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I want to find whether a given number is a power of two in a mathematical way, not with a bitwise approach. Here is my code:
private static double logBaseTwo(final double x) {
return Math.log(x) / Math.log(2);
}
private static double roundToNearestHundredThousandth(final double x) {
return Math.round(x * 100000.0) / 100000.0;
}
private static boolean isInteger(final double x) {
return (int)(Math.ceil(x)) == (int)(Math.floor(x));
}
public static boolean isPowerOfTwo(final int n) {
return isInteger(roundToNearestHundredThousandth(logBaseTwo(n)));
}
It incorrectly returns true for certain numbers, such as 524287. Why is that?
Your code fails because you may need more precision than you allow to capture the difference between the logs of BIG_NUMBER and BIG_NUMBER+1
The bitwise way is really best, but if you really want to use only "mathy" operations, then the best you can do is probably:
public static boolean isPowerOfTwo(final int n) {
int exp = (int)Math.round(logBaseTwo(n));
int test = (int)Math.round(Math.pow(2.0,exp));
return test == n;
}
This solution does not require any super-fine precision, and will work fine for all positive ints.
This is truly horrifyingly bad code, and I have no idea what you are trying to do. You seem to be trying to check if the log base 2 of n is an integer. Instead I would write a loop:
while (n>1) {
m = (n/2) * 2
if (n!=m){
return false;
}
n /=2;
}
return true;
The solution seems more complicated than it should be. I don't get the 100000d parts - seems to potentially cause problems when converting to ceiling.
This is the simple solution that works for all cases:
public static boolean isPowerOfTwo(int n) {
return Math.ceil(Math.log(n)/Math.log(2)) == Math.floor(Math.log(n)/Math.log(2));
}
Originally I had a problem using Math.log in my computations. I switched to Math.log10 and the problem went away. Although mathematically, any logB of base B should work, the nature of floating point math can be unpredictable.
Try this.
public static boolean isPowerOfTwo(int n) {
return n > 0 && Integer.highestOneBit(n) == Integer.lowestOneBit(n);
}
If you prefer to use logs you can do it this way.
public static boolean isPowerOfTwo(int n) {
return n > 0 && (Math.log10(n)/Math.log10(2))%1 == 0;
}
doubles and floats have, respectively, 64-bit and 32-bit precision. That means they can hold at the very most 18446744073709551616 unique numbers. That's a lot of numbers, but not an infinite amount of them. At some point (in fact, that point occurs about at 2^52), the 'gap' between any 2 numbers which are part of the 18446744073709551616 representable ones becomes larger than 1.000. Similar rules apply to small numbers. Math.log does double based math.
Secondarily, ints are similarly limited. They can hold up to 4294967296 different numbers. For ints it's much simpler: Ints can hold from -2147483648 up to 2147483647. If you try to add 1 to 2147483647, you get -2147483648 (it silently wraps around). It's quite possible you're running into that with trying to convert such a large number (your double times 10000d) to an int first.
Note that ? true : false (as in the original version of the question) is literally completely useless. the thing to the left of the question mark must be a boolean, and booleans are already true or false, that's their nature.
See the other answers for simpler approaches to this problem. Although, of course, the simplest solution is to simply count bits in the number. If it's precisely 1 bit, it's a power of 2. If it's 0 bits, well, you tell me if you consider '0' a power of 2 :)
This question already has answers here:
How can I perform multiplication without the '*' operator?
(31 answers)
Closed 4 years ago.
I had an interesting interview yesterday where the interviewer asked me a classic question: How can we multiply two numbers in Java without using the * operator. Honestly, I don't know if it's the stress that comes with interviews, but I wasn't able to come up with any solution.
After the interview, I went home and breezed through SO for answers. So far, here are the ones I have found:
First Method: Using a For loop
// Using For loop
public static int multiplierLoop(int a, int b) {
int resultat = 0;
for (int i = 0; i < a; i++) {
resultat += b;
}
return resultat;
}
Second Method: Using Recursion
// using Recursion
public static int multiplier(int a, int b) {
if ((a == 0) || (b == 0))
return 0;
else
return (a + multiplier(a, b - 1));
}
Third Method: Using Log10
**// Using Math.Log10
public static double multiplierLog(int a, int b) {
return Math.pow(10, (Math.log10(a) + Math.log10(b)));
}**
So now I have two questions for you:
Is there still another method I'm missing?
Does the fact that I wasn't able to come up with the answer proves that my logical reasoning isn't strong enough to come up with solutions and that I'm not "cut out" to be a programmer? Cause let's be honest, the question didn't seem that difficult and I'm pretty sure most programmers would easily and quickly find an answer.
I don't know whether that has to be a strictly "programming question". But in Maths:
x * y = x / (1 / y) #divide by inverse
So:
Method 1:
public static double multiplier(double a, double b) {
// return a / (1 / b);
// the above may be too rough
// Java doesn't know that "(a / (b / 0)) == 0"
// a special case for zero should probably be added:
return 0 == b ? 0 : a / (1 / b);
}
Method 2 (a more "programming/API" solution):
Use big decimal, big integer:
new BigDecimal("3").multiply(new BigDecimal("9"))
There are probably a few more ways.
There is a method called [Russian Peasant Multiplication][1]. Demonstrate this with the help of a shift operator,
public static int multiply(int n, int m)
{
int ans = 0, count = 0;
while (m > 0)
{
if (m % 2 == 1)
ans += n << count;
count++;
m /= 2;
}
return ans;
}
The idea is to double the first number and halve the second number repeatedly till the second number doesn’t become 1. In the process, whenever the second number become odd, we add the first number to result (result is initialized as 0) One other implementation is,
static int russianPeasant(int n, int m) {
int ans = 0;
while (m > 0) {
if ((m & 1) != 0)
ans = ans + n;
n = n << 1;
m = m >> 1;
}
return ans;
}
refer :
https://www.geeksforgeeks.org/russian-peasant-multiply-two-numbers-using-bitwise-operators/
https://www.geeksforgeeks.org/multiplication-two-numbers-shift-operator/
[1]: https://web.archive.org/web/20180101093529/http://mathforum.org/dr.math/faq/faq.peasant.html
Others have hit on question 1 sufficiently that I'm not going to rehash it here, but I did want to hit on question 2 a little, because it seems (to me) the more interesting one.
So, when someone is asking you this type of question, they are less concerned with what your code looks like, and more concerned with how you are thinking. In the real world, you won't ever actually have to write multiplication without the * operator; every programming language known to man (with the exception of Brainfuck, I guess) has multiplication implemented, almost always with the * operator. The point is, sometimes you are working with code, and for whatever reason (maybe due to library bloat, due to configuration errors, due to package incompatibility, etc), you won't be able to use a library you are used to. The idea is to see how you function in those situations.
The question isn't whether or not you are "cut out" to be a programmer; skills like these can be learned. A trick I use personally is to think about what, exactly, is the expected result for the question they're asking? In this particular example, as I (and I presume you as well) learned in grade 4 in elementary school, multiplication is repeated addition. Therefore, I would implement it (and have in the past; I've had this same question in a few interviews) with a for loop doing repeated addition.
The thing is, if you don't realize that multiplication is repeated addition (or whatever other question you're being asked to answer), then you'll just be screwed. Which is why I'm not a huge fan of these types of questions, because a lot of them boil down to trivia that you either know or don't know, rather than testing your true skills as a programmer (the skills mentioned above regarding libraries etc can be tested much better in other ways).
TL;DR - Inform the interviewer that re-inventing the wheel is a bad idea
Rather than entertain the interviewer's Code Golf question, I would have answered the interview question differently:
Brilliant engineers at Intel, AMD, ARM and other microprocessor manufacturers have agonized for decades as how to multiply 32 bit integers together in the fewest possible cycles, and in fact, are even able to produce the correct, full 64 bit result of multiplication of 32 bit integers without overflow.
(e.g. without pre-casting a or b to long, a multiplication of 2 ints such as 123456728 * 23456789 overflows into a negative number)
In this respect, high level languages have only one job to do with integer multiplications like this, viz, to get the job done by the processor with as little fluff as possible.
Any amount of Code Golf to replicate such multiplication in software IMO is insanity.
There's undoubtedly many hacks which could simulate multiplication, although many will only work on limited ranges of values a and b (in fact, none of the 3 methods listed by the OP perform bug-free for all values of a and b, even if we disregard the overflow problem). And all will be (orders of magnitude) slower than an IMUL instruction.
For example, if either a or b is a positive power of 2, then bit shifting the other variable to the left by log can be done.
if (b == 2)
return a << 1;
if (b == 4)
return a << 2;
...
But this would be really tedious.
In the unlikely event of the * operator really disappearing overnight from the Java language spec, next best, I would be to use existing libraries which contain multiplication functions, e.g. BigInteger.multiply(), for the same reasons - many years of critical thinking by minds brighter than mine has gone into producing, and testing, such libraries.
BigInteger.multiply would obviously be reliable to 64 bits and beyond, although casting the result back to a 32 bit int would again invite overflow problems.
The problem with playing operator * Code Golf
There's inherent problems with all 3 of the solutions cited in the OP's question:
Method A (loop) won't work if the first number a is negative.
for (int i = 0; i < a; i++) {
resultat += b;
}
Will return 0 for any negative value of a, because the loop continuation condition is never met
In Method B, you'll run out of stack for large values of b in method 2, unless you refactor the code to allow for Tail Call Optimisation
multiplier(100, 1000000)
"main" java.lang.StackOverflowError
And in Method 3, you'll get rounding errors with log10 (not to mention the obvious problems with attempting to take a log of any number <= 0). e.g.
multiplier(2389, 123123);
returns 294140846, but the actual answer is 294140847 (the last digits 9 x 3 mean the product must end in 7)
Even the answer using two consecutive double precision division operators is prone to rounding issues when re-casting the double result back to an integer:
static double multiply(double a, double b) {
return 0 == (int)b
? 0.0
: a / (1 / b);
}
e.g. for a value (int)multiply(1, 93) returns 92, because multiply returns 92.99999.... which is truncated with the cast back to a 32 bit integer.
And of course, we don't need to mention that many of these algorithms are O(N) or worse, so the performance will be abysmal.
For completeness:
Math.multiplyExact(int,int):
Returns the product of the arguments, throwing an exception if the result overflows an int.
if throwing on overflow is acceptable.
If you don't have integer values, you can take advantage of other mathematical properties to get the product of 2 numbers. Someone has already mentioned log10, so here's a bit more obscure one:
public double multiply(double x, double y) {
Vector3d vx = new Vector3d(x, 0, 0);
Vector3d vy = new Vector3d(0, y, 0);
Vector3d result = new Vector3d().cross(vx, vy);
return result.length();
}
One solution is to use bit wise operations. That's a bit similar to an answer presented before, but eliminating division also. We can have something like this. I'll use C, because I don't know Java that well.
uint16_t multiply( uint16_t a, uint16_t b ) {
uint16_t i = 0;
uint16_t result = 0;
for (i = 0; i < 16; i++) {
if ( a & (1<<i) ) {
result += b << i;
}
}
return result;
}
The questions interviewers ask reflect their values. Many programmers prize their own puzzle-solving skills and mathematical acumen, and they think those skills make the best programmers.
They are wrong. The best programmers work on the most important thing rather than the most interesting bit; make simple, boring technical choices; write clearly; think about users; and steer away from stupid detours. I wish I had these skills and tendencies!
If you can do several of those things and also crank out working code, many programming teams need you. You might be a superstar.
But what should you do in an interview when you're stumped?
Ask clarifying questions. ("What kind of numbers?" "What kind of programming language is this that doesn't have multiplication?" And without being rude: "Why am I doing this?") If, as you suspect, the question is just a dumb puzzle with no bearing on reality, these questions will not produce useful answers. But common sense and a desire to get at "the problem behind the problem" are important engineering virtues.
The best you can do in a bad interview is demonstrate your strengths. Recognizing them is up to your interviewer; if they don't, that's their loss. Don't be discouraged. There are other companies.
Use BigInteger.multiply or BigDecimal.multiply as appropriate.
so I have a question about an algorithm I'm supposed to "invent"/"find". It's an algorithm which calculates 2^(n) - 1 for Ө(n^n) and Ө(1) and Ө(n).
I was thinking for several hours but I couldn't find any solution for both tasks (the first ones while the last one was the easist imo, I posted the algorithm below). But I'm not skilled enough to "invent"/"find" one for a very slow and very fast algorithm.
So far my algorithms are (In Pseudocode):
The one for Ө(n)
int f(int n) {
int number = 2
if(n = 0) then return 0
if(n==1) then return 1
while(n > 1)
number = number * 2
n--
number = number - 1
return number
A simple one and kinda obvious one which uses recursion though I don't know how fast it is (It would be nice if someone could tell me that):
int f(int n) {
if(n==0) then return 0
if(n==1) then return 1
return 3*f(n-1) - 2*f(n-2)
}
Assuming n is not bounded by any constant (and output should not be a simple int, but a data type that can contain large integers to allow it) - there is no algorithm
to yield 2^n -1 in Ө(1), since the size of the output itself is
Ө(log(n)), so if we assume there is such algorithm, and let it
run in constant time and makes less than C operations, for n =
2^(C+1), you will require C+1 operations only to print the
output, which contradicts the assumption that C is the upper bound, so
there is no such algorithm.
For Ө(n^n), if you have a more efficient algorithm (Ө(n) for example), you can make a pointless loop that runs extra n^n iterations and do nothing important, it will make your algorithm Ө(n^n).
There is also a Ө(log(n)*M(logn)) algorithm, using exponent by squaring, and then simply reducing 1 from this value. In here M(x) is complexity of your multiplying operator for number containing x digits.
As commented by #kajacx, you can even improve (3) by applying Fourier transform
Something like:
HugeInt h = 1;
h = h << n;
h = h - 1;
Obviously HugeInt is pseudo-code for an integer type that can be of arbitrary size allowing for any n.
=====
Look at amit's answer instead!
the Ө(n^n) is too tricky for me, but a real Ө(1) algorithm on any "binary" architecture would be:
return n-1 bits filled with 1
(assuming your architecture can allocate and fill n-1 bits in constant time)
;)
I have a Java method that repeatedly evaluates the following expression in a very tight loop with a large number of repetitions:
Math.abs(a - b) - Math.abs(c - d)
a, b, c and d are long values that can span the whole range of their type. They are different in each loop iteration and they do not satisfy any invariant that I know of.
The profiler indicates that a significant portion of the processor time is spent in this method. While I will pursue other avenues of optimization first, I was wondering if there is a smarter way to calculate the aforementioned expression.
Apart from inlining the Math.abs() calls manually for a very slight (if any) performance gain, is there any mathematical trick that I could use to speed-up the evaluation of this expression?
I suspect the profiler isn't giving you a true result as it trying to profile (and thus adding over head to) such a trivial "method". Without the profile Math.abs can be turned into a small number of machine code instructions, and you won't be able to make it faster than that.
I suggest you do a micro-benchmark to confirm this. I would expect loading the data to be an order of magnitude more expensive.
long a = 10, b = 6, c = -2, d = 3;
int runs = 1000 * 1000 * 1000;
long start = System.nanoTime();
for (int i = 0; i < runs; i += 2) {
long r = Math.abs(i - a) - Math.abs(c - i);
long r2 = Math.abs(i - b) - Math.abs(d - i);
if (r + r2 < Integer.MIN_VALUE) throw new AssertionError();
}
long time = System.nanoTime() - start;
System.out.printf("Took an average of %.1f ns per abs-abs. %n", (double) time / runs);
prints
Took an average of 0.9 ns per abs-abs.
I ended up using this little method:
public static long diff(final long a, final long b, final long c, final long d) {
final long a0 = (a < b)?(b - a):(a - b);
final long a1 = (c < d)?(d - c):(c - d);
return a0 - a1;
}
I experienced a measurable performance increase - about 10-15% for the whole application. I believe this is mostly due to:
The elimination of a method call: Rather than calling Math.abs() twice, I call this method once. Sure, static method calls are not inordinately expensive, but they still have an impact.
The elimination of a couple of negation operations: This may be offset by the slightly increased size of the code, but I'll happily fool myself into believing that it actually made a difference.
EDIT:
It seems that it's actually the other way around. Explicitly inlining the code does not seem to impact the performance in my micro-benchmark. Changing the way the absolute values are calculated does...
You can always try to unroll the functions and hand optimize, if you don't get more cache misses it might be faster.
If I got the unrolling right it could be something like this:
if(a<b)
{
if(c<d)
{
r=b-a-d+c;
}
else
{
r=b-a+d-c;
}
}
else
{
if(c<d)
{
r=a-b-d+c;
}
else
{
r=a-b+d-c;
}
}
are you sure its the method itself causes the problem? Maybe its an enormous amount of invocation of this method and you just see the aggregated results (like TIME_OF_METHOD_EXECUTION X NUMBER_OF_INVOCATIONS) in your profiler?
I am reading a text file which contains numbers in the range [1, 10^100]. I am then performing a sequence of arithmetic operations on each number. I would like to use a BigInteger only if the number is out of the int/long range. One approach would be to count how many digits there are in the string and switch to BigInteger if there are too many. Otherwise I'd just use primitive arithmetic as it is faster. Is there a better way?
Is there any reason why Java could not do this automatically i.e. switch to BigInteger if an int was too small? This way we would not have to worry about overflows.
I suspect the decision to use primitive values for integers and reals (done for performance reasons) made that option not possible. Note that Python and Ruby both do what you ask.
In this case it may be more work to handle the smaller special case than it is worth (you need some custom class to handle the two cases), and you should just use BigInteger.
Is there any reason why Java could not do this automatically i.e. switch to BigInteger if an int was too small?
Because that is a higher level programming behavior than what Java currently is. The language is not even aware of the BigInteger class and what it does (i.e. it's not in JLS). It's only aware of Integer (among other things) for boxing and unboxing purposes.
Speaking of boxing/unboxing, an int is a primitive type; BigInteger is a reference type. You can't have a variable that can hold values of both types.
You could read the values into BigIntegers, and then convert them to longs if they're small enough.
private final BigInteger LONG_MAX = BigInteger.valueOf(Long.MAX_VALUE);
private static List<BigInteger> readAndProcess(BufferedReader rd) throws IOException {
List<BigInteger> result = new ArrayList<BigInteger>();
for (String line; (line = rd.readLine()) != null; ) {
BigInteger bignum = new BigInteger(line);
if (bignum.compareTo(LONG_MAX) > 0) // doesn't fit in a long
result.add(bignumCalculation(bignum));
else result.add(BigInteger.valueOf(primitiveCalculation(bignum.longValue())));
}
return result;
}
private BigInteger bignumCalculation(BigInteger value) {
// perform the calculation
}
private long primitiveCalculation(long value) {
// perform the calculation
}
(You could make the return value a List<Number> and have it a mixed collection of BigInteger and Long objects, but that wouldn't look very nice and wouldn't improve performance by a lot.)
The performance may be better if a large amount of the numbers in the file are small enough to fit in a long (depending on the complexity of calculation). There's still risk for overflow depending on what you do in primitiveCalculation, and you've now repeated the code, (at least) doubling the bug potential, so you'll have to decide if the performance gain really is worth it.
If your code is anything like my example, though, you'd probably have more to gain by parallelizing the code so the calculations and the I/O aren't performed on the same thread - you'd have to do some pretty heavy calculations for an architecture like that to be CPU-bound.
The impact of using BigDecimals when something smaller will suffice is surprisingly, err, big: Running the following code
public static class MyLong {
private long l;
public MyLong(long l) { this.l = l; }
public void add(MyLong l2) { l += l2.l; }
}
public static void main(String[] args) throws Exception {
// generate lots of random numbers
long ls[] = new long[100000];
BigDecimal bds[] = new BigDecimal[100000];
MyLong mls[] = new MyLong[100000];
Random r = new Random();
for (int i=0; i<ls.length; i++) {
long n = r.nextLong();
ls[i] = n;
bds[i] = new BigDecimal(n);
mls[i] = new MyLong(n);
}
// time with longs & Bigints
long t0 = System.currentTimeMillis();
for (int j=0; j<1000; j++) for (int i=0; i<ls.length-1; i++) {
ls[i] += ls[i+1];
}
long t1 = Math.max(t0 + 1, System.currentTimeMillis());
for (int j=0; j<1000; j++) for (int i=0; i<ls.length-1; i++) {
bds[i].add(bds[i+1]);
}
long t2 = System.currentTimeMillis();
for (int j=0; j<1000; j++) for (int i=0; i<ls.length-1; i++) {
mls[i].add(mls[i+1]);
}
long t3 = System.currentTimeMillis();
// compare times
t3 -= t2;
t2 -= t1;
t1 -= t0;
DecimalFormat df = new DecimalFormat("0.00");
System.err.println("long: " + t1 + "ms, bigd: " + t2 + "ms, x"
+ df.format(t2*1.0/t1) + " more, mylong: " + t3 + "ms, x"
+ df.format(t3*1.0/t1) + " more");
}
produces, on my system, this output:
long: 375ms, bigd: 6296ms, x16.79 more, mylong: 516ms, x1.38 more
The MyLong class is there only to look at the effects of boxing, to compare against what you would get with a custom BigOrLong class.
Java is Fast--really really Fast. It's only 2-4x slower than c and sometimes as fast or a tad faster where most other languages are 10x (python) to 100x (ruby) slower than C/Java. (Fortran is also hella-fast, by the way)
Part of this is because it doesn't do things like switch number types for you. It could, but currently it can inline an operation like "a*5" in just a few bytes, imagine the hoops it would have to go through if a was an object. It would at least be a dynamic call to a's multiply method which would be a few hundred / thousand times slower than it was when a was simply an integer value.
Java probably could, these days, actually use JIT compiling to optimize the call better and inline it at runtime, but even then very few library calls support BigInteger/BigDecimal so there would be a LOT of native support, it would be a completely new language.
Also imagine how switching from int to BigInteger instead of long would make debugging video games crazy-hard! (Yeah, every time we move to the right side of the screen the game slows down by 50x, the code is all the same! How is this possible?!??)
Would it have been possible? Yes. But there are many problems with it.
Consider, for instance, that Java stores references to BigInteger, which is actually allocated on the heap, but store int literals. The difference can be made clear in C:
int i;
BigInt* bi;
Now, to automatically go from a literal to a reference, one would necessarily have to annotate the literal somehow. For instance, if the highest bit of the int was set, then the other bits could be used as a table lookup of some sort to retrieve the proper reference. That also means you'd get a BigInt** bi whenever it overflowed into that.
Of course, that's the bit usually used for sign, and hardware instructions pretty much depend on it. Worse still, if we do that, then the hardware won't be able to detect overflow and set the flags to indicate it. As a result, each operation would have to be accompanied by some test to see if and overflow has happened or will happen (depending on when it can be detected).
All that would add a lot of overhead to basic integer arithmetic, which would in practice negate any benefits you had to begin with. In other words, it is faster to assume BigInt than it is to try to use int and detect overflow conditions while at the same time juggling with the reference/literal problem.
So, to get any real advantage, one would have to use more space to represent ints. So instead of storing 32 bits in the stack, in the objects, or anywhere else we use them, we store 64 bits, for example, and use the additional 32 bits to control whether we want a reference or a literal. That could work, but there's an obvious problem with it -- space usage. :-) We might see more of it with 64 bits hardware, though.
Now, you might ask why not just 40 bits (32 bits + 1 byte) instead of 64? Basically, on modern hardware it is preferable to store stuff in 32 bits increments for performance reasons, so we'll be padding 40 bits to 64 bits anyway.
EDIT
Let's consider how one could go about doing this in C#. Now, I have no programming experience with C#, so I can't write the code to do it, but I expect I can give an overview.
The idea is to create a struct for it. It should look roughly like this:
public struct MixedInt
{
private int i;
private System.Numeric.BigInteger bi;
public MixedInt(string s)
{
bi = BigInteger.Parse(s);
if (parsed <= int.MaxValue && parsed => int.MinValue)
{
i = (int32) parsed;
bi = 0;
}
}
// Define all required operations
}
So, if the number is in the integer range we use int, otherwise we use BigInteger. The operations have to ensure transition from one to another as required/possible. From the client point of view, this is transparent. It's just one type MixedInt, and the class takes care of using whatever fits better.
Note, however, that this kind of optimization may well be part of C#'s BigInteger already, given it's implementation as a struct.
If Java had something like C#'s struct, we could do something like this in Java as well.
Is there any reason why Java could not
do this automatically i.e. switch to
BigInteger if an int was too small?
This is one of the advantage of dynamic typing, but Java is statically typed and prevents this.
In a dynamically type language when two Integer which are summed together would produce an overflow, the system is free to return, say, a Long. Because dynamically typed language rely on duck typing, it's fine. The same can not happen in a statically typed language; it would break the type system.
EDIT
Given that my answer and comment was not clear, here I try to provide more details why I think that static typing is the main issue:
1) the very fact that we speak of primitive type is a static typing issue; we wouldn't care in a dynamically type language.
2) with primitive types, the result of the overflow can not be converted to another type than an int because it would not be correct w.r.t static typing
int i = Integer.MAX_VALUE + 1; // -2147483648
3) with reference types, it's the same except that we have autoboxing. Still, the addition could not return, say, a BigInteger because it would not match the static type sytem (A BigInteger can not be casted to Integer).
Integer j = new Integer( Integer.MAX_VALUE ) + 1; // -2147483648
4) what could be done is to subclass, say, Number and implement at type UnboundedNumeric that optimizes the representation internally (representation independence).
UnboundedNum k = new UnboundedNum( Integer.MAX_VALUE ).add( 1 ); // 2147483648
Still, it's not really the answer to the original question.
5) with dynamic typing, something like
var d = new Integer( Integer.MAX_VALUE ) + 1; // 2147483648
would return a Long which is ok.