Given an unsorted array of positive integers. Find the number of triangles that can be formed with three different array elements as three sides of triangles. For a triangle to be possible from 3 values, the sum of any two values (or sides) must be greater than the third value (or third side).

For example, if the input array is {4, 6, 3, 7}, the output should be 3. There are three triangles possible {3, 4, 6}, {4, 6, 7} and {3, 6, 7}. Note that {3, 4, 7} is not a possible triangle.

As another example, consider the array {10, 21, 22, 100, 101, 200, 300}. There can be 6 possible triangles: {10, 21, 22}, {21, 100, 101}, {22, 100, 101}, {10, 100, 101}, {100, 101, 200} and {101, 200, 300}

**Method 1 (Brute force)**

The brute force method is to run three loops and keep track of the number of triangles possible so far. The three loops select three different values from array, the innermost loop checks for the triangle property ( the sum of any two sides must be greater than the value of third side).

Time Complexity: O(N^3) where N is the size of input array.

**Method 2 (Tricky and Efficient)**

Let a, b and c be three sides. The below condition must hold for a triangle (Sum of two sides is greater than the third side)

i) a + b > c

ii) b + c > a

iii) a + c > b

Following are steps to count triangle.

**1.** Sort the array in non-decreasing order.

**2.** Initialize two pointers ‘i’ and ‘j’ to first and second elements respectively, and initialize count of triangles as 0.

**3.** Fix ‘i’ and ‘j’ and find the rightmost index ‘k’ (or largest ‘arr[k]’) such that ‘arr[i] + arr[j] > arr[k]’. The number of triangles that can be formed with ‘arr[i]’ and ‘arr[j]’ as two sides is ‘k – j’. Add ‘k – j’ to count of triangles.

Let us consider ‘arr[i]’ as ‘a’, ‘arr[j]’ as b and all elements between ‘arr[j+1]’ and ‘arr[k]’ as ‘c’. The above mentioned conditions (ii) and (iii) are satisfied because ‘arr[i] < arr[j] < arr[k]'. And we check for condition (i) when we pick 'k'
**4.** Increment ‘j’ to fix the second element again.

Note that in step 3, we can use the previous value of ‘k’. The reason is simple, if we know that the value of ‘arr[i] + arr[j-1]’ is greater than ‘arr[k]’, then we can say ‘arr[i] + arr[j]’ will also be greater than ‘arr[k]’, because the array is sorted in increasing order.

**5.** If ‘j’ has reached end, then increment ‘i’. Initialize ‘j’ as ‘i + 1’, ‘k’ as ‘i+2’ and repeat the steps 3 and 4.

Following is implementation of the above approach.

## C/C++

// C/C++ program to count number of triangles that can be // formed from given array #include <stdio.h> #include <stdlib.h> /* Following function is needed for library function qsort(). Refer http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */ int comp(const void* a, const void* b) { return *(int*)a > *(int*)b ; } // Function to count all possible triangles with arr[] // elements int findNumberOfTriangles(int arr[], int n) { // Sort the array elements in non-decreasing order qsort(arr, n, sizeof( arr[0] ), comp); // Initialize count of triangles int count = 0; // Fix the first element. We need to run till n-3 // as the other two elements are selected from // arr[i+1...n-1] for (int i = 0; i < n-2; ++i) { // Initialize index of the rightmost third // element int k = i+2; // Fix the second element for (int j = i+1; j < n; ++j) { // Find the rightmost element which is // smaller than the sum of two fixed elements // The important thing to note here is, we // use the previous value of k. If value of // arr[i] + arr[j-1] was greater than arr[k], // then arr[i] + arr[j] must be greater than k, // because the array is sorted. while (k < n && arr[i] + arr[j] > arr[k]) ++k; // Total number of possible triangles that can // be formed with the two fixed elements is // k - j - 1. The two fixed elements are arr[i] // and arr[j]. All elements between arr[j+1]/ to // arr[k-1] can form a triangle with arr[i] and arr[j]. // One is subtracted from k because k is incremented // one extra in above while loop. // k will always be greater than j. If j becomes equal // to k, then above loop will increment k, because arr[k] // + arr[i] is always greater than arr[k] count += k - j - 1; } } return count; } // Driver program to test above functionarr[j+1] int main() { int arr[] = {10, 21, 22, 100, 101, 200, 300}; int size = sizeof( arr ) / sizeof( arr[0] ); printf("Total number of triangles possible is %d ", findNumberOfTriangles( arr, size ) ); return 0; }

## Java

// Java program to count number of triangles that can be // formed from given array import java.io.*; import java.util.*; class CountTriangles { // Function to count all possible triangles with arr[] // elements static int findNumberOfTriangles(int arr[]) { int n = arr.length; // Sort the array elements in non-decreasing order Arrays.sort(arr); // Initialize count of triangles int count = 0; // Fix the first element. We need to run till n-3 as // the other two elements are selected from arr[i+1...n-1] for (int i = 0; i < n-2; ++i) { // Initialize index of the rightmost third element int k = i + 2; // Fix the second element for (int j = i+1; j < n; ++j) { /* Find the rightmost element which is smaller than the sum of two fixed elements The important thing to note here is, we use the previous value of k. If value of arr[i] + arr[j-1] was greater than arr[k], then arr[i] + arr[j] must be greater than k, because the array is sorted. */ while (k < n && arr[i] + arr[j] > arr[k]) ++k; /* Total number of possible triangles that can be formed with the two fixed elements is k - j - 1. The two fixed elements are arr[i] and arr[j]. All elements between arr[j+1] to arr[k-1] can form a triangle with arr[i] and arr[j]. One is subtracted from k because k is incremented one extra in above while loop. k will always be greater than j. If j becomes equal to k, then above loop will increment k, because arr[k] + arr[i] is always/ greater than arr[k] */ count += k - j - 1; } } return count; } public static void main (String[] args) { int arr[] = {10, 21, 22, 100, 101, 200, 300}; System.out.println("Total number of triangles is " + findNumberOfTriangles(arr)); } } /*This code is contributed by Devesh Agrawal*/

## Python

# Python function to count all possible triangles with arr[] # elements def findnumberofTriangles(arr): # Sort array and initialize count as 0 n = len(arr) arr.sort() count = 0 # Fix the first element. We need to run till n-3 as # the other two elements are selected from arr[i+1...n-1] for i in range(0,n-2): # Initialize index of the rightmost third element k = i + 2 # Fix the second element for j in range(i+1,n): # Find the rightmost element which is smaller # than the sum of two fixed elements # The important thing to note here is, we use # the previous value of k. If value of arr[i] + # arr[j-1] was greater than arr[k], then arr[i] + # arr[j] must be greater than k, because the array # is sorted. while (k < n and arr[i] + arr[j] > arr[k]): k += 1 # Total number of possible triangles that can be # formed with the two fixed elements is k - j - 1. # The two fixed elements are arr[i] and arr[j]. All # elements between arr[j+1] to arr[k-1] can form a # triangle with arr[i] and arr[j]. One is subtracted # from k because k is incremented one extra in above # while loop. k will always be greater than j. If j # becomes equal to k, then above loop will increment k, # because arr[k] + arr[i] is always greater than arr[k] count += k - j - 1 return count # Driver function to test above function arr = [10, 21, 22, 100, 101, 200, 300] print "Number of Triangles:",findnumberofTriangles(arr) # This code is contributed by Devesh Agrawal

Output:

Total number of triangles possible is 6

Time Complexity: O(n^2). The time complexity looks more because of 3 nested loops. If we take a closer look at the algorithm, we observe that k is initialized only once in the outermost loop. The innermost loop executes at most O(n) time for every iteration of outer most loop, because k starts from i+2 and goes upto n for all values of j. Therefore, the time complexity is O(n^2).

Source: http://stackoverflow.com/questions/8110538/total-number-of-possible-triangles-from-n-numbers

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