Maximum Perimeter Triangle from array

Given an Array of non-negative integers. Find out three elements from this array which form a triangle of maximum perimeter.

Examples :

Input : {6, 1, 6, 5, 8, 4}
Output : 20

Input : {2, 20, 7, 55, 1, 33, 12, 4}
Output : Triangle formation is not possible.

Input: {33, 6, 20, 1, 8, 12, 5, 55, 4, 9}
Output: 41

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Solution:
The brute force solution is: check for all combination of 3 elements whether it forms a triangle or not and update the maximum perimeter if it forms a triangle. Complexity of naive solution is O(n³). Below is the code for it.

C++

// Brute force solution to find 
// out maximum perimeter triangle which 
// can be formed using the elements 
// of the given array 
#include <iostream> 
#include <algorithm> 
  
using namespace std; 
  
// Function to find out maximum perimeter 
void maxPerimeter(int arr[], int n){ 
  
    // initialize maximum perimeter 
    // as 0. 
    int maxi = 0; 
  
    // pick up 3 different elements 
    // from the array. 
    for (int i = 0; i < n - 2; i++){ 
        for (int j = i + 1; j < n - 1; j++){ 
            for (int k = j + 1; k < n; k++){ 
  
                // a, b, c are 3 sides of the triangle 
                int a = arr[i]; 
                int b = arr[j]; 
                int c = arr[k]; 
  
                // check whether a, b, c forms 
                // a triangle or not. 
                if (a < b+c && b < c+a && c < a+b){ 
  
                    // if it forms a triangle 
                    // then update the maximum value. 
                    maxi = max(maxi, a+b+c); 
                } 
            } 
        } 
    } 
  
    // If maximum perimeter is non-zero 
    // then print it. 
    if (maxi) cout << "Maximum Perimeter is: " 
                   << maxi << endl; 
  
    // otherwise no triangle formation 
    // is possible. 
    else cout << "Triangle formation "
        << "is not possible." << endl; 
} 
  
// Driver Program 
int main() 
{ 
    // test case 1 
    int arr1[6] = {6, 1, 6, 5, 8, 4}; 
    maxPerimeter(arr1, 6); 
  
    // test case 2 
    int arr2[8] = {2, 20, 7, 55, 1, 
                    33, 12, 4}; 
    maxPerimeter(arr2, 8); 
  
    // test case 3 
    int arr3[10] = {33, 6, 20, 1, 8, 
                    12, 5, 55, 4, 9}; 
    maxPerimeter(arr3, 10); 
  
    return 0; 
} 

Java

// Brute force solution to find out maximum  
// perimeter triangle which can be formed  
// using the elements of the given array 
import java.io.*; 
  
class GFG { 
  
    // Function to find out maximum perimeter 
    static void maxPerimeter(int arr[], int n) 
    { 
      
        // initialize maximum perimeter as 0. 
        int maxi = 0; 
      
        // pick up 3 different elements 
        // from the array. 
        for (int i = 0; i < n - 2; i++) 
        { 
            for (int j = i + 1; j < n - 1; j++) 
            { 
                for (int k = j + 1; k < n; k++)  
                { 
      
                    // a, b, c are 3 sides of 
                    // the triangle 
                    int a = arr[i]; 
                    int b = arr[j]; 
                    int c = arr[k]; 
      
                    // check whether a, b, c  
                    // forms a triangle or not. 
                    if (a < b+c && b < c+a && c < a+b) 
                    { 
      
                        // if it forms a triangle 
                        // then update the maximum  
                        // value. 
                        maxi = Math.max(maxi, a+b+c); 
                    } 
                } 
            } 
        } 
      
        // If maximum perimeter is non-zero 
        // then print it. 
        if (maxi > 0)  
        System.out.println( "Maximum Perimeter is: "
                                             + maxi); 
      
        // otherwise no triangle formation 
        // is possible. 
        else
        System.out.println( "Triangle formation "
                              + "is not possible." ); 
    } 
      
    // Driver Program 
    public static void main (String[] args) 
    { 
          
        // test case 1 
        int arr1[] = {6, 1, 6, 5, 8, 4}; 
        maxPerimeter(arr1, 6); 
      
        // test case 2 
        int arr2[] = {2, 20, 7, 55, 1, 33, 12, 4}; 
        maxPerimeter(arr2, 8); 
      
        // test case 3 
        int arr3[] = {33, 6, 20, 1, 8, 
                                12, 5, 55, 4, 9}; 
        maxPerimeter(arr3, 10); 
    } 
} 
  
// This code is contributed by anuj_67. 

Python

# Brute force solution to find 
# out maximum perimeter triangle  
# which can be formed using the  
# elements of the given array 
  
# Function to find out  
# maximum perimeter 
def maxPerimeter(arr): 
    maxi = 0
    n = len(arr) 
      
    # pick up 3 different  
    # elements from the array. 
    for i in range(n - 2): 
        for j in range(i + 1, n - 1): 
            for k in range(j + 1, n): 
                  
                # a, b, c are 3 sides  
                # of the triangle 
                a = arr[i] 
                b = arr[j] 
                c = arr[k] 
                if(a < b + c and b < a + c  
                             and c < a + b): 
                    maxi = max(maxi, a + b + c) 
  
    if(maxi == 0): 
        return "Triangle formation is not possible"
    else: 
        return "Maximum Perimeter is: "+ str(maxi) 
  
# Driver code 
def main(): 
    arr1 = [6, 1, 6, 5, 8, 4] 
    a = maxPerimeter(arr1) 
    print(a) 
  
    arr2 = [2, 20, 7, 55,  
            1, 33, 12, 4] 
    a = maxPerimeter(arr2) 
    print(a) 
  
    arr3 = [33, 6, 20, 1, 8,  
            12, 5, 55, 4, 9] 
    a = maxPerimeter(arr3) 
    print(a) 
  
if __name__=='__main__': 
    main() 
  
# This code is contributed  
# by Pritha Updhayay 

C#

// Brute force solution to find out  
// maximum perimeter triangle which  
// can be formed using the elements  
// of the given array 
using System; 
  
class GFG  
{ 
  
    // Function to find out  
    // maximum perimeter 
    static void maxPerimeter(int []arr,      
                             int n) 
    { 
      
        // initialize maximum  
        // perimeter as 0. 
        int maxi = 0; 
      
        // pick up 3 different elements 
        // from the array. 
        for (int i = 0; i < n - 2; i++) 
        { 
            for (int j = i + 1; j < n - 1; j++) 
            { 
                for (int k = j + 1; k < n; k++)  
                { 
      
                    // a, b, c are 3 sides of 
                    // the triangle 
                    int a = arr[i]; 
                    int b = arr[j]; 
                    int c = arr[k]; 
      
                    // check whether a, b, c  
                    // forms a triangle or not. 
                    if (a < b + c &&  
                        b < c + a &&  
                        c < a + b) 
                    { 
      
                        // if it forms a triangle 
                        // then update the maximum  
                        // value. 
                        maxi = Math.Max(maxi, a + b + c); 
                    } 
                } 
            } 
        } 
      
        // If maximum perimeter is  
        // non-zero then print it. 
        if (maxi > 0)  
        Console.WriteLine("Maximum Perimeter is: "+ maxi); 
      
        // otherwise no triangle  
        // formation is possible. 
        else
        Console.WriteLine("Triangle formation "+  
                          "is not possible."); 
    } 
      
    // Driver Code 
    public static void Main () 
    { 
          
        // test case 1 
        int []arr1 = {6, 1, 6,  
                      5, 8, 4}; 
        maxPerimeter(arr1, 6); 
      
        // test case 2 
        int []arr2 = {2, 20, 7, 55,  
                      1, 33, 12, 4}; 
        maxPerimeter(arr2, 8); 
      
        // test case 3 
        int []arr3 = {33, 6, 20, 1, 8, 
                      12, 5, 55, 4, 9}; 
        maxPerimeter(arr3, 10); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// Brute force solution to find 
// out maximum perimeter triangle which 
// can be formed using the elements 
// of the given array 
  
// Function to find out 
// maximum perimeter 
function maxPerimeter($arr, $n) 
{ 
  
    // initialize maximum  
    // perimeter as 0. 
    $maxi = 0; 
  
    // pick up 3 different  
    // elements from the array. 
    for ($i = 0; $i < $n - 2; $i++) 
    { 
        for ( $j = $i + 1; $j < $n - 1; $j++) 
        { 
            for ( $k = $j + 1; $k < $n; $k++) 
            { 
  
                // a, b, c are 3 sides  
                // of the triangle 
                $a = $arr[$i]; 
                $b = $arr[$j]; 
                $c = $arr[$k]; 
  
                // check whether a, b, c  
                // forms a triangle or not. 
                if ($a < $b + $c and 
                    $b < $c + $a and 
                    $c < $a + $b) 
                { 
  
                    // if it forms a triangle 
                    // then update the maximum value. 
                    $maxi = max($maxi, $a + $b + $c); 
                } 
            } 
        } 
    } 
  
    // If maximum perimeter is  
    // non-zero then print it. 
    if ($maxi)  
    { 
    echo "Maximum Perimeter is: "; 
    echo $maxi ,"\n"; 
    } 
  
    // otherwise no triangle  
    // formation is possible. 
    else
    { 
    echo "Triangle formation "; 
    echo "is not possible. \n"; 
    } 
} 
  
// Driver Code 
  
// test case 1 
$arr1 = array(6, 1, 6, 5, 8, 4); 
maxPerimeter($arr1, 6); 
  
// test case 2 
$arr2 = array(2, 20, 7, 55,  
              1, 33, 12, 4); 
maxPerimeter($arr2, 8); 
  
// test case 3 
$arr3 = array(33, 6, 20, 1, 8, 
              12, 5, 55, 4, 9); 
maxPerimeter($arr3, 10); 
  
// This code is contributed by anuj_67. 
?> 

Output :

Maximum Perimeter is: 20
Triangle formation is not possible.
Maximum Perimeter is: 41

Efficient Approach:
First we can sort the array in non-increasing order. So, the first element will be maximum and the last will be minimum. Now if the first 3 elements of this sorted array forms a triangle then it will be the maximum perimeter triangle, as for all other combination the sum of elements(i.e. the perimeter of that triangle) will be = b >= c). a, b,c can not form a triangle, so a >= b + c. As, b and c = c+d (if we drop b and take d) or a >= b+d (if we drop c and take d). So, we have to drop a and pick up d.
Again same set of analysis for b, c and d. We can continue this till last and whenever we find a triangle forming triple then we can stop checking, as this triple gives maximum perimeter.
Hence, if arr[i] < arr[i+1] + arr[i+2] (0 <= i <= n-3)in the sorted array then arr[i], arr[i+1] and arr[i+2] forms a triangle.
Below is the simple implementation of this concept:

C++

// Efficient solution to find 
// out maximum perimeter triangle which 
// can be formed using the elements 
// of the given array 
#include <iostream> 
#include <algorithm> 
  
using namespace std; 
  
// Function to find out maximum perimeter 
void maxPerimeter(int arr[], int n){ 
  
    // sort the array elements 
    // in reversed order 
    sort(arr, arr+n, greater<int>()); 
  
    // initialize maximum 
    // perimeter to 0 
    int maxi = 0; 
  
    // loop through the sorted array 
    // and check whether it forms a 
    // triangle or not. 
    for (int i = 0; i < n-2; i++){ 
  
        // Check whether arr[i], arr[i+1] 
        // and arr[i+2] forms a triangle 
        // or not. 
        if (arr[i] < arr[i+1] + arr[i+2]){ 
  
            // if it forms a triangle then 
            // it is the triangle with 
            // maximum perimeter. 
            maxi = max(maxi, arr[i] + arr[i+1] + arr[i+2]); 
            break; 
        } 
    } 
  
    // If maximum perimeter is non-zero 
    // then print it. 
    if (maxi) 
        cout << "Maximum Perimeter is: "
        << maxi << endl; 
  
    // otherwise no triangle formation 
    // is possible. 
    else
        cout << "Triangle formation"
        << "is not possible." << endl; 
} 
  
// Driver Program 
int main() 
{ 
    // test case 1 
    int arr1[6] = {6, 1, 6, 5, 8, 4}; 
    maxPerimeter(arr1, 6); 
  
    // test case 2 
    int arr2[8] = {2, 20, 7, 55, 1, 
                    33, 12, 4}; 
    maxPerimeter(arr2, 8); 
  
    // test case 3 
    int arr3[10] = {33, 6, 20, 1, 8, 
                    12, 5, 55, 4, 9}; 
    maxPerimeter(arr3, 10); 
  
    return 0; 
} 

Java

// Efficient solution to find  
// out maximum perimeter triangle which  
// can be formed using the elements  
// of the given array  
  
import java.util.Arrays; 
  
class GFG { 
  
// Function to find out maximum perimeter  
    static void maxPerimeter(int arr[], int n) { 
  
        // sort the array elements  
        // in reversed order  
        arr = arrRevSort(arr); 
        //sort(arr, arr+n, greater<int>());  
  
        // initialize maximum  
        // perimeter to 0  
        int maxi = 0; 
  
        // loop through the sorted array  
        // and check whether it forms a  
        // triangle or not.  
        for (int i = 0; i < n - 2; i++) { 
  
            // Check whether arr[i], arr[i+1]  
            // and arr[i+2] forms a triangle  
            // or not.  
            if (arr[i] < arr[i + 1] + arr[i + 2]) { 
  
                // if it forms a triangle then  
                // it is the triangle with  
                // maximum perimeter.  
                maxi = Math.max(maxi, arr[i] + arr[i + 1] + arr[i + 2]); 
                break; 
            } 
        } 
  
        // If maximum perimeter is non-zero  
        // then print it.  
        if (maxi > 0) { 
            System.out.println("Maximum Perimeter is: " + maxi); 
        } // otherwise no triangle formation  
        // is possible.  
        else { 
            System.out.println("Triangle formation is not possible."); 
        } 
    } 
    //Function return sorted array in Decreasing  
  
    static int[] arrRevSort(int[] arr) { 
        Arrays.sort(arr, 0, arr.length); 
        int j = arr.length - 1; 
        for (int i = 0; i < arr.length / 2; i++, j--) { 
            int temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        return arr; 
    } 
  
// Driver Program  
    public static void main(String[] args) { 
        // test case 1  
        int arr1[] = {6, 1, 6, 5, 8, 4}; 
        maxPerimeter(arr1, 6); 
  
        // test case 2  
        int arr2[] = {2, 20, 7, 55, 1, 33, 12, 4}; 
        maxPerimeter(arr2, 8); 
  
        // test case 3  
        int arr3[] = {33, 6, 20, 1, 8, 12, 5, 55, 4, 9}; 
        maxPerimeter(arr3, 10); 
    } 
} 
/*This Java code is contributed by 29AjayKumar*/

Python3

# Efficient solution to find 
# out maximum perimeter triangle which 
# can be formed using the elements 
# of the given array 
  
# Function to find the  
# maximum perimeter 
def maxPerimeter(arr): 
    maxi = 0
    n = len(arr) 
    arr.sort(reverse = True) 
  
    for i in range(0, n - 2): 
        if arr[i] < (arr[i + 1] + arr[i + 2]): 
            maxi = max(maxi, arr[i] + 
                       arr[i + 1] + arr[i + 2]) 
            break
  
    if(maxi == 0): 
        return "Triangle formation is not possible"
    else: 
        return "Maximum Perimeter is: "+ str(maxi) 
  
# Driver Code 
def main(): 
    arr1 = [6, 1, 6, 5, 8, 4] 
    a = maxPerimeter(arr1) 
    print(a) 
  
    arr2 = [2, 20, 7, 55,  
            1, 33, 12, 4] 
    a = maxPerimeter(arr2) 
    print(a) 
  
    arr3 = [33, 6, 20, 1, 8, 
            12, 5, 55, 4, 9] 
    a = maxPerimeter(arr3) 
    print(a) 
  
if __name__=='__main__': 
    main() 
  
# This code is contributed  
# by Pritha Upadhyay 

C#

// Efficient solution to find  
// out maximum perimeter triangle which  
// can be formed using the elements  
// of the given array  
  
using System; 
  
class GFG { 
  
// Function to find out maximum perimeter  
    static void maxPerimeter(int[] arr, int n) { 
  
        // sort the array elements  
        // in reversed order  
        arr = arrRevSort(arr); 
        //sort(arr, arr+n, greater<int>());  
  
        // initialize maximum  
        // perimeter to 0  
        int maxi = 0; 
  
        // loop through the sorted array  
        // and check whether it forms a  
        // triangle or not.  
        for (int i = 0; i < n - 2; i++) { 
  
            // Check whether arr[i], arr[i+1]  
            // and arr[i+2] forms a triangle  
            // or not.  
            if (arr[i] < arr[i + 1] + arr[i + 2]) { 
  
                // if it forms a triangle then  
                // it is the triangle with  
                // maximum perimeter.  
                maxi = Math.Max(maxi, arr[i] + arr[i + 1] + arr[i + 2]); 
                break; 
            } 
        } 
  
        // If maximum perimeter is non-zero  
        // then print it.  
        if (maxi > 0) { 
            Console.WriteLine("Maximum Perimeter is: " + maxi); 
        } // otherwise no triangle formation  
        // is possible.  
        else { 
            Console.WriteLine("Triangle formation is not possible."); 
        } 
    } 
    //Function return sorted array in Decreasing  
  
    static int[] arrRevSort(int[] arr) { 
        Array.Sort(arr); 
        int j = arr.Length - 1; 
        for (int i = 0; i < arr.Length / 2; i++, j--) { 
            int temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        return arr; 
    } 
  
// Driver Program  
    public static void Main() { 
        // test case 1  
        int[] arr1 = {6, 1, 6, 5, 8, 4}; 
        maxPerimeter(arr1, 6); 
  
        // test case 2  
        int[] arr2 = {2, 20, 7, 55, 1, 33, 12, 4}; 
        maxPerimeter(arr2, 8); 
  
        // test case 3  
        int[] arr3 = {33, 6, 20, 1, 8, 12, 5, 55, 4, 9}; 
        maxPerimeter(arr3, 10); 
    } 
} 
/*This Java code is contributed by mits*/

PHP

<?php 
// Efficient solution to find out maximum  
// perimeter triangle which can be formed  
// using the elements of the given array 
  
// Function to find out maximum perimeter 
function maxPerimeter(&$arr, $n) 
{ 
  
    // sort the array elements in  
    // reversed order 
    rsort($arr); 
  
    // initialize maximum perimeter to 0 
    $maxi = 0; 
  
    // loop through the sorted array 
    // and check whether it forms a 
    // triangle or not. 
    for ($i = 0; $i < $n - 2; $i++) 
    { 
  
        // Check whether arr[i], arr[i+1] 
        // and arr[i+2] forms a triangle 
        // or not. 
        if ($arr[$i] < $arr[$i + 1] +  
                       $arr[$i + 2]) 
        { 
  
            // if it forms a triangle then 
            // it is the triangle with 
            // maximum perimeter. 
            $maxi = max($maxi, $arr[$i] + 
                               $arr[$i + 1] +  
                               $arr[$i + 2]); 
            break; 
        } 
    } 
  
    // If maximum perimeter is non-zero 
    // then print it. 
    if ($maxi) 
    { 
        echo ("Maximum Perimeter is: "); 
        echo ($maxi) ; 
        echo ("\n"); 
    } 
  
    // otherwise no triangle formation 
    // is possible. 
    else
    { 
        echo ("Triangle formation "); 
        echo ("is not possible."); 
        echo ("\n"); 
    } 
} 
  
// Driver Code 
  
// test case 1 
$arr1 = array(6, 1, 6, 5, 8, 4); 
$s = sizeof($arr1); 
maxPerimeter($arr1, $s); 
  
// test case 2 
$arr2 = array(2, 20, 7, 55, 1,33, 12, 4); 
$st = sizeof($arr2); 
maxPerimeter($arr2, $st); 
  
// test case 3 
$arr3 = array(33, 6, 20, 1, 8, 
              12, 5, 55, 4, 9); 
$st1 = sizeof($arr3); 
maxPerimeter($arr3, $st1); 
  
// This code is contributed 
// by Shivi_Aggarwal  
?> 

Output :

Maximum Perimeter is: 20
Triangle formation is not possible.
Maximum Perimeter is: 41

Time complexity of this approach is O(n*log(n)). This much time is required to sort the array.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

PHP

C++

Java

Python3

C#

PHP

Recommended Posts: