Maximum Perimeter Triangle


Problem Statement :


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 .


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.


Solution :



title-img 


                            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])


View More Similar Problems

Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

View Solution →

Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

View Solution →

Insert a Node at the head of a Linked List

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

View Solution →

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →