;
Transitivity defines as :
compare(a, b) == -compare(b, a)
Which implies compare(a, a) == -compare(a, a) == 0
Now, considering your code,
compare(a, a) == 1
A java.lang.IllegalArgumentException: Comparison method violates its general contract! is thrown when I do a Collections.sort() with a List of ISimulationResultSet.
I have not found the reason why contract is not respected.
If someone has a idea of the reason, it will be great to explain.
This is the Comparator I am using :
public int compare(ISimulationResultSet r1, ISimulationResultSet r2) {
final float r1Esperance = r1.getResults().getEsperanceGainOuPerte();
final float r2Esperance = r2.getResults().getEsperanceGainOuPerte();
final float r1PrctCibleAtteinte = r1.getResults().getPrctCibleAtteinte();
final float r2PrctCibleAtteinte = r2.getResults().getPrctCibleAtteinte();
if (r1Esperance / r2Esperance > 1.05F)
return -1;
else if (r1Esperance / r2Esperance < 0.95F) {
return 1;
}
else {
if (r1PrctCibleAtteinte == r2PrctCibleAtteinte) {
if (r1Esperance > r2Esperance)
return -1;
else if (r1Esperance < r2Esperance)
return 1;
return 0;
}
else if (r1PrctCibleAtteinte > r2PrctCibleAtteinte)
return -1;
else if (r1PrctCibleAtteinte < r2PrctCibleAtteinte)
return 1;
}
return 0;
}
The comparator has to be symetric, i.e. sgn(compare(x, y)) == -sgn(compare(y, x)) (sgn being the signum function here). This is not the case for your comparator:
Let a1 and a2 denote the values of x.getResults().getEsperanceGainOuPerte() and y.getResults().getEsperanceGainOuPerte() respectively and let b1 and b2 denote the values of x.getResults().getPrctCibleAtteinte() and y.getResults().getPrctCibleAtteinte() respectively.
Now consider the following:
1.05 < a1 < 1.052
a2 = 1
b2 > b1
Therefore a2 / a1 > 0.95
compare(x, y) == -1;// first (r1Esperance / r2Esperance > 1.05F) is true
compare(y, x) == -1; // first 3 conditions false, (r1PrctCibleAtteinte > r2PrctCibleAtteinte) is true
That violates the contract.
I need to write a method that will take a base and raises it to any integer power, positive or negative. It can be assumed that the base will not be 0.
In the method I need to call a recursive method and use it.
Here is the previous recursive method I need to use:
public static double nonNegInt(double base, int pow)
{
if (pow == 0)
return 1;
else
return base * nonNegInt(base,pow-1);
}
So my question is, can someone please help or show me how to write the method I need?
I know the current method is fine, but I need to call it in another method. When I do this I am getting a runtime error
Your method is a good start, although you will need to handle negative exponents as stated in your requirements. Take advantage of the fact that x^(-n) = 1.0 / x^n.
This is how you handle negative values too:
public static double nonNegInt(double base, int pow)
{
if (pow == 0)
return 1;
else if(pow < 0)
return (1 / nonNegInt(base, -pow));
else
return base * nonNegInt(base,pow-1);
}
Running it:
public static void main(String args[])
{
double result = nonNegInt(4,-1);
System.out.println(result); //Will print 0.25
}
Of course you should give it a meaningful name, since now it does handle negative cases.
public BigDecimal exp(BigDecimal base, BigInteger pow) {
if(base == null || base.intValue() == 0 ) return BigDecimal.ZERO;
BigInteger absPow = pow.abs();
if(absPow.intValue() == 0) return BigDecimal.ONE;
if(absPow.intValue() == 1) return pow.intValue() > 0 ? base :
BigDecimal.ONE.divide(base, MathContext.DECIMAL128);
if(absPow.intValue() == 2) return pow.intValue() > 0 ? base.multiply(base):
BigDecimal.ONE.divide(base.multiply(base), MathContext.DECIMAL128);
BigInteger i = BigInteger.ONE;
BigDecimal result = base;
HashMap<BigInteger, BigDecimal> history = new HashMap<>();
history.put(i, result);
while (i.compareTo(absPow) < 0) {
if(i.add(i).compareTo(absPow) <= 0) {
i = i.add(i);
result = result.multiply(result);
history.put(i, result);
} else {
BigInteger diff = absPow.subtract(i);
for (; diff.intValue() > 0 && !history.containsKey(diff); diff = diff.subtract(BigInteger.ONE));
i = i.add(diff);
result = result.multiply(history.get(diff));
history.put(i, result);
}
}
return pow.intValue() > 0 ? result : BigDecimal.ONE.divide(result, MathContext.DECIMAL128);
}
I was asked in an interview, how to determine whether a number is positive or negative. The rules are that we should not use relational operators such as <, and >, built in java functions (like substring, indexOf, charAt, and startsWith), no regex, or API's.
I did some homework on this and the code is given below, but it only works for integer type. But they asked me to write a generic code that works for float, double, and long.
// This might not be better way!!
S.O.P ((( number >> 31 ) & 1) == 1 ? "- ve number " : "+ve number );
any ideas from your side?
The integer cases are easy. The double case is trickier, until you remember about infinities.
Note: If you consider the double constants "part of the api", you can replace them with overflowing expressions like 1E308 * 2.
int sign(int i) {
if (i == 0) return 0;
if (i >> 31 != 0) return -1;
return +1;
}
int sign(long i) {
if (i == 0) return 0;
if (i >> 63 != 0) return -1;
return +1;
}
int sign(double f) {
if (f != f) throw new IllegalArgumentException("NaN");
if (f == 0) return 0;
f *= Double.POSITIVE_INFINITY;
if (f == Double.POSITIVE_INFINITY) return +1;
if (f == Double.NEGATIVE_INFINITY) return -1;
//this should never be reached, but I've been wrong before...
throw new IllegalArgumentException("Unfathomed double");
}
The following is a terrible approach that would get you fired at any job...
It depends on you getting a Stack Overflow Exception [or whatever Java calls it]... And it would only work for positive numbers that don't deviate from 0 like crazy.
Negative numbers are fine, since you would overflow to positive, and then get a stack overflow exception eventually [which would return false, or "yes, it is negative"]
Boolean isPositive<T>(T a)
{
if(a == 0) return true;
else
{
try
{
return isPositive(a-1);
}catch(StackOverflowException e)
{
return false; //It went way down there and eventually went kaboom
}
}
}
This will only works for everything except [0..2]
boolean isPositive = (n % (n - 1)) * n == n;
You can make a better solution like this (works except for [0..1])
boolean isPositive = ((n % (n - 0.5)) * n) / 0.5 == n;
You can get better precision by changing the 0.5 part with something like 2^m (m integer):
boolean isPositive = ((n % (n - 0.03125)) * n) / 0.03125 == n;
You can do something like this:
((long) (num * 1E308 * 1E308) >> 63) == 0 ? "+ve" : "-ve"
The main idea here is that we cast to a long and check the value of the most significant bit. As a double/float between -1 and 0 will round to zero when cast to a long, we multiply by large doubles so that a negative float/double will be less than -1. Two multiplications are required because of the existence of subnormals (it doesn't really need to be that big though).
What about this?
return ((num + "").charAt(0) == '-');
// Returns 0 if positive, nonzero if negative
public long sign(long value) {
return value & 0x8000000000000000L;
}
Call like:
long val1 = ...;
double val2 = ...;
float val3 = ...;
int val4 = ...;
sign((long) valN);
Casting from double / float / integer to long should preserve the sign, if not the actual value...
You say
we should not use conditional operators
But this is a trick requirement, because == is also a conditional operator. There is also one built into ? :, while, and for loops. So nearly everyone has failed to provide an answer meeting all the requirements.
The only way to build a solution without a conditional operator is to use lookup table vs one of a few other people's solutions that can be boiled down to 0/1 or a character, before a conditional is met.
Here are the answers that I think might work vs a lookup table:
Nabb
Steven Schlansker
Dennis Cheung
Gary Rowe
This solution uses modulus. And yes, it also works for 0.5 (tests are below, in the main method).
public class Num {
public static int sign(long x) {
if (x == 0L || x == 1L) return (int) x;
return x == Long.MIN_VALUE || x % (x - 1L) == x ? -1 : 1;
}
public static int sign(double x) {
if (x != x) throw new IllegalArgumentException("NaN");
if (x == 0.d || x == 1.d) return (int) x;
if (x == Double.POSITIVE_INFINITY) return 1;
if (x == Double.NEGATIVE_INFINITY) return -1;
return x % (x - 1.d) == x ? -1 : 1;
}
public static int sign(int x) {
return Num.sign((long)x);
}
public static int sign(float x) {
return Num.sign((double)x);
}
public static void main(String args[]) {
System.out.println(Num.sign(Integer.MAX_VALUE)); // 1
System.out.println(Num.sign(1)); // 1
System.out.println(Num.sign(0)); // 0
System.out.println(Num.sign(-1)); // -1
System.out.println(Num.sign(Integer.MIN_VALUE)); // -1
System.out.println(Num.sign(Long.MAX_VALUE)); // 1
System.out.println(Num.sign(1L)); // 1
System.out.println(Num.sign(0L)); // 0
System.out.println(Num.sign(-1L)); // -1
System.out.println(Num.sign(Long.MIN_VALUE)); // -1
System.out.println(Num.sign(Double.POSITIVE_INFINITY)); // 1
System.out.println(Num.sign(Double.MAX_VALUE)); // 1
System.out.println(Num.sign(0.5d)); // 1
System.out.println(Num.sign(0.d)); // 0
System.out.println(Num.sign(-0.5d)); // -1
System.out.println(Num.sign(Double.MIN_VALUE)); // -1
System.out.println(Num.sign(Double.NEGATIVE_INFINITY)); // -1
System.out.println(Num.sign(Float.POSITIVE_INFINITY)); // 1
System.out.println(Num.sign(Float.MAX_VALUE)); // 1
System.out.println(Num.sign(0.5f)); // 1
System.out.println(Num.sign(0.f)); // 0
System.out.println(Num.sign(-0.5f)); // -1
System.out.println(Num.sign(Float.MIN_VALUE)); // -1
System.out.println(Num.sign(Float.NEGATIVE_INFINITY)); // -1
System.out.println(Num.sign(Float.NaN)); // Throws an exception
}
}
This code covers all cases and types:
public static boolean isNegative(Number number) {
return (Double.doubleToLongBits(number.doubleValue()) & Long.MIN_VALUE) == Long.MIN_VALUE;
}
This method accepts any of the wrapper classes (Integer, Long, Float and Double) and thanks to auto-boxing any of the primitive numeric types (int, long, float and double) and simply checks it the high bit, which in all types is the sign bit, is set.
It returns true when passed any of:
any negative int/Integer
any negative long/Long
any negative float/Float
any negative double/Double
Double.NEGATIVE_INFINITY
Float.NEGATIVE_INFINITY
and false otherwise.
Untested, but illustrating my idea:
boolean IsNegative<T>(T v) {
return (v & ((T)-1));
}
It seems arbitrary to me because I don't know how you would get the number as any type, but what about checking Abs(number) != number? Maybe && number != 0
Integers are trivial; this you already know. The deep problem is how to deal with floating-point values. At that point, you've got to know a bit more about how floating point values actually work.
The key is Double.doubleToLongBits(), which lets you get at the IEEE representation of the number. (The method's really a direct cast under the hood, with a bit of magic for dealing with NaN values.) Once a double has been converted to a long, you can just use 0x8000000000000000L as a mask to select the sign bit; if zero, the value is positive, and if one, it's negative.
If it is a valid answer
boolean IsNegative(char[] v) throws NullPointerException, ArrayIndexOutOfBoundException
{
return v[0]=='-';
}
one more option I could think of
private static boolean isPositive(Object numberObject) {
Long number = Long.valueOf(numberObject.toString());
return Math.sqrt((number * number)) != number;
}
private static boolean isPositive(Object numberObject) {
Long number = Long.valueOf(numberObject.toString());
long signedLeftShifteredNumber = number << 1; // Signed left shift
long unsignedRightShifterNumber = signedLeftShifteredNumber >>> 1; // Unsigned right shift
return unsignedRightShifterNumber == number;
}
This one is roughly based on ItzWarty's answer, but it runs in logn time! Caveat: Only works for integers.
Boolean isPositive(int a)
{
if(a == -1) return false;
if(a == 0) return false;
if(a == 1) return true;
return isPositive(a/2);
}
I think there is a very simple solution:
public boolean isPositive(int|float|double|long i){
return (((i-i)==0)? true : false);
}
tell me if I'm wrong!
Try this without the code: (x-SQRT(x^2))/(2*x)
Write it using the conditional then take a look at the assembly code generated.
Why not get the square root of the number? If its negative - java will throw an error and we will handle it.
try {
d = Math.sqrt(THE_NUMBER);
}
catch ( ArithmeticException e ) {
console.putln("Number is negative.");
}
I don't know how exactly Java coerces numeric values, but the answer is pretty simple, if put in pseudocode (I leave the details to you):
sign(x) := (x == 0) ? 0 : (x/x)
If you are allowed to use "==" as seems to be the case, you can do something like that taking advantage of the fact that an exception will be raised if an array index is out of bounds. The code is for double, but you can cast any numeric type to a double (here the eventual loss of precision would not be important at all).
I have added comments to explain the process (bring the value in ]-2.0; -1.0] union [1.0; 2.0[) and a small test driver as well.
class T {
public static boolean positive(double f)
{
final boolean pos0[] = {true};
final boolean posn[] = {false, true};
if (f == 0.0)
return true;
while (true) {
// If f is in ]-1.0; 1.0[, multiply it by 2 and restart.
try {
if (pos0[(int) f]) {
f *= 2.0;
continue;
}
} catch (Exception e) {
}
// If f is in ]-2.0; -1.0] U [1.0; 2.0[, return the proper answer.
try {
return posn[(int) ((f+1.5)/2)];
} catch (Exception e) {
}
// f is outside ]-2.0; 2.0[, divide by 2 and restart.
f /= 2.0;
}
}
static void check(double f)
{
System.out.println(f + " -> " + positive(f));
}
public static void main(String args[])
{
for (double i = -10.0; i <= 10.0; i++)
check(i);
check(-1e24);
check(-1e-24);
check(1e-24);
check(1e24);
}
The output is:
-10.0 -> false
-9.0 -> false
-8.0 -> false
-7.0 -> false
-6.0 -> false
-5.0 -> false
-4.0 -> false
-3.0 -> false
-2.0 -> false
-1.0 -> false
0.0 -> true
1.0 -> true
2.0 -> true
3.0 -> true
4.0 -> true
5.0 -> true
6.0 -> true
7.0 -> true
8.0 -> true
9.0 -> true
10.0 -> true
-1.0E24 -> false
-1.0E-24 -> false
1.0E-24 -> true
1.0E24 -> true
Well, taking advantage of casting (since we don't care what the actual value is) perhaps the following would work. Bear in mind that the actual implementations do not violate the API rules. I've edited this to make the method names a bit more obvious and in light of #chris' comment about the {-1,+1} problem domain. Essentially, this problem does not appear to solvable without recourse to API methods within Float or Double that reference the native bit structure of the float and double primitives.
As everybody else has said: Stupid interview question. Grr.
public class SignDemo {
public static boolean isNegative(byte x) {
return (( x >> 7 ) & 1) == 1;
}
public static boolean isNegative(short x) {
return (( x >> 15 ) & 1) == 1;
}
public static boolean isNegative(int x) {
return (( x >> 31 ) & 1) == 1;
}
public static boolean isNegative(long x) {
return (( x >> 63 ) & 1) == 1;
}
public static boolean isNegative(float x) {
return isNegative((int)x);
}
public static boolean isNegative(double x) {
return isNegative((long)x);
}
public static void main(String[] args) {
// byte
System.out.printf("Byte %b%n",isNegative((byte)1));
System.out.printf("Byte %b%n",isNegative((byte)-1));
// short
System.out.printf("Short %b%n",isNegative((short)1));
System.out.printf("Short %b%n",isNegative((short)-1));
// int
System.out.printf("Int %b%n",isNegative(1));
System.out.printf("Int %b%n",isNegative(-1));
// long
System.out.printf("Long %b%n",isNegative(1L));
System.out.printf("Long %b%n",isNegative(-1L));
// float
System.out.printf("Float %b%n",isNegative(Float.MAX_VALUE));
System.out.printf("Float %b%n",isNegative(Float.NEGATIVE_INFINITY));
// double
System.out.printf("Double %b%n",isNegative(Double.MAX_VALUE));
System.out.printf("Double %b%n",isNegative(Double.NEGATIVE_INFINITY));
// interesting cases
// This will fail because we can't get to the float bits without an API and
// casting will round to zero
System.out.printf("{-1,1} (fail) %b%n",isNegative(-0.5f));
}
}
This solution uses no conditional operators, but relies on catching two excpetions.
A division error equates to the number originally being "negative". Alternatively, the number will eventually fall off the planet and throw a StackOverFlow exception if it is positive.
public static boolean isPositive( f)
{
int x;
try {
x = 1/((int)f + 1);
return isPositive(x+1);
} catch (StackOverFlow Error e) {
return true;
} catch (Zero Division Error e) {
return false;
}
}
What about the following?
T sign(T x) {
if(x==0) return 0;
return x/Math.abs(x);
}
Should work for every type T...
Alternatively, one can define abs(x) as Math.sqrt(x*x),
and if that is also cheating, implement your own square root function...
if (((Double)calcYourDouble()).toString().contains("-"))
doThis();
else doThat();
Combined generics with double API. Guess it's a bit of cheating, but at least we need to write only one method:
static <T extends Number> boolean isNegative(T number)
{
return ((number.doubleValue() * Double.POSITIVE_INFINITY) == Double.NEGATIVE_INFINITY);
}
Two simple solutions. Works also for infinities and numbers -1 <= r <= 1
Will return "positive" for NaNs.
String positiveOrNegative(double number){
return (((int)(number/0.0))>>31 == 0)? "positive" : "negative";
}
String positiveOrNegative(double number){
return (number==0 || ((int)(number-1.0))>>31==0)? "positive" : "negative";
}
There is a function is the math library called signnum.
http://www.tutorialspoint.com/java/lang/math_signum_float.htm
http://www.tutorialspoint.com/java/lang/math_signum_double.htm
It's easy to do this like
private static boolean isNeg(T l) {
return (Math.abs(l-1)>Math.abs(l));
}
static boolean isNegative(double v) {
return new Double(v).toString().startsWith("-");
}