Maximum Perimeter Triangle from array

Try it on GfG Practice

Given an array arr[] of positive integers. Find out the maximum perimeter of the triangle from the array.

Note: Return -1, if it is not possible to construct a triangle.

Examples:

Input: arr[] = [6, 1, 6, 5, 8, 4]
Output: 20
Explanation: Triangle formed by  8,6 & 6 has perimeter 20, which is the max possible.

Input: arr[] = [7, 55, 20, 1, 4, 33, 12]
Output: -1
Explanation: The triangle is not possible because the condition: the sum of two sides should be greater than third is not fulfilled here.

[Naive Approach] – Using Nested Loops – O(n ^ 3) Time and O(1) Space

The idea is to check for all possible combinations of three integers using nested loops, whether it forms a valid triangle (the sum of any two must be more than the third), if so, update the maximum possible perimeter.

#include <bits/stdc++.h>
using namespace std;

int maxPerimeter(vector<int> &arr) {
    int n = arr.size();

    // to store the result
    int ans = -1;

    // nested loops to generall
    // all possible combinations
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            for(int k = j + 1; k < n; k++) {

                // check if the three sides
                // can form a triangle or not
                if(arr[i] + arr[j] > arr[k] &&
                   arr[j] + arr[k] > arr[i] &&
                   arr[k] + arr[i] > arr[j]) {
                    ans = max(ans, arr[i] + arr[j] + arr[k]);
                }
            }
        }
    }

    return ans;
}

int main() {
    vector<int> arr = {6, 1, 6, 5, 8, 4};
    cout << maxPerimeter(arr);
    return 0;
}

import java.util.*;

class GfG {

    // to store the result
    static int maxPerimeter(int[] arr) {
        int n = arr.length;

        // to store the result
        int ans = -1;

        // nested loops to generall
        // all possible combinations
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {

                    // check if the three sides
                    // can form a triangle or not
                    if (arr[i] + arr[j] > arr[k] &&
                        arr[j] + arr[k] > arr[i] &&
                        arr[k] + arr[i] > arr[j]) {
                        ans = Math.max(ans, arr[i] + arr[j] + arr[k]);
                    }
                }
            }
        }

        return ans;
    }

    public static void main(String[] args) {
        int[] arr = {6, 1, 6, 5, 8, 4};
        System.out.println(maxPerimeter(arr));
    }
}

# to store the result
def maxPerimeter(arr):
    n = len(arr)

    # to store the result
    ans = -1

    # nested loops to generall
    # all possible combinations
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):

                # check if the three sides
                # can form a triangle or not
                if arr[i] + arr[j] > arr[k] and arr[j] + arr[k] > arr[i] and arr[k] + arr[i] > arr[j]:
                    ans = max(ans, arr[i] + arr[j] + arr[k])
    return ans

arr = [6, 1, 6, 5, 8, 4]
print(maxPerimeter(arr))

using System;
using System.Collections.Generic;

class GfG {

    // to store the result
    static int maxPerimeter(int[] arr) {
        int n = arr.Length;

        // to store the result
        int ans = -1;

        // nested loops to generall
        // all possible combinations
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {

                    // check if the three sides
                    // can form a triangle or not
                    if (arr[i] + arr[j] > arr[k] &&
                        arr[j] + arr[k] > arr[i] &&
                        arr[k] + arr[i] > arr[j]) {
                        ans = Math.Max(ans, arr[i] + arr[j] + arr[k]);
                    }
                }
            }
        }

        return ans;
    }

    public static void Main(string[] args) {
        int[] arr = {6, 1, 6, 5, 8, 4};
        Console.WriteLine(maxPerimeter(arr));
    }
}

// to store the result
function maxPerimeter(arr) {
    let n = arr.length;

    // to store the result
    let ans = -1;

    // nested loops to generall
    // all possible combinations
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            for (let k = j + 1; k < n; k++) {

                // check if the three sides
                // can form a triangle or not
                if (arr[i] + arr[j] > arr[k] &&
                    arr[j] + arr[k] > arr[i] &&
                    arr[k] + arr[i] > arr[j]) {
                    ans = Math.max(ans, arr[i] + arr[j] + arr[k]);
                }
            }
        }
    }
    return ans;
}

function main() {
    let arr = [6, 1, 6, 5, 8, 4];
    console.log(maxPerimeter(arr));
}

main();

20

[Expected Approach] – Using Sorting – (n * log n) Time and O(1) Space

The idea is to sort the array in non-increasing order so that the first element is the maximum and the last is the minimum. Then for every three consecutive elements, check if they form a triangle, the first such combination is the required answer.

Please note that, we pick the first arr[i] such that arr[i] < arr[i+1] + arr[i+2] (0 <= i <= n-3).

How does this work? For any arr[i] to be part of the triangle, it must be smaller than sum of other two. In a decreasing order sorted array if arr[i] is greater than or equal to arr[i+1] + arr[i+2], then it would be greater than all other pairs after arr[i+1]. So there is no point checking other pairs. So after sorting, we only need to check the next 2, Since the array is sorted in decreasing order, the first arr[i] satisfying the condition from left to right would give us the max result.

Below is given the step-by-step approach:

• First, sort the array in non-increasing order.
• Then, check the first three elements of the sorted array. If they satisfy the triangle inequality (i.e. arr[i] < arr[i+1] + arr[i+2]), they form a triangle with the maximum possible perimeter because any other combination will result in a lower sum.
• If the first three elements do not form a triangle, it implies that the largest element is too large (i.e., a ≥ b + c), so drop the largest element and consider the next triple.
• Repeat the process by checking consecutive triples (i.e., b, c, d; then c, d, e; and so on) until a valid triangle is found.
• Once a valid triple is identified, it gives the maximum perimeter, and further checking is unnecessary.

Below is given the implementation:

#include <bits/stdc++.h>
using namespace std;

int maxPerimeter(vector<int> &arr) {
    int n = arr.size();

    // sort the array in descending order
    sort(arr.begin(), arr.end(), greater<int>());

    // loop through the array
    for(int i = 0; i < n - 2; i++) {

        // check if the three sides can 
        // form a triangle or not
        if(arr[i] < arr[i + 1] + arr[i + 2]) {
            return arr[i] + arr[i + 1] + arr[i + 2];
        }
    }

    return -1;
}

int main() {
    vector<int> arr = {6, 1, 6, 5, 8, 4};
    cout << maxPerimeter(arr);
    return 0;
}

import java.util.*;

class GfG {

    public static int maxPerimeter(int[] arr) {
        int n = arr.length;

        // sort the array in descending order
        Arrays.sort(arr);
        for (int i = 0; i < n / 2; i++) {
            int temp = arr[i];
            arr[i] = arr[n - 1 - i];
            arr[n - 1 - i] = temp;
        }

        // loop through the array
        for (int i = 0; i < n - 2; i++) {

            // check if the three sides can 
            // form a triangle or not
            if (arr[i] < arr[i + 1] + arr[i + 2]) {
                return arr[i] + arr[i + 1] + arr[i + 2];
            }
        }

        return -1;
    }

    public static void main(String[] args) {
        int[] arr = {6, 1, 6, 5, 8, 4};
        System.out.println(maxPerimeter(arr));
    }
}

def maxPerimeter(arr):
    n = len(arr)

    # sort the array in descending order
    arr.sort(reverse=True)

    # loop through the array
    for i in range(n - 2):

        # check if the three sides can 
        # form a triangle or not
        if arr[i] < arr[i + 1] + arr[i + 2]:
            return arr[i] + arr[i + 1] + arr[i + 2]

    return -1

arr = [6, 1, 6, 5, 8, 4]
print(maxPerimeter(arr))

using System;
using System.Collections.Generic;

class GfG {

    public static int maxPerimeter(int[] arr) {
        int n = arr.Length;

        // sort the array in descending order
        Array.Sort(arr);
        Array.Reverse(arr);

        // loop through the array
        for (int i = 0; i < n - 2; i++) {

            // check if the three sides can 
            // form a triangle or not
            if (arr[i] < arr[i + 1] + arr[i + 2]) {
                return arr[i] + arr[i + 1] + arr[i + 2];
            }
        }

        return -1;
    }

    public static void Main() {
        int[] arr = {6, 1, 6, 5, 8, 4};
        Console.WriteLine(maxPerimeter(arr));
    }
}

function maxPerimeter(arr) {
    let n = arr.length;

    // sort the array in descending order
    arr.sort((a, b) => b - a);

    // loop through the array
    for (let i = 0; i < n - 2; i++) {

        // check if the three sides can 
        // form a triangle or not
        if (arr[i] < arr[i + 1] + arr[i + 2]) {
            return arr[i] + arr[i + 1] + arr[i + 2];
        }
    }

    return -1;
}

let arr = [6, 1, 6, 5, 8, 4];
console.log(maxPerimeter(arr));

20

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Article Tags :

• Arrays
• DSA
• Mathematical
• triangle

Practice Tags :

• Arrays
• Mathematical