Given four integers A, B, C, and D, the task is to find the number of distinct sets (X, Y, and Z) where X, Y and Z denotes the length of sides forming a valid triangle. A ? X ? B, B ? Y ? C, and C ? Z ? D.
Examples:
Input: A = 2, B = 3, C = 4, D = 5
Output: 7
Explanation:
Possible Length of Side of Triangles are –
{(2, 3, 4), (2, 4, 4), (2, 4, 5), (3, 3, 4), (3, 3, 5), (3, 4, 4) and (3, 4, 5)}
Input: A = 1, B = 1, C = 2, D = 2
Output: 1
Explanation:
Only possible length of sides we can choose triangle is (1, 2, 2)
Naive Approach: The key observation in the problem is that, If X, Y and Z are the valid sides of a triangle and X ? Y ? Z, then sufficient conditions for these sides to form a valid triangle will be X+Y > Z.
Finally, the count of the possible Z value for the given X and Y can be computed as –
- If X+Y is greater than D, for this case Z can be chosen from [C, D], Total possible values of Z will be (D-C+1).
- If X+Y is less than D and greater than C, then Z can be chosen from [C, X+Y-1].
- If X+Y is less than or equal to C, then we cannot choose Z as these sides will not form a valid triangle.
Time Complexity: 
Efficient Approach: The idea is to iterate for all the possible values of A and then compute the number of possible Y and Z values possible for the given X using mathematical computations.
For a given X, the value of X+Y will be in the range of . If we compute the number of possible value greater than D, then the total number of possible values of Y and Z will be –
// Number of possible values of Y and Z // If num_greater is the number of possible // Y values which is greater than D
Similarly, Let R and L is the upper bound and Lower Bound of the values of X+Y in the range of C and D. Then, total combinations for Y and Z will be –
Below is the implementation of the above approach:
C++
// C++ implementation to count the// number of possible triangles// for the given sides ranges#include <bits/stdc++.h>using namespace std;// Function to count the number of// possible triangles for the given// sides rangesint count_triangles(int a, int b, int c, int d){ int ans = 0; // Iterate for every possible of x for (int x = a; x <= b; ++x) { // Range of y is [b, c] // From this range First // we will find the number // of x + y greater than d int num_greater_than_d = max(d, c + x) - max(d, b + x - 1); // For x+y greater than d // we can choose all z from [c, d] // Total permutation will be ans += num_greater_than_d * (d - c + 1); // Now we will find the number // of x+y in between the [c, d] int r = min(max(c, c + x), d) - c; int l = min(max(c, b + x - 1), d) - c; // [l, r] will be the range // from total [c, d] x+y belongs // For any r such that r = x+y // We can choose z, in the range // [c, d] only less than r, // Thus total permutation be int x1 = (r * (r + 1)) / 2; int x2 = (l * (l + 1)) / 2; ans += x1 - x2; } return ans;}// Driver Codeint main(){ int a = 2, b = 3, c = 4, d = 5; cout << count_triangles(a, b, c, d) << endl; return 0;} |
Java
// Java implementation to count the// number of possible triangles// for the given sides rangesimport java.util.Scanner;import java.util.Arrays; class GFG{// Function to count the number of// possible triangles for the given// sides rangespublic static int count_triangles(int a, int b, int c, int d){ int ans = 0; // Iterate for every possible of x for(int x = a; x <= b; ++x) { // Range of y is [b, c] // From this range First // we will find the number // of x + y greater than d int num_greater_than_d = Math.max(d, c + x) - Math.max(d, b + x - 1); // For x+y greater than d // we can choose all z from [c, d] // Total permutation will be ans += num_greater_than_d * (d - c + 1); // Now we will find the number // of x+y in between the [c, d] int r = Math.min(Math.max(c, c + x), d) - c; int l = Math.min(Math.max(c, b + x - 1), d) - c; // [l, r] will be the range // from total [c, d] x+y belongs // For any r such that r = x+y // We can choose z, in the range // [c, d] only less than r, // Thus total permutation be int x1 = (r * (r + 1)) / 2; int x2 = (l * (l + 1)) / 2; ans += x1 - x2; } return ans;}// Driver Codepublic static void main(String args[]){ int a = 2, b = 3, c = 4, d = 5; System.out.println(count_triangles(a, b, c, d));}}// This code is contributed by SoumikMondal |
Python3
# Python3 implementation to count # the number of possible triangles # for the given sides ranges # Function to count the number of # possible triangles for the given # sides ranges def count_triangles(a, b, c, d): ans = 0 # Iterate for every possible of x for x in range(a, b + 1): # Range of y is [b, c] # From this range First # we will find the number # of x + y greater than d num_greater_than_d = (max(d, c + x) - max(d, b + x - 1)) # For x+y greater than d we # can choose all z from [c, d] # Total permutation will be ans = (ans + num_greater_than_d * (d - c + 1)) # Now we will find the number # of x+y in between the [c, d] r = min(max(c, c + x), d) - c; l = min(max(c, b + x - 1), d) - c; # [l, r] will be the range # from total [c, d] x+y belongs # For any r such that r = x+y # We can choose z, in the range # [c, d] only less than r, # Thus total permutation be x1 = int((r * (r + 1)) / 2) x2 = int((l * (l + 1)) / 2) ans = ans + (x1 - x2) return ans# Driver Code a = 2b = 3c = 4d = 5print (count_triangles(a, b, c, d), end = '\n') # This code is contributed by PratikBasu |
C#
// C# implementation to count the// number of possible triangles// for the given sides rangesusing System;class GFG{// Function to count the number of// possible triangles for the given// sides rangespublic static int count_triangles(int a, int b, int c, int d){ int ans = 0; // Iterate for every possible of x for(int x = a; x <= b; ++x) { // Range of y is [b, c] // From this range First // we will find the number // of x + y greater than d int num_greater_than_d = Math.Max(d, c + x) - Math.Max(d, b + x - 1); // For x+y greater than d // we can choose all z from [c, d] // Total permutation will be ans += num_greater_than_d * (d - c + 1); // Now we will find the number // of x+y in between the [c, d] int r = Math.Min(Math.Max(c, c + x), d) - c; int l = Math.Min(Math.Max(c, b + x - 1), d) - c; // [l, r] will be the range // from total [c, d] x+y belongs // For any r such that r = x+y // We can choose z, in the range // [c, d] only less than r, // Thus total permutation be int x1 = (r * (r + 1)) / 2; int x2 = (l * (l + 1)) / 2; ans += x1 - x2; } return ans;}// Driver Codepublic static void Main(String []args){ int a = 2, b = 3, c = 4, d = 5; Console.WriteLine(count_triangles(a, b, c, d));}}// This code is contributed by gauravrajput1 |
Javascript
<script>// JavaScript implementation to count the// number of possible triangles// for the given sides ranges// Function to count the number of// possible triangles for the given// sides rangesfunction count_triangles(a , b, c , d){ var ans = 0; // Iterate for every possible of x for(x = a; x <= b; ++x) { // Range of y is [b, c] // From this range First // we will find the number // of x + y greater than d var num_greater_than_d = Math.max(d, c + x) - Math.max(d, b + x - 1); // For x+y greater than d // we can choose all z from [c, d] // Total permutation will be ans += num_greater_than_d * (d - c + 1); // Now we will find the number // of x+y in between the [c, d] var r = Math.min(Math.max(c, c + x), d) - c; var l = Math.min(Math.max(c, b + x - 1), d) - c; // [l, r] will be the range // from total [c, d] x+y belongs // For any r such that r = x+y // We can choose z, in the range // [c, d] only less than r, // Thus total permutation be var x1 = (r * (r + 1)) / 2; var x2 = (l * (l + 1)) / 2; ans += x1 - x2; } return ans;}// Driver Codevar a = 2, b = 3, c = 4, d = 5;document.write(count_triangles(a, b, c, d));// This code contributed by shikhasingrajput </script> |
Output:
7
Time Complexity : O(b-a)
Space Complexity : O(1) ,as we are not using any extra space
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