Given three sides of a triangle. Write the Program to
determine whether the triangle is :
1) Invalid
2) Right Angled
3) Isoscales
4) Equilateral
5) Not Special
An Isoscales right angled triangle should be taken as a
Right Angled Triangle
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Answer / sunil
if a, b and c are the three sides of a triangle, then a + b > c
if this is not satisfied, then its not a valid triangle.
To check for right angle, use Pythagoras theorem. Assume
that the longest side is the hypotenuse.
Issosless and Equilateral can be found by simply comparing
the sides.
| Is This Answer Correct ? | 65 Yes | 31 No |
Answer / giri
One should check for valid sides also. Side values dhould
be greated than ZERO.
Here is the correct routine:
public static String checkTriangle(int[]
triangleSide){
boolean validTriangle = false;
boolean validSides = true;
String result = "NOT VALID TRIANGLE";
for(int side: triangleSide)
if(side <= 0)
validSides = false;
if(validSides){
for(int count= 0; count< 3 ;
count++){
if(((triangleSide[count%3]
+ triangleSide[(count+1)%3]) > triangleSide[(count+2)%3]))
validTriangle =
true;
}
if(validTriangle){
if( triangleSide[0] ==
triangleSide[1] && triangleSide[2] == triangleSide[1])
result
= "EQUILATERAL";
else{
for(int count= 0;
count< 3 ; count++){
if( (
triangleSide[count%3] * triangleSide[count%3] + triangleSide
[(count+1)%3] * triangleSide[(count+1)%3]) == (triangleSide
[(count+2)%3] * triangleSide[(count+2)%3])){
result = "RIGHANGLED";
break;
}else
if
((triangleSide[count%3] == triangleSide[(count+1)%3]))
result = "ISOSCALAUS";
}
}
if("NOT VALID
TRIANGLE".equals(result))
result = "NOT
SPECIAL";
}
}
System.out.println(result);
return result;
}
| Is This Answer Correct ? | 32 Yes | 29 No |
Answer / ganesh bhat
I believe, this is more efficient one. solves all the cases.. comments please. Written in java
public static Map validateTriange(int a,int b,int c)
{
Map props = new HashMap();
boolean isValidTriangle;
int bigSide = a;
if(bigSide<b){bigSide = b;}
if(bigSide<c){bigSide = c;}
boolean isSpecial = false;
if((a+b+c-bigSide)>bigSide)
{
props.put("VALID","YES");
}
else
{
props.put("VALID","NO");
return props;
}
if(a == b||b==c||c == a)
{
props.put("ISOSCELES","YES");
isSpecial = true;
}
if(a == b && b == c)
{
props.put("EQUALATERAL","YES");
isSpecial = true;
}
if(((a*a+b*b+c*c)-bigSide*bigSide) == bigSide*bigSide)
{
props.put("RIGHT_ANGLED","YES");
isSpecial = true;
}
return props;
}
| Is This Answer Correct ? | 22 Yes | 22 No |
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In Bioinformatics, a DNA sequence is made up of a combination of 4 characters, namely “A,C,G,T”. A subsequence of a given sequence of characters a0, a1, …an- 1, is any subset of the characters taken in order, of the form ai0 , ai1 ,…..aik-1 where 0 ≤ i0 <i1….< ik-1 ≤ n-1. For example in the sequence “A,C,G,T,G,T,C,A,A,A,A,T,C,G”, we can have subsequences “A,G,T”, “A,C,A,A” and many more. A subsequence is palindromic if it is the same whether read left to right or right to left. For instance, the sequence “A,C,G,T,G,T,C,A,A,A,A,T,C,G”, has many palindromic subsequences, including “A,C,G,C,A” and “A,A,A,A” (on the other hand, the subsequence “A,C,T” is not palindromic). Devise an algorithm (using dynamic programming) that takes a sequence of characters X[0 … n-1] from the alphabet set (A,C,G,T) and returns the (length of the) longest palindromic subsequence. Implement the algorithm in an appropriate language.
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