In this Maximum Perimeter Triangle HackerRank Problem, Given an array of stick lengths, use of them to construct a non-degenerate triangle with the maximum possible perimeter. Return an array of the lengths of its sides as integers in non-decreasing order.
If there are several valid triangles having the maximum perimeter:
- Choose the one with the longest maximum side.
- If more than one has that maximum, choose from them the one with the longest minimum side.
- If more than one has that maximum as well, print any one them.
If no non-degenerate triangle exists, return .
Example
The triplet will not form a triangle. Neither will or , so the problem is reduced to and . The longer perimeter is .
Function Description
Complete the maximumPerimeterTriangle function in the editor below.
maximumPerimeterTriangle has the following parameter(s):
- int sticks[n]: the lengths of sticks available
Returns
- int[3] or int[1]: the side lengths of the chosen triangle in non-decreasing order or -1
Input Format
The first line contains single integer , the size of array .
The second line contains space-separated integers , each a stick length.
Constraints
Explanation
Sample Case 0:
There are possible unique triangles:
The second triangle has the largest perimeter, so we print its side lengths on a new line in non-decreasing order.
Sample Case 1:
The triangle is degenerate and thus can’t be constructed, so we print -1 on a new line.
Maximum Perimeter Triangle HackerRank Problem Solutions
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in JavaScript
function isNonDegenerate(a, b, c) {
if (a + b > c && b + c > a && a + c > b) {
return true;
} else {
return false;
}
}
function maximumPerimeterTriangle(sticks) {
sticks = sticks.sort((a, b) => a - b);
let triangles = [];
for (let i = 0; i < sticks.length - 2; i++) {
if (isNonDegenerate(sticks[i], sticks[i + 1], sticks[i+2])) {
triangles.push([sticks[i], sticks[i + 1], sticks[i+2]])
}
}
if(triangles.length === 0){
return [-1];
}else{
return triangles.pop();
}
}Problem Solution in C#
public static List<int> maximumPerimeterTriangle(List<int> sticks)
{
sticks.Sort();
for (int i = sticks.Count - 3; i >= 0; i--)
{
if (sticks[i] + sticks[i + 1] > sticks[i + 2])
{
return new List<int> { sticks[i], sticks[i + 1], sticks[i + 2] };
}
}
return new List<int> { -1 };
}Maximum Perimeter Triangle Problem Solution in C++
vector<int> maximumPerimeterTriangle(vector<int> sticks) {
int len = sticks.size();
vector<int>triangle;
unsigned long long int p = 0,s1,s2,s3;
unsigned long long int result = 0;
sort(sticks.begin(),sticks.end());
for (int i = 0; i < len; i++) {
for (int j = i+1; j < len; j++) {
for (int k = j+1; k < len; k++) {
if (sticks[i] + sticks[j] > sticks[k] && sticks[i] + sticks[k] > sticks[j] && sticks[j] + sticks[k] > sticks[i]) {
p = sticks[i] + sticks[j] + sticks[k];
if (result < p) {
result = p;
s1 = i;
s2 = j;
s3 = k;
}
}
}
}
}
if (result != 0) {
triangle.push_back(sticks[s1]);
triangle.push_back(sticks[s2]);
triangle.push_back(sticks[s3]);
}else {
triangle.push_back(-1);
}
return triangle;
}Problem Solution in Python
def maximumPerimeterTriangle(sticks):
sticks.sort(reverse = True)
for i in range(len(sticks)-2):
a = sticks[i]
b = sticks[i+1]
c = sticks[i+2]
if a < b + c:
return (c, b, a)
return [-1]