How to get rid of the arithmetic overflow

Question

This is the code for the Triangle problem in codility that is throwing me an arithmetic overflow error.

int solution(vector<int> &A) {

    int i, n;
    n=A.size();
    sort(A.begin(), A.end());
    for(i=0; i<n-2; i++)
    {
        if((A[i]+A[i+1]>A[i+2])&&(A[i]+A[i+2]>A[i+1])&&(A[i+1]+A[i+2]>A[i]))
        {
            return 1;
        }
    }
    return 0;
}

It passes all the tests except for the 'extreme_arith_overflow1 overflow test, 3 MAXINTs' saying the code returns 0 but it expects 1. Anybody have any idea on how to fix this?

First of all make sure that you don't go out of bounds of your vector. (which you will do with the code you currently show) Secondly perhaps use `unsigned int` instead, (if no value can be negative) or maybe `long long` (or `unsigned long long`) for a 64-bit type? Also, you should probaby be doing some input validation to make sure the input you read is valid to begin with. — Some programmer dude, Jun 03 '20 at 09:37
@Someprogrammerdude yesss thanks i fixed the out of bounds issue and validated input :) — monkikat, Jun 03 '20 at 12:43

Thomas Sablik · Accepted Answer · 2020-06-03T13:27:07.297

0

You store A.size() in n and then you loop until i<n and access A[i+2]. In the error cases this is A[A.size()] or even A[A.size()+1]. It's out of bounds. Fix the range of the loop.

The next problem occurs when the sum is larger than INT_MAX. Use the difference instead of the sum to avoid overflow. Remember that the elements are sorted with A[i] <= A[i+1] <= A[i+2]

int solution(vector<int> &A) {
    if (A.size() < 3) return 0;
    const auto n = A.size() - 2;
    std::sort(A.begin(), A.end());
    for(decltype(n) i = 0; i < n; ++i) {
        if((A[i]>A[i+2]-A[i+1])&&(A[i+2]>A[i+1]-A[i])&&A[i]>0) {
            return 1;
        }
    }
    return 0;
}

edited Jun 03 '20 at 13:27

answered Jun 03 '20 at 09:45

Thomas Sablik

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Hey thanks for your answer! Fixed the out of bounds issue but i still seem to be getting an arithmetic overflow error. – monkikat Jun 03 '20 at 09:51
Now prone to underflow. :-) Consider `INT_MIN` - 1. – jvd Jun 03 '20 at 11:59
@jvd Where do you see underflow? `A` is sorted. `A[i+2]-A[i+1]` and `A[i+1]-A[i]` return positive values. – Thomas Sablik Jun 03 '20 at 13:12
@ThomasSablik You are correct. I forgot that `A` was sorted. – jvd Jun 03 '20 at 13:14
The correct way to write the loop is 'for (size_t i = 0; i < A.size()-2; ++i)' – Uri Raz Jun 03 '20 at 13:14
1

@UriRaz `int i = 0; i < A.size()-2` is an implicit check that `A.size()` is at least 3 before the loop. Of course you can add redundant checks. – Thomas Sablik Jun 03 '20 at 13:16
@Thomas Sablik I tried to compile that on Visual Studio, and I get a warning - "warning C4018: '<': signed/unsigned mismatch". Also, vector's size() returns size_t. If sizeof(int) <= sizeof(size_t), this might cause an infinite loop. – Uri Raz Jun 03 '20 at 13:24
@UriRaz Fixed it. – Thomas Sablik Jun 03 '20 at 13:27
If numbers can be negative, you still have overflows for the data `{ -2, -1, INT_MAX }`. – j6t Jun 03 '20 at 13:34
@j6t What would be a triangle with negative side length? Of course, you are right but this code solves the problem. – Thomas Sablik Jun 03 '20 at 13:37

How to get rid of the arithmetic overflow

1 Answers1