Given three integers A, B and C. The task is to count the number of triples (a, b, c) such that a * c > b2, where 0 < a <= A, 0 < b <= B and 0 < c <= C.
Examples:
Input: A = 3, B = 2, C = 2
Output: 6
Following triples are counted :
(1, 1, 2), (2, 1, 1), (2, 1, 2), (3, 1, 1), (3, 1, 2) and (3, 2, 2).
Input: A = 3, B = 3, C = 3
Output: 11
Naive approach:
The brute force approach is to consider all possible triples (a, b, c) and count those triples that satisfy the constraint a*c > b2.
Below is the implementation of the given approach.
C++
// C++ implementation#include <bits/stdc++.h>using namespace std;// function to return the count// of the valid tripletslong long countTriplets(int A, int B, int C){ long long ans = 0; for (int i = 1; i <= A; i++) { for (int j = 1; j <= B; j++) { for (int k = 1; k <= C; k++) { if (i * k > j * j) ans++; } } } return ans;}// Driver Codeint main(){ int A, B, C; A = 3, B = 2, C = 2; // function calling cout << countTriplets(A, B, C);} |
Java
// Java implementation of above approachimport java.util.*;class GFG{// function to return the count// of the valid tripletsstatic long countTriplets(int A, int B, int C){ long ans = 0; for (int i = 1; i <= A; i++) { for (int j = 1; j <= B; j++) { for (int k = 1; k <= C; k++) { if (i * k > j * j) ans++; } } } return ans;}// Driver Codepublic static void main (String[] args){ int A = 3, B = 2, C = 2; // function calling System.out.println(countTriplets(A, B, C));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation for above approach# function to return the count# of the valid tripletsdef countTriplets(A, B, C): ans = 0 for i in range(1, A + 1): for j in range(1, B + 1): for k in range(1, C + 1): if (i * k > j * j): ans += 1 return ans# Driver CodeA = 3B = 2C = 2# function callingprint(countTriplets(A, B, C))# This code is contributed by Mohit Kumar |
C#
// C# implementation of above approachusing System;class GFG{// function to return the count// of the valid tripletsstatic long countTriplets(int A, int B, int C){ long ans = 0; for (int i = 1; i <= A; i++) { for (int j = 1; j <= B; j++) { for (int k = 1; k <= C; k++) { if (i * k > j * j) ans++; } } } return ans;}// Driver Codepublic static void Main (String[] args){ int A = 3, B = 2, C = 2; // function calling Console.WriteLine(countTriplets(A, B, C));}} // This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation// function to return the count// of the valid tripletsfunction countTriplets(A, B, C){ let ans = 0; for (let i = 1; i <= A; i++) { for (let j = 1; j <= B; j++) { for (let k = 1; k <= C; k++) { if (i * k > j * j) ans++; } } } return ans;}// Driver Code let A, B, C; A = 3, B = 2, C = 2; // function calling document.write(countTriplets(A, B, C));</script> |
Output:
6
Time Complexity: 
since three nested loops are used the time taken by the algorithm to complete all operations is O(A*B*C).
Auxiliary Space: O(1), since no extra array is used so the space taken by the algorithm is constant
Efficient approach:
Let us count all triplets for a given value of b = k for all k from 1 to B.
- For a given b = k we need to find all a = i and c = j that satisfy i * j > k2
- For a = i, find smallest c = j that satisfies the condition.
Since c = j satisfies this condition therefore c = j + 1, c = j + 2, … and so on, will also satisfy the condition.
So we can easily count all triples in which a = i and b = k. - Also if for some a = i, c = j is the smallest value such that the given condition is satisfied so it can be observed that a = j and all c >= i also satisfy the condition.
The condition is also satisfied by a = j + 1 and c >= i that is all values a >= j and c >= i also satisfy the condition. - The above observation helps us to count all triples in which b = k and a >= j easily.
- Now we need to count all triples in which b = k and i < a < j.
- Thus for a given value of b = k we only need to go upto a = square root of k.
Below is the implementation of the above approach:
C++
// C++ implementation#include <bits/stdc++.h>using namespace std;// Counts the number of triplets// for a given value of blong long getCount(int A, int B2, int C){ long long count = 0; // Count all triples in which a = i for (int i = 1; i <= A; i++) { // Smallest value j // such that i*j > B2 long long j = (B2 / i) + 1; // Count all (i, B2, x) // such that x >= j if (C >= j) count = (count + C - j + 1); // count all (x, B2, y) such // that x >= j this counts // all such triples in // which a >= j if (A >= j && C >= i) count = (count + (C - i + 1) * (A - j + 1)); // As all triples with a >= j // have been counted reduce // A to j - 1. if (A >= j) A = j - 1; } return count;}// Counts the number of triples that// satisfy the given constraintslong long countTriplets(int A, int B, int C){ long long ans = 0; for (int i = 1; i <= B; i++) { // GetCount of triples in which b = i ans = (ans + getCount(A, i * i, C)); } return ans;}// Driver Codeint main(){ int A, B, C; A = 3, B = 2, C = 2; // Function calling cout << countTriplets(A, B, C);} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Counts the number of triplets// for a given value of bstatic long getCount(int A, int B2, int C){ long count = 0; // Count all triples in which a = i for (int i = 1; i <= A; i++) { // Smallest value j // such that i*j > B2 long j = (B2 / i) + 1; // Count all (i, B2, x) // such that x >= j if (C >= j) count = (count + C - j + 1); // count all (x, B2, y) such // that x >= j this counts // all such triples in // which a >= j if (A >= j && C >= i) count = (count + (C - i + 1) * (A - j + 1)); // As all triples with a >= j // have been counted reduce // A to j - 1. if (A >= j) A = (int) (j - 1); } return count;}// Counts the number of triples that// satisfy the given constraintsstatic long countTriplets(int A, int B, int C){ long ans = 0; for (int i = 1; i <= B; i++) { // GetCount of triples in which b = i ans = (ans + getCount(A, i * i, C)); } return ans;}// Driver Codepublic static void main(String[] args){ int A, B, C; A = 3; B = 2; C = 2; // Function calling System.out.println(countTriplets(A, B, C));}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation# Counts the number of triplets# for a given value of bdef getCount(A, B2, C): count = 0 # Count all triples in which a = i i=1 while(i<A): # Smallest value j # such that i*j > B2 j = (B2 // i) + 1 # Count all (i, B2, x) # such that x >= j if (C >= j): count = count + C - j + 1 # count all (x, B2, y) such # that x >= j this counts # all such triples in # which a >= j if (A>= j and C >= i): count = count+ (C - i + 1) * (A - j + 1) # As all triples with a >= j # have been counted reduce # A to j - 1. if (A >= j): A = j - 1 i+=1 return count# Counts the number of triples that# satisfy the given constraintsdef countTriplets(A, B, C): ans = 0 for i in range(1,B+1): # GetCount of triples in which b = i ans = (ans+ getCount(A, i * i, C)) return ans# Driver CodeA = 3B = 2C = 2# Function callingprint(countTriplets(A, B, C))# This code is contributed by shubhamsingh10 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {// Counts the number of triplets// for a given value of bstatic long getCount(int A, int B2, int C){ long count = 0; // Count all triples in which a = i for (int i = 1; i <= A; i++) { // Smallest value j // such that i*j > B2 long j = (B2 / i) + 1; // Count all (i, B2, x) // such that x >= j if (C >= j) count = (count + C - j + 1); // count all (x, B2, y) such // that x >= j this counts // all such triples in // which a >= j if (A >= j && C >= i) count = (count + (C - i + 1) * (A - j + 1)); // As all triples with a >= j // have been counted reduce // A to j - 1. if (A >= j) A = (int) (j - 1); } return count;}// Counts the number of triples that// satisfy the given constraintsstatic long countTriplets(int A, int B, int C){ long ans = 0; for (int i = 1; i <= B; i++) { // GetCount of triples in which b = i ans = (ans + getCount(A, i * i, C)); } return ans;}// Driver Codepublic static void Main(String[] args){ int A, B, C; A = 3; B = 2; C = 2; // Function calling Console.WriteLine(countTriplets(A, B, C));}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript implementation// Counts the number of triplets// for a given value of bfunction getCount(A, B2, C){ let count = 0; // Count all triples in which a = i for (let i = 1; i <= A; i++) { // Smallest value j // such that i*j > B2 let j = parseInt(B2 / i) + 1; // Count all (i, B2, x) // such that x >= j if (C >= j) count = (count + C - j + 1); // count all (x, B2, y) such // that x >= j this counts // all such triples in // which a >= j if (A >= j && C >= i) count = (count + (C - i + 1) * (A - j + 1)); // As all triples with a >= j // have been counted reduce // A to j - 1. if (A >= j) A = j - 1; } return count;}// Counts the number of triples that// satisfy the given constraintsfunction countTriplets(A, B, ){ let ans = 0; for (let i = 1; i <= B; i++) { // GetCount of triples in which b = i ans = (ans + getCount(A, i * i, C)); } return ans;}// Driver Code let A, B, C; A = 3, B = 2, C = 2; // Function calling document.write(countTriplets(A, B, C));</script> |
Output:
6
Time Complexity: O(A*B), since two nested loops are used the time taken by the algorithm to complete all operations is O(A*B).
Auxiliary Space: O(1), since no extra array is used so the space taken by the algorithm is constant
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