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 of the two values (or sides) must be greater than the third value (or third side).
Examples:
Input: arr= {4, 6, 3, 7} Output: 3 Explanation: 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. Input: arr= {10, 21, 22, 100, 101, 200, 300}. Output: 6 Explanation: 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)
 Approach: 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 an array. The innermost loop checks for the triangle property which specifies the sum of any two sides must be greater than the value of the third side).

Algorithm:
 Run three nested loops each loop starting from the index of the previous loop to end of array i.e run first loop from 0 to n, loop j from i to n and k from j to n.
 Check if array[i] + array[j] > array[k], array[i] + array[k] > array[j], array[k] + array[j] > array[i], i.e. sum of two sides is greater than the third
 if all three conditions match, increase the count.
 Print the count
 Implementation:
// Java code to count the number of
// possible triangles using brute
// force approach
import
java.io.*;
import
java.util.*;
class
GFG
{
// Function to count all possible
// triangles with arr[] elements
static
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
public
static
void
main(String[] args)
{
int
arr[] = {
10
,
21
,
22
,
100
,
101
,
200
,
300
};
int
size = arr.length;
System.out.println(
"Total number of triangles possible is "
+
findNumberOfTriangles(arr, size));
}
}
Output:
Total number of triangles possible is 6

Complexity Analysis:
 Time Complexity: O(N^3) where N is the size of input array.
 Space Complexity: O(1)
Method 2: This is a tricky and efficient approach to reduce the time complexity from O(n^3) to **O(n^2)**where two sides of the triangles are fixed and the count can be found using those two sides.
 Approach: First sort the array in ascending order. Then use two loops. The outer loop to fix the first side and inner loop to fix the second side and then find the farthest index of the third side (greater than indices of both sides) whose length is less than sum of the other two sides. So a range of values third sides can be found, where it is guaranteed that its length if greater than the other individual sides but less than the sum of both sides.

Algorithm: 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. Sort the array in ascending order.
 Now run a nested loop. The outer loop runs from start to end and the innner loop runs from index + 1 of the first loop to the end. Take the loop counter of first loop as i and second loop as j. Take another variable k = i + 2
 Now there is two pointers i and j, where array[i] and array[j] represents two sides of the triangles. For a fixed i and j, find the count of third sides which will satisfy the conditions of a triangle. i.e find the largest value of array[k] such that array[i] + array[j] > array[k]
 So when we get the largest value, then the count of third side is k – j, add it to the total count.
 Now sum up for all valid pairs of i and j where i < j
 Implementation:
// 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 nondecreasing order
Arrays.sort(arr);
// Initialize count of triangles
int
count =
0
;
// Fix the first element. We need to run till n3 as
// the other two elements are selected from arr[i+1...n1]
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[j1] 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[k1] 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] */
if
(k > j)
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));
}
}
Output:
Total number of triangles possible is 6

Complexity Analysis:

Time Complexity: O(n^2).
The time complexity looks more because of 3 nested loops. It can be observed that k is initialized only once in the outermost loop. The innermost loop executes at most O(n) time for every iteration of the outermost loop, because k starts from i+2 and goes up to n for all values of j. Therefore, the time complexity is O(n^2). 
Space Complexity: O(1).
No extra space is required. So space complexity is constant

Time Complexity: O(n^2).
Method 3: The time complexity can be greatly reduced using Two Pointer methods in just two nested loops.
 Approach: First sort the array, and run a nested loop, fix an index and then try to fix an upper and lower index within which we can use all the lengths to form a triangle with that fixed index.

Algorithm:

Sort the array and then take three variables l, r and i, pointing to start, end1 and array element starting from end of the array.

Traverse the array from end (n1 to 1), and for each iteration keep the value of l = 0 and r = i1

Now if a triangle can be formed using arr[l] and arr[r] then triangles can obviously formed
from a[l+1], a[l+2]……a[r1], arr[r] and a[i], because the array is sorted , which can be directly calculated using (rl). and then decrement the value of r and continue the loop till l is less than r 
If a triangle cannot be formed using arr[l] and arr[r] then increment the value of l and continue the loop till l is less than r

So the overall complexity of iterating
through all array elements reduces.

 Implementation:
// Java implementation of the above approach
import
java.util.*;
class
GFG {
static
void
CountTriangles(
int
[] A)
{
int
n = A.length;
Arrays.sort(A);
int
count =
0
;
for
(
int
i = n 
1
; i >=
1
; i) {
int
l =
0
, r = i 
1
;
while
(l < r) {
if
(A[l] + A[r] > A[i]) {
// If it is possible with a[l], a[r]
// and a[i] then it is also possible
// with a[l+1]..a[r1], a[r] and a[i]
count += r  l;
// checking for more possible solutions
r;
}
else
// if not possible check for
// higher values of arr[l]
{
l++;
}
}
}
System.out.print(
"No of possible solutions: "
+ count);
}
// Driver Code
public
static
void
main(String[] args)
{
int
[] A = {
4
,
3
,
5
,
7
,
6
};
CountTriangles(A);
}
}
Output:
No of possible solutions: 9

Complexity Analysis:

Time complexity: O(n^2).
As two nested loops are used, but overall iterations in comparison to the above method reduces greatly. 
Space Complexity: O(1).
As no extra space is required, so space complexity is constant

Time complexity: O(n^2).