Analysis of Algorithms

Count Possible Triangles

Last Updated : 01 Aug, 2025

Try it on GfG Practice

Given an unsorted array of positive integers, count the number of triangles that can be formed with three different array elements as three sides of triangles.
Note: The sum of any two sides of a triangle must be greater than the 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]
Input: arr[] = [1, 2, 3]
Output: 0
Examples: No triangles are possible.

Table of Content

[Naive Approach] Checking all Triplets - O(n^3) Time and O(1) Space
[Better Approach] Using Binary Search - O((n^2)*log n) Time and O(1) Space
[Expected Approach] Using Two Pointer - O(n^2) Time and O(1) Space

[Naive Approach] Checking all Triplets - O(n^3) Time and O(1) Space

A simple approach is to run three nested loops that select three different values from an array. And in the innermost loop, we checks for the triangle property which specifies the sum of any two sides must be greater than the value of the third side.

C++

#include <iostream>
#include <vector>
using namespace std;

int countTriangles(vector<int>& arr) {
    int res = 0;

    // The three loops select three 
    // different values from array
    for (int i = 0; i < arr.size(); i++) {
        for (int j = i + 1; j < arr.size(); j++) {
            for (int k = j + 1; k < arr.size(); 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])
                    res++;
        }
    }
    return res;
}

int main() {
    vector<int> arr = {4, 6, 3, 7};
    cout << countTriangles(arr);
    return 0;
}

C

#include <stdio.h>
#include <stdlib.h>

int countTriangles(int arr[], int n) {
    int res = 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++) {
            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])
                    res++;
        }
    }
    return res;
}

int main() {
    int arr[] = {4, 6, 3, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d", countTriangles(arr, n));
    return 0;
}

Java

import java.util.ArrayList;

class GfG {
    static int countTriangles(int[] arr) {
        int res = 0;

        // The three loops select three 
        // different values from array
        for (int i = 0; i < arr.length; i++) {
            for (int j = i + 1; j < arr.length; j++) {
                for (int k = j + 1; k < arr.length; 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]) {
                        res++;
                    }
                }
            }
        }
        return res;
    }

    public static void main(String[] args) {
        int[] arr = {4, 6, 3, 7};
        System.out.println(countTriangles(arr));
    }
}

Python

def countTriangles(arr):
    res = 0

    # The three loops select three 
    # different values from array
    for i in range(len(arr)):
        for j in range(i + 1, len(arr)):
            for k in range(j + 1, len(arr)):

                # Sum of two sides is greater than the third
                if arr[i] + arr[j] > arr[k] and \
                   arr[i] + arr[k] > arr[j] and \
                   arr[k] + arr[j] > arr[i]:
                    res += 1
    return res


if __name__ == "__main__":
    arr = [4, 6, 3, 7]
    print(countTriangles(arr))

C#

using System;

class GfG {
    
    static int countTriangles(int[] arr) {
        int res = 0;

        // The three loops select three 
        // different values from array
        for (int i = 0; i < arr.Length; i++) {
            for (int j = i + 1; j < arr.Length; j++) {
                for (int k = j + 1; k < arr.Length; 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]) {
                        res++;
                    }
                }
            }
        }
        return res;
    }

    static void Main(string[] args) {
        int[] arr = { 4, 6, 3, 7 };
        Console.WriteLine(countTriangles(arr));
    }
}

JavaScript

function countTriangles(arr) {
    let res = 0;

    // The three loops select three 
    // different values from array
    for (let i = 0; i < arr.length; i++) {
        for (let j = i + 1; j < arr.length; j++) {
            for (let k = j + 1; k < arr.length; 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]) {
                    res++;
                }
            }
        }
    }
    return res;
}

// Driver Code
const arr = [4, 6, 3, 7];
console.log(countTriangles(arr));

Output

[Better Approach] Using Binary Search - O((n^2)*log n) Time and O(1) Space

The idea is to sort the array in ascending order. Then, use two nested loops: the outer loop to fix the first side, and the inner loop to fix the second side.
Next, we find the farthest index for the third side (beyond the indices of the first two sides), such that its value is less than the sum of the first two sides, using Binary Search. So a range of values for third side can be found, where it is guaranteed that its length is greater than the other individual sides but less than the sum of both sides. Add this range size to the result.

C++

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int countTriangles(vector<int> &arr) {
    int res = 0;

    // Sort the array to apply the 
    // triangle inequality efficiently
    sort(arr.begin(), arr.end());

    // Iterate through pairs of sides (arr[i], arr[j])
    for (int i = 0; i < arr.size(); i++) {
        for (int j = i + 1; j < arr.size(); j++) {
          
            // Find the first index where the 
            // sum of two sides is not valid
            int k = lower_bound(arr.begin() + j + 1, arr.end(), arr[i] + arr[j])
                - arr.begin();
            
            // Count the number of valid third sides
            int cnt = k - j - 1;
            res += cnt;
        }
    }
    return res;
}

int main() {
    vector<int> arr = {4, 6, 3, 7};
    cout << countTriangles(arr);
    return 0;
}

C

#include <stdio.h>
#include <stdlib.h>

// Comparison function for qsort
int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

// Lower bound function to find the first index 
// where the sum of two sides is not valid
int lower_bound(int arr[], int n, int value, int start) {
    int left = start, right = n;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] < value)
            left = mid + 1;
        else
            right = mid;
    }
    return left;
}

// Function to count the number of valid triangles.
int countTriangles(int arr[], int n) {
    int res = 0;

    // Sort the array to apply the 
    // triangle inequality efficiently
    qsort(arr, n, sizeof(int), compare);

    // Iterate through pairs of sides (arr[i], arr[j])
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
          
            // Find the first index where the 
            // sum of two sides is not valid
            int k = lower_bound(arr, n, arr[i] + arr[j], j + 1);
            
            // Count the number of valid third sides
            int cnt = k - j - 1;
            res += cnt;
        }
    }
    return res;
}

int main() {
    int arr[] = {4, 6, 3, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d\n", countTriangles(arr, n));
    return 0;
}

Java

import java.util.Arrays;

class GfG {
    
    static int countTriangles(int[] arr) {
        int res = 0;

        // Sort the array to apply the 
        // triangle inequality efficiently
        Arrays.sort(arr);

        // Iterate through pairs of sides (arr[i], arr[j])
        for (int i = 0; i < arr.length; i++) {
            for (int j = i + 1; j < arr.length; j++) {

                // Find the first index where the 
                // sum of two sides is not valid
                int lo = j + 1, hi = arr.length;
                int target = arr[i] + arr[j];
                while (lo < hi) {
                    int mid = lo + (hi - lo) / 2;
                    if (arr[mid] < target) {
                        lo = mid + 1;
                    } else {
                        hi = mid;
                    }
                }
                // Count the number of valid third sides
                int cnt = lo - j - 1;
                res += cnt;
            }
        }
        return res;
    }

    public static void main(String[] args) {
        int[] arr = {4, 6, 3, 7};
        System.out.println(countTriangles(arr));
    }
}

Python

from bisect import bisect_left

def countTriangles(arr):
    res = 0

    # Sort the array to apply the 
    # triangle inequality efficiently
    arr.sort()

    # Iterate through pairs of sides (arr[i], arr[j])
    for i in range(len(arr)):
        for j in range(i + 1, len(arr)):

            # Find the first index where the 
            # sum of two sides is not valid
            k = bisect_left(arr, arr[i] + arr[j], j + 1)

            # Count the number of valid third sides
            cnt = k - j - 1
            res += cnt

    return res

if __name__ == "__main__":
    arr = [4, 6, 3, 7]
    print(countTriangles(arr))

C#

using System;
using System.Linq;

class GfG {
    
    static int countTriangles(int[] arr) {
        int res = 0;

        // Sort the array to apply the 
        // triangle inequality efficiently
        Array.Sort(arr);

        // Iterate through pairs of sides (arr[i], arr[j])
        for (int i = 0; i < arr.Length; i++) {
            for (int j = i + 1; j < arr.Length; j++) {

                // Find the first index where the 
                // sum of two sides is not valid
                int lo = j + 1, hi = arr.Length;
                int target = arr[i] + arr[j];
                while (lo < hi) {
                    int mid = lo + (hi - lo) / 2;
                    if (arr[mid] < target) {
                        lo = mid + 1;
                    } else {
                        hi = mid;
                    }
                }

                // Count the number of valid third sides
                int cnt = lo - j - 1;
                res += cnt;
            }
        }
        return res;
    }

    static void Main(string[] args) {
        int[] arr = { 4, 6, 3, 7 };
        Console.WriteLine(countTriangles(arr));
    }
}

JavaScript

function countTriangles(arr) {
    let res = 0;

    // Sort the array to apply the 
    // triangle inequality efficiently
    arr.sort((a, b) => a - b);

    // Iterate through pairs of sides (arr[i], arr[j])
    for (let i = 0; i < arr.length; i++) {
        for (let j = i + 1; j < arr.length; j++) {

            // Use binary search to find the first index 
            // where the sum of two sides is not valid
            let lo = j + 1, hi = arr.length;
            while (lo < hi) {
                let mid = Math.floor((lo + hi) / 2);
                if (arr[mid] < arr[i] + arr[j]) {
                    lo = mid + 1;
                } else {
                    hi = mid;
                }
            }
            
            // Count the number of valid third sides
            let cnt = lo - (j + 1);
            res += cnt;
        }
    }
    return res;
}

// Driver Code
const arr = [4, 6, 3, 7];
console.log(countTriangles(arr));

Output

[Expected Approach] Using Two Pointers Technique - O(n^2) Time and O(1) Space

The idea is to sort the array to simplify checking the triangle inequality. Then, for each element (treated as the largest side), use two pointers technique to find count of pairs of smaller sides that can form a triangle with it.
For this, the two pointers are initialized as: one pointer (left) starts at index 0, and the other pointer (right) is positioned just before the current largest side (arr[i]).
Now, compare the sum of arr[left] + arr[right] with the current largest side (arr[i]):
If the sum is greater than or equal to arr[i], a valid triangle can be formed. Count all valid pairs between left and right, then move the right pointer to the left to explore smaller side values.
If the sum is less than arr[i], increment the left pointer to increase the sum and check larger values.

C++

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int countTriangles(vector<int> &arr) {
    int res = 0;
    sort(arr.begin(), arr.end());

    // Iterate through the array, fixing 
    // the largest side at arr[i]
    for (int i = 2; i < arr.size(); ++i) {
      
      	// Initialize pointers for the two smaller sides
        int left = 0, right = i - 1; 

        while (left < right) {
          
            if (arr[left] + arr[right] > arr[i]) {
              
                // arr[left] + arr[right] satisfies 
                // the triangle inequality, so all pairs
				// (x, right) with (left <= x < right) are valid
                res += right - left; 
              
              	// Move the right pointer to check smaller pairs
                right--; 
            } 
          	else {
              
              	// Move the left pointer to increase the sum
                left++; 
            }
        }
    }

    return res;
}

int main() {
    vector<int> arr = {4, 6, 3, 7};
    cout << countTriangles(arr);
    return 0;
}

C

#include <stdio.h>
#include <stdlib.h>

// Function to compare two integers for qsort
int compare(const void *a, const void *b) {
    return (*(int *)a - *(int *)b);
}

// Function to count the number of valid triangles
int countTriangles(int arr[], int n) {
    int res = 0;
    qsort(arr, n, sizeof(int), compare);

    // Iterate through the array, fixing 
    // the largest side at arr[i]
    for (int i = 2; i < n; ++i) {
        // Initialize pointers for the two smaller sides
        int left = 0, right = i - 1;

        while (left < right) {
            if (arr[left] + arr[right] > arr[i]) {
                // arr[left] + arr[right] satisfies the triangle inequality,
                // so all pairs (x, right) with (left <= x < right) are valid
                res += right - left;

                // Move the right pointer to check smaller pairs
                right--;
            } else {
                // Move the left pointer to increase the sum
                left++;
            }
        }
    }

    return res;
}

int main() {
    int arr[] = {4, 6, 3, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d\n", countTriangles(arr, n));
    return 0;
}

Java

import java.util.Arrays;

class GfG {
    
    static int countTriangles(int[] arr) {
        int res = 0;
        Arrays.sort(arr);

        // Iterate through the array, fixing 
        // the largest side at arr[i]
        for (int i = 2; i < arr.length; ++i) {
            // Initialize pointers for the two smaller sides
            int left = 0, right = i - 1;

            while (left < right) {
                if (arr[left] + arr[right] > arr[i]) {
                    // arr[left] + arr[right] satisfies the
                    // triangle inequality,so all pairs (x, right)
                    // with (left <= x < right) are valid
                    res += right - left;

                    // Move the right pointer to check smaller pairs
                    right--;
                } else {
                    // Move the left pointer to increase the sum
                    left++;
                }
            }
        }

        return res;
    }

    public static void main(String[] args) {
        int[] arr = {4, 6, 3, 7};
        System.out.println(countTriangles(arr));
    }
}

Python

def countTriangles(arr):
    res = 0
    arr.sort()

    # Iterate through the array, fixing 
    # the largest side at arr[i]
    for i in range(2, len(arr)):
      
        # Initialize pointers for the two smaller sides
        left, right = 0, i - 1

        while left < right:
            if arr[left] + arr[right] > arr[i]:
                # arr[left] + arr[right] satisfies the triangle inequality,
                # so all pairs (x, right) with (left <= x < right) are valid
                res += right - left

                # Move the right pointer to check smaller pairs
                right -= 1
                
            else:
                # Move the left pointer to increase the sum
                left += 1

    return res


if __name__ == "__main__":
    arr = [4, 6, 3, 7]
    print(countTriangles(arr))

C#

using System;

class GfG {
    
    static int countTriangles(int[] arr) {
        int res = 0;
        Array.Sort(arr);

        // Iterate through the array, fixing 
        // the largest side at arr[i]
        for (int i = 2; i < arr.Length; ++i) {
            // Initialize pointers for the two smaller sides
            int left = 0, right = i - 1;

            while (left < right) {
                if (arr[left] + arr[right] > arr[i]) {
                    // arr[left] + arr[right] satisfies the 
                    // triangle inequality, so all pairs (x, right) 
                    // with (left <= x < right) are valid
                    res += right - left;

                    // Move the right pointer to check smaller pairs
                    right--;
                } else {
                    // Move the left pointer to increase the sum
                    left++;
                }
            }
        }

        return res;
    }

    static void Main(string[] args) {
        int[] arr = {4, 6, 3, 7};
        Console.WriteLine(countTriangles(arr));
    }
}

JavaScript

function countTriangles(arr) {
    let res = 0;
    arr.sort((a, b) => a - b);

    // Iterate through the array, fixing 
    // the largest side at arr[i]
    for (let i = 2; i < arr.length; ++i) {
    
        // Initialize pointers for the two smaller sides
        let left = 0, right = i - 1;

        while (left < right) {
            if (arr[left] + arr[right] > arr[i]) {
                // arr[left] + arr[right] satisfies the 
                // triangle inequality, so all pairs pairs
                // (x, right) with (left <= x < right) are valid
                res += right - left;

                // Move the right pointer to check smaller pairs
                right--;
            } else {
                // Move the left pointer to increase the sum
                left++;
            }
        }
    }

    return res;
}

// Driver Code
const arr = [4, 6, 3, 7];
console.log(countTriangles(arr));

Output

Count Possible Triangles

Analysis of Algorithms

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