Minimum and maximum possible length of the third side of a triangle

Given two sides of a triangle s1 and s2, the task is to find the minimum and maximum possible length of the third side of the given triangle. Print -1 if it is not possible to make a triangle with the given side lengths. Note that the length of all the sides must be integers.

Examples:

Input: s1 = 3, s2 = 6
Output:
Max = 8
Min = 4

Input: s1 = 5, s2 = 8
Output:
Max = 12
Min = 4

Approach: Let s1, s2 and s3 be the sides of the given triangle where s1 and s2 are given. As we know that in a triangle, the sum of two sides must always be greater than the third side. So, the following equations must be satisfied:

  1. s1 + s2 > s3
  2. s1 + s3 > s2
  3. s2 + s3 > s1

Solving for s3, we get s3 < s1 + s2, s3 > s2 – s1 and s3 > s1 – s2.
It is clear now that the length of the third side must lie in the range (max(s1, s2) – min(s1, s2), s1 + s2)
So, the minimum possible value will be max(s1, s2) – min(s1, s2) + 1 and the maximum possible value will be s1 + s2 – 1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function to find the minimum and the
// maximum possible length of the third
// side of the given triangle
void find_length(int s1, int s2)
{
  
    // Not a valid triangle
    if (s1 <= 0 || s2 <= 0) {
        cout << -1;
        return;
    }
    int max_length = s1 + s2 - 1;
    int min_length = max(s1, s2) - min(s1, s2) + 1;
  
    // Not a valid triangle
    if (min_length > max_length) {
        cout << -1;
        return;
    }
  
    cout << "Max = " << max_length << endl;
    cout << "Min = " << min_length;
}
  
// Driver code
int main()
{
    int s1 = 8, s2 = 5;
    find_length(s1, s2);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
  
// Function to find the minimum and the
// maximum possible length of the third
// side of the given triangle
static void find_length(int s1, int s2)
{
  
    // Not a valid triangle
    if (s1 <= 0 || s2 <= 0)
    {
        System.out.print(-1);
        return;
    }
    int max_length = s1 + s2 - 1;
    int min_length = Math.max(s1, s2) - Math.min(s1, s2) + 1;
  
    // Not a valid triangle
    if (min_length > max_length) 
    {
        System.out.print(-1);
        return;
    }
  
    System.out.println("Max = " + max_length);
    System.out.print("Min = " + min_length);
}
  
// Driver code
public static void main (String[] args) 
{
    int s1 = 8, s2 = 5;
    find_length(s1, s2);
}
}
  
// This code is contributed by anuj_67..

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Python3

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# Python3 implementation of the approach 
  
# Function to find the minimum and the 
# maximum possible length of the third 
# side of the given triangle 
def find_length(s1, s2) : 
  
    # Not a valid triangle 
    if (s1 <= 0 or s2 <= 0) : 
        print(-1, end = ""); 
        return
          
    max_length = s1 + s2 - 1
    min_length = max(s1, s2) - min(s1, s2) + 1
  
    # Not a valid triangle 
    if (min_length > max_length) :
        print(-1, end = ""); 
        return
  
    print("Max =", max_length); 
    print("Min =", min_length); 
  
  
# Driver code 
if __name__ == "__main__"
  
    s1 = 8;
    s2 = 5;
      
    find_length(s1, s2); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to find the minimum and the
// maximum possible length of the third
// side of the given triangle
static void find_length(int s1, int s2)
{
  
    // Not a valid triangle
    if (s1 <= 0 || s2 <= 0)
    {
        Console.Write(-1);
        return;
    }
    int max_length = s1 + s2 - 1;
    int min_length = Math.Max(s1, s2) - Math.Min(s1, s2) + 1;
  
    // Not a valid triangle
    if (min_length > max_length) 
    {
        Console.WriteLine(-1);
        return;
    }
  
    Console.WriteLine("Max = " + max_length);
    Console.WriteLine("Min = " + min_length);
}
  
// Driver code
public static void Main () 
{
    int s1 = 8, s2 = 5;
    find_length(s1, s2);
}
}
  
// This code is contributed by anuj_67..

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Output:

Max = 12
Min = 4


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Improved By : AnkitRai01, vt_m



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