Asked in AMAZON and MICROSOFT
Given an unsorted array arr[] of n positive integers. Find the number of triangles that can be formed with three different array elements as lengths of three sides of triangles.
Example 1:
Input:
n = 3
arr[] = {3, 5, 4}
Output:
1
Explanation:
A triangle is possible
with all the elements 5, 3 and 4.Example 2:
Input:
n = 5
arr[] = {6, 4, 9, 7, 8}
Output:
10
Explanation:
There are 10 triangles
possible with the given elements like
(6,4,9), (6,7,8),...
Your Task:
This is a function problem. You only need to complete the function findNumberOfTriangles() that takes arr[] and n as input parameters and returns the count of total possible triangles.
Expected Time Complexity: O(n2).
Expected Space Complexity: O(1).
Constraints:
3 <= n <= 103
1 <= arr[i] <= 103
Code:
#include <bits/stdc++.h>
using namespace std;
// Function to count all possible
// triangles with arr[] elements
int findNumberOfTriangles(int arr[], int n)
{
// Count of triangles
int count = 0;
// The three loops select three
// different values from array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// The innermost loop checks for
// the triangle property
for (int k = j + 1; k < n; k++)
// Sum of two sides is greater
// than the third
if (
arr[i] + arr[j] > arr[k]
&& arr[i] + arr[k] > arr[j]
&& arr[k] + arr[j] > arr[i])
count++;
}
}
return count;
}
// Driver code
int main()
{
int arr[] = { 10, 21, 22, 100, 101, 200, 300 };
int size = sizeof(arr) / sizeof(arr[0]);
cout
<< "Total number of triangles possible is "
<< findNumberOfTriangles(arr, size);
return 0;
}
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