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
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.
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