Data Modelling & AI Data Structure & Algorithm

Count of triples (A, B, C) where AC is greater than BB

23 June 2025

0

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 triplets
long 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 Code
int main()
{
    int A, B, C;
    A = 3, B = 2, C = 2;
 
    // function calling
    cout << countTriplets(A, B, C);
}

Java

// Java implementation of above approach
import java.util.*;
 
class GFG
{
 
// function to return the count
// of the valid triplets
static 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 Code
public 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 triplets
def 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 Code
A = 3
B = 2
C = 2
 
# function calling
print(countTriplets(A, B, C))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of above approach
using System;
 
class GFG
{
 
// function to return the count
// of the valid triplets
static 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 Code
public 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 triplets
function 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:

Time Complexity: $O(A*B*C)$ 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 b
long 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 constraints
long 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 Code
int main()
{
    int A, B, C;
    A = 3, B = 2, C = 2;
 
    // Function calling
    cout << countTriplets(A, B, C);
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG 
{
 
// Counts the number of triplets
// for a given value of b
static 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 constraints
static 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 Code
public 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 b
def 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 constraints
def 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 Code
 
A = 3
B = 2
C = 2
 
# Function calling
print(countTriplets(A, B, C))
 
# This code is contributed by shubhamsingh10

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;                 
     
class GFG 
{
 
// Counts the number of triplets
// for a given value of b
static 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 constraints
static 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 Code
public 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 b
function 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 constraints
function 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:

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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Count of triples (A, B, C) where AC is greater than BB

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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Count of triples (A, B, C) where A*C is greater than B*B

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Javascript

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Java

Python3

C#

Javascript

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Count of triples (A, B, C) where AC is greater than BB