Count the number of possible triangles

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:
    1. 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.
    2. 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
    3. if all three conditions match, increase the count.
    4. Print the count
  • Implementation:

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C++ code to count the number of
    // possible triangles using brute
    // force approach
    #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;
    }

    chevron_right

    
    

  • 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 some 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.

    1. Sort the array in ascending order.
    2. 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
    3. 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]
    4. So when we get the largest value, then the count of third side is k – j, add it to the total count.
    5. Now sum up for all valid pairs of i and j where i < j
  • Implementation:

    C++

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C++ program to count number of triangles that can be
    // formed from given array
    #include <bits/stdc++.h>
    using namespace std;
      
    /* 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]
                if (k > j)
                    count += k - j - 1;
            }
        }
      
        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;
    }
      
    // This code is contributed by rathbhupendra

    chevron_right

    
    

    C

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // 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]
                if (k > j)
                    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;
    }

    chevron_right

    
    

    Java

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // 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] */
                    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));
        }
    }
    /*This code is contributed by Devesh Agrawal*/

    chevron_right

    
    

    Python

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    # 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]
                if(k>j):
                    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

    chevron_right

    
    

    C#

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C# program to count number
    // of triangles that can be
    // formed from given array
    using System;
      
    class GFG {
        // 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
            Array.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] */
                    if (k > j)
                        count += k - j - 1;
                }
            }
            return count;
        }
      
        // Driver Code
        public static void Main()
        {
            int[] arr = { 10, 21, 22, 100,
                          101, 200, 300 };
            Console.WriteLine("Total number of triangles is " + findNumberOfTriangles(arr));
        }
    }
      
    // This code is contributed by anuj_67.

    chevron_right

    
    

    PHP

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    <?php
    // PHP program to count number 
    // of triangles that can be
    // formed from given array
      
    // Function to count all 
    // possible triangles with 
    // arr[] element
    function findNumberOfTriangles($arr)
    {
        $n = count($arr);
          
        // Sort the array elements 
        // in non-decreasing order
        sort($arr);
      
        // Initialize count 
        // of triangles
        $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 = 0; $i < $n - 2; ++$i)
        {
            // Initialize index of the 
            // rightmost third element
            $k = $i + 2;
      
            // Fix the second element
            for ($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] */
                if($k>$j)
                $count += $k - $j - 1;
            }
        }
        return $count;
    }
      
    // Driver code
    $arr = array(10, 21, 22, 100,
                101, 200, 300);
    echo"Total number of triangles is ",
            findNumberOfTriangles($arr);
      
    // This code is contributed by anuj_67.
    ?>

    chevron_right

    
    

  • 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

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:
    1. Sort the array and then take three variables l, r and i, pointing to start, end-1 and array element starting from end of the array.
    2. Traverse the array from end (n-1 to 1), and for each iteration keep the value of l = 0 and r = i-1
    3. 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[r-1], arr[r] and a[i], because the array is sorted , which can be directly calculated using (r-l). and then decrement the value of r and continue the loop till l is less than r
    4. If triangle cannot be formed using arr[l] and arr[r] then increment the value of r and continue the loop till l is less than r
    5. So the overall complexity of iterating
      through all array elements reduces.
  • Implementation:

    C++

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C++ implementation of the above approach
    #include <bits/stdc++.h>
    using namespace std;
      
    void CountTriangles(vector<int> A)
    {
      
        int n = A.size();
      
        sort(A.begin(), A.end());
      
        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[r-1], 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++;
            }
        }
        cout << "No of possible solutions: " << count;
    }
    int main()
    {
      
        vector<int> A = { 4, 3, 5, 7, 6 };
      
        CountTriangles(A);
    }

    chevron_right

    
    

    Java

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // 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[r-1], 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);
        }
    }
      
    // This code is contributed by PrinciRaj1992

    chevron_right

    
    

    Python3

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    # Python implementation of the above approach
    def CountTriangles( A):
      
        n = len(A);
      
        A.sort(); 
      
        count = 0;
          
        for i in range(n - 1, 0, -1):
            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[r-1], a[r] and a[i]
                    count += r - l; 
      
                    # checking for more possible solutions
                    r -= 1
                  
                else:
      
                    # if not possible check for 
                    # higher values of arr[l]
                    l += 1
        print("No of possible solutions: ", count);
      
    # Driver Code
    if __name__ == '__main__':
      
        A = [ 4, 3, 5, 7, 6 ]; 
      
        CountTriangles(A);
          
    # This code is contributed by PrinciRaj1992

    chevron_right

    
    

    C#

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C# implementation of the above approach
    using System;
      
    class GFG {
        static void CountTriangles(int[] A)
        {
            int n = A.Length;
      
            Array.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[r-1], 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++;
                    }
                }
            }
            Console.Write("No of possible solutions: " + count);
        }
      
        // Driver Code
        public static void Main(String[] args)
        {
            int[] A = { 4, 3, 5, 7, 6 };
      
            CountTriangles(A);
        }
    }
      
    // This code is contributed by Rajput-Ji

    chevron_right

    
    

  • 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 above method reduces greatly.
    • Space Complexity: O(1).
      As no extra space is required, so space complexity is constant
  • Source: http://stackoverflow.com/questions/8110538/total-number-of-possible-triangles-from-n-numbers

    https://www.geeksforgeeks.org/two-pointers-technique/

    Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.




    My Personal Notes arrow_drop_up