HackerRank Maximum Perimeter Triangle problem solution

In this HackerRank Maximum Perimeter Triangle problem solution we have given an array of stick lengths, use 3 of them to construct a non-degenerate triangle with the maximum possible perimeter. Return an array of the lengths of its sides as 3 integers in non-decreasing order.

Problem solution in Python.

n = int(input())
a = list(map(int, input().split()))
a = sorted(a)
for i in range(0,n-2):
    x = a[n-1-i]
    y = a[n-2-i]
    z = a[n-3-i]
    if y + z > x:
        print(str(z) + " "+ str(y) + " " + str(x))
        exit()
print(-1)

Problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] sticks = new int[n];
        for(int i=0; i < n; i++){
            sticks[i] = in.nextInt();
        }
        Arrays.sort(sticks);
        int trianglePosition = n-3;
        while ((trianglePosition>=0) && (sticks[trianglePosition] + sticks[trianglePosition+1] <= sticks[trianglePosition+2])) {
            trianglePosition--;
        }
        if (trianglePosition < 0) {
            System.out.println(-1);
        } else {
            System.out.println(sticks[trianglePosition] + " " + sticks[trianglePosition + 1] + " " + sticks[trianglePosition + 2]);
        }
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
    }
}

Problem solution in C++.

#include<bits/stdc++.h>

using namespace std;
#define foreach(i,x) for(type(x)i=x.begin();i!=x.end();i++)
#define FOR(ii,aa,bb) for(int ii=aa;ii<=bb;ii++)
#define FORE(ii,aa,bb) for(int ii=aa;ii<bb;ii++)
#define ROF(ii,aa,bb) for(int ii=aa;ii>=bb;ii--)
#define type(x) __typeof(x.begin())

#define bit(x,y) ((x>>y)&1)
#define y1 fkfrgff
#define ll long long
#define pii pair<int,int>
#define mod 1000000007
#define N (int)(2e5+10)
#define mp make_pair
#define pb push_back
#define sd second
#define ft first
#define endll puts("")
#define endl 'n'
#define inf mod
#define ort ((sol+sag)/2)
int n,a[N];
vector<pair<pii,pii > >ans;
int main(){
    cin >> n;
    FOR(i,1,n) scanf("%d",&a[i]);
    FOR(i,1,n)
        FOR(j,i+1,n)
            FOR(k,j+1,n){
                vector<int>v;
                v.pb(a[i]);
                v.pb(a[j]);
                v.pb(a[k]);
                sort(v.rbegin(),v.rend());
                if(v[0] < v[1] + v[2])
                    ans.pb(mp(mp(v[0]+v[1]+v[2],v[0]),mp(v[2],v[1])));

            }
    if(!ans.size()) return puts("-1"),0;
    sort(ans.rbegin(),ans.rend());
    printf("%d %d %d",ans[0].sd.ft , ans[0].sd.sd , ans[0].ft.sd);
}

Problem solution in C.

#include <stdio.h>
void sort(int *a, int n)
{
	
	int *i,*j,*max_i, temp;
	for (i=a;i<a+n;i++)
	{
		max_i=i;
		for(j=i+1;j<a+n;j++)
		{	
			if(*max_i<*j)
			max_i=j;
	    }
		temp=*i;
		*i=*max_i;
		*max_i=temp;
	}
}
int main()
{
	int i,n;
	scanf("%d",&n);
	int array[n];
	for(i=0;i<n;i++)
	{
		scanf("%d",&array[i]);
	}
	
	sort(array,n);
	int c=0;
	for(i=0;i<n-2;i++)
	{
		if(array[i]<array[i+1]+array[i+2])
		{
		printf("%d %d %d",array[i+2],array[i+1],array[i]);
		c++;
		break;
	    }
	}
	if(c==0)
	printf("%d",-1);
	return 0;
	
}