Maximum Perimeter Triangle

                            Solution in C :

In  C  :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int comp(void *a,void *b)
    {
    return (*(int *)a-*(int *)b);
}


int main() {

    int n;
    scanf("%d",&n);
    int i,a[n];
    for(i=0;i<n;i++)
        scanf("%d",&a[i]);
    qsort((void *)a,n,sizeof(a[0]),comp);
    int x,y,z;
    for(i=n-1;i>=2;i--)
        {
        if(a[i-2]+a[i-1]>a[i])
            break;
    }
    if(i==1)
        printf("-1");
    else
        printf("%d %d %d",a[i-2],a[i-1],a[i]);
    return 0;
}
                        

                        Solution in C++ :

In  C++ :







/*
*/


#define _CRT_SECURE_NO_WARNINGS

#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <memory.h>
#include <assert.h>

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash
#define norm asdfasdgasdgsd

#define eps 1e-9
#define M_PI 3.141592653589793
#define bs 1000000007
#define bsize 256

using namespace std;

const int INF = 1e9;
const int N = 500031;

int n;
vector<int> v;
int ar[100];
vector<vector<int> > res;

int main(){
	//freopen("fabro.in","r",stdin);
	//freopen("fabro.out","w",stdout);
	//freopen("F:/in.txt", "r", stdin);
	//freopen("F:/output.txt", "w", stdout);
	ios_base::sync_with_stdio(0);
	//cin.tie(0);
	
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> ar[i];
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			for (int q = 0; q < n; q++)
			{
				if (i == j || i == q || j == q)
					continue;
				if (ar[i] + ar[j] <= ar[q] || ar[i] + ar[q] <= ar[j] || ar[j] + ar[q] <= ar[i])
					continue;
				vector<int> v;
				v.push_back(ar[i] + ar[j] + ar[q]);
				v.push_back(ar[i]);
				v.push_back(ar[j]);
				v.push_back(ar[q]);
				res.push_back(v);
			}
		}
	}
	sort(res.begin(), res.end());
	if (res.size() == 0)
	{
		cout << -1 << endl;
		return 0;
	}
	vector<int> v = res.back();
	cout << v[3] << " " << v[2] << " " << v[1] << endl;

	cin.get(); cin.get();
	return 0;
}
                    

                        Solution in Java :

In  Java :






import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        int n = sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        Arrays.sort(a);
        
        int ans = -1;
        for (int i = 2; i < n; i++) {
            if (a[i-2]+a[i-1]>a[i])
                ans = i;
        }
        if (ans == -1)
            System.out.println(-1);
        else
            System.out.println(a[ans-2]+" "+a[ans-1]+" "+a[ans]);
    }
}
                    

                        Solution in Python : 
                            
In  Python3  :






from itertools import *
n = int(input())
l = sorted(map(int, input().split()))
ans = (-1, -1, -1)
for i, j, k in product(*repeat(range(n), 3)):
    if i < j < k and l[i] + l[j] > l[k]:
        ans = max(ans, (l[k], l[i], l[j]))
if ans[0] == -1:
    print(-1)
else:    
    print(ans[1], ans[2], ans[0])