Maximum Perimeter Triangle Hacker Rank Solution Best & Easiest

Posted by admin October 5, 2022
Maximum Perimeter Triangle Problem Solution
Maximum Perimeter Triangle Problem Solution

In this post, we will solve the Maximum Perimeter Triangle HackerRank Solution. This problem (Maximum Perimeter Triangle) is a part of the HackerRank Problem Solving series.

Problem: https://www.hackerrank.com/challenges/three-month-preparation-kit-maximum-perimeter-triangle/problem

Maximum Perimeter Triangle Hacker Rank Solution

Problem solution in Python programming:

def maximumPerimeterTriangle(sticks):
    arr = sorted(sticks)
    res = [-1]
    for i in range(len(arr)-2):
        cut = arr[i:i+3]
        if cut[0] + cut[1] > cut[2]:
            res = cut
    return res

Problem solution in C++ programming:

vector<int> maximumPerimeterTriangle(vector<int> sticks) {
    vector<int> maxtriangle;
    sort(sticks.begin(),sticks.end(),greater<int>());
    {
        for(int i=0;i<=sticks.size()-3;i++)
        {
            if(sticks[i+1]+sticks[i+2]>sticks[i])
            {
                maxtriangle.push_back(sticks[i+2]);
                maxtriangle.push_back(sticks[i+1]);
                maxtriangle.push_back(sticks[i]);
                return maxtriangle;
            }
        }
        maxtriangle.push_back(-1);
    }
    return maxtriangle;
}

Problem solution in JavaScript programming:

function maximumPerimeterTriangle(sticks) { 
     let triangleVal=[];
let sorted = sticks.sort((a,b)=> a-b);

let final = []; 

for(var i = 0; i < sticks.length; i++){
    let a = sticks[i];
    let b = sticks[i+1];
    let c = sticks[i+2];


    if(a + b > c && a + c > b && b + c > a){
        final.push([a,b,c])
    }

}

return final.length > 0 ? final[final.length-1] : [-1]
}

Problem solution in Swift programming:

func maximumPerimeterTriangle(sticks: [Int]) -> [Int] {
    let sorted: [Int] = sticks.sorted().reversed()
    print(sorted)
    for i in 0...sorted.count - 3{
        let L3 = sorted[i]
        let L2 = sorted[i + 1]
        let L1 = sorted[i + 2]
        if L2 + L1 > L3 {
            return [L1,L2,L3]
        }
    }
    return [-1]
}