Count Possible Triangles (original) (raw)
Last Updated : 27 Jan, 2026
Given an array of positive integers **arr[], 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 #include using namespace std;
int countTriangles(vector& 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 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 resif 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));
`
[Better Approach] Using Binary Search - O((n^2)*log n) Time and O(1) Space
The idea is to first sort the array in ascending order. Then, use two nested loops where the outer loop fixes the first side and the inner loop fixes the second side. For each such pair, perform a binary search to find the farthest valid index for the third side (with an index greater than the first two) whose value is less than the sum of the first two sides. This gives a valid range of choices for the third side that satisfies the required condition. Finally, add the size of this range to the result.
C++ `
#include #include #include using namespace std;
int countTriangles(vector &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 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 resif 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));
`
[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 strictly greater than 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 #include #include using namespace std;
int countTriangles(vector &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 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 resif 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));
`