Brute force Approach would be
int checkForTriangle(int a, int b, int c) {
return (a + b > c) && (b + c > a) && (a + c > b);
}
int triangle (int a[], int n) {
if (a == null || n <= 0) return 0;
for (int i=0 ; i < n ; i++) {
for (int j=i + 1; j < n; j++ ) {
for (int k=j + 1; k < n; k++ ) {
if (checkForTriangle(a[i], a[j], a[k])) return 1;
}
}
}
return 0;
}
Thanks,
Immanuel
On Wed, May 25, 2011 at 11:23 PM, Piyush Sinha <ecstasy.piy...@gmail.com>wrote:
> Write a function
>
> int triangle( int A [ ] )
>
> that given a zero-indexed array A consisting of N integers returns 1
> if there exists a triple (P, Q, R) such that 0 <= P < Q < R < N and
> A[P] + A[Q] > A[R],
> A[Q] + A[R] > A[P],
> A[R] + A[P] > A[Q].
> The function should return 0 if such triple does not exist.
>
> For example, given array A such that
>
> A[0]=10, A[1]=2, A[2]=5, A[3]=1, A[4]=8, A[5]=20
>
> the function should return 1, because the triple (0, 2, 4) fulfills
> all of the required conditions.
>
> For array A such that
>
> A[0]=10, A[1]=50, A[2]=5, A[3]=1
>
> the function should return 0
>
> --
> *Piyush Sinha*
> *IIIT, Allahabad*
> *+91-8792136657*
> *+91-7483122727*
> *https://www.facebook.com/profile.php?id=100000655377926 *
>
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