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Find a triplet (A, B, C) such that 3*A + 5*B + 7*C is equal to N

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Given an integer N, the task is to find three positive integers A, B, and C such that the value of the expression (3*A + 5*B + 7*C) is equal to N. If no such triplet exists, then print “-1”.

Examples:

Input: N = 19
Output:
A = 3
B = 2
C = 0
Explanation: Setting A, B, and C equal to 0, 1, and 2 respectively, the evaluated value of the expression = 3 * A + 5 * B + 7 * C = 3 * 3 + 5 * 2 + 7 * 0 = 19, which is the same as N (= 19).

Input: N = 4
Output: -1

Naive Approach: The simplest approach to solve the problem is to generate all possible triplets with integers up to N and check if there exists any triplet (A, B, C), such that the value of (3*A + 5*B + 7*C) is equal to N. If found to be true, then print that triplet. Otherwise, print “-1”
Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observation that the value of A lies over the range [0, N / 3], the value of B lies over the range [0, N / 5], and the value of C lies over the range [0, N / 7]. Follow the steps below to solve the problem:

  • Iterate over the range [0, N/7] and perform the following operations:
  • After completing the above steps, if there doesn’t exist any such triplet then print “-1”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find a triplet (A, B, C)
// such that 3 * A + 5 * B + 7 * C is N
void CalculateValues(int N){
 
  int A = 0, B = 0, C = 0;
 
  // Iterate over the range [0, N//7]
  for (C = 0; C < N/7; C++)
  {
 
    // Iterate over the range [0, N//5]
    for ( B = 0; B < N/5; B++)
    {
 
      // Find the value of A
      int A = N - 7 * C - 5 * B;
 
      // If A is greater than or equal
      // to 0 and divisible by 3
      if (A >= 0 && A % 3 == 0)
      {
        cout << "A = " << A / 3 << ", B = " << B << ", C = "<< C << endl;
 
        return;
      }
    }
  }
 
  // Otherwise, print -1
  cout << -1 << endl;
 
}
 
// Driver Code
int main()
{
  int N = 19;
  CalculateValues(19);
 
  return 0;
}
 
// This code is contributed by susmitakundugoaldanga.


Java




// Java program to implement
// the above approach
import java.util.*;
 
class GFG
{
 
  // Function to find a triplet (A, B, C)
  // such that 3 * A + 5 * B + 7 * C is N
  static void CalculateValues(int N)
  {
    int A = 0, B = 0, C = 0;
 
    // Iterate over the range [0, N//7]
    for (C = 0; C < N/7; C++)
    {
 
      // Iterate over the range [0, N//5]
      for ( B = 0; B < N/5; B++)
      {
 
        // Find the value of A
        A = N - 7 * C - 5 * B;
 
        // If A is greater than or equal
        // to 0 and divisible by 3
        if (A >= 0 && A % 3 == 0)
        {
          System.out.print("A = " + A / 3 + ", B = " + B + ", C = "+ C);
 
          return;
        }
      }
    }
 
    // Otherwise, print -1
    System.out.println(-1);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 19;
    CalculateValues(19);
  }
}
 
// This code is contributed by souravghosh0416.


Python3




# Python program for the above approach
 
# Function to find a triplet (A, B, C)
# such that 3 * A + 5 * B + 7 * C is N
def CalculateValues(N):
   
    # Iterate over the range [0, N//7]
    for C in range(0, N//7 + 1):
         
        # Iterate over the range [0, N//5]
        for B in range(0, N//5 + 1):
 
            # Find the value of A
            A = N - 7 * C - 5 * B
 
            # If A is greater than or equal
            # to 0 and divisible by 3
            if (A >= 0 and A % 3 == 0):
                print("A =", A / 3, ", B =", B, ", \
                       C =", C, sep =" ")
                return
     
    # Otherwise, print -1
    print(-1)
    return
 
 
# Driver Code
if __name__ == '__main__':
   
    N = 19
    CalculateValues(19)


C#




// C# program for the above approach
using System;
 
class GFG{
 
  // Function to find a triplet (A, B, C)
  // such that 3 * A + 5 * B + 7 * C is N
  static void CalculateValues(int N)
  {
    int A = 0, B = 0, C = 0;
 
    // Iterate over the range [0, N//7]
    for (C = 0; C < N/7; C++)
    {
 
      // Iterate over the range [0, N//5]
      for ( B = 0; B < N/5; B++)
      {
 
        // Find the value of A
        A = N - 7 * C - 5 * B;
 
        // If A is greater than or equal
        // to 0 and divisible by 3
        if (A >= 0 && A % 3 == 0)
        {
          Console.Write("A = " + A / 3 + ", B = " + B + ", C = "+ C);
 
          return;
        }
      }
    }
 
    // Otherwise, print -1
    Console.WriteLine(-1);
  }
 
  // Driver Code
  static public void Main()
  {
    int N = 19;
    CalculateValues(19);
  }
}
 
// This code is contributed by splevel62.


Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find a triplet (A, B, C)
// such that 3 * A + 5 * B + 7 * C is N
function CalculateValues(N){
 
var A = 0, B = 0, C = 0;
 
// Iterate over the range [0, N//7]
for (C = 0; C < N/7; C++)
{
 
    // Iterate over the range [0, N//5]
    for ( B = 0; B < N/5; B++)
    {
 
    // Find the value of A
    var A = N - 7 * C - 5 * B;
 
    // If A is greater than or equal
    // to 0 and divisible by 3
    if (A >= 0 && A % 3 == 0)
    {
        document.write( "A = " + A / 3 + ", B = " + B + ", C = "+ C );
 
        return;
    }
    }
}
 
// Otherwise, print -1
document.write( -1 );
 
}
 
// Driver Code
 
var N = 19;
CalculateValues(19);
 
 
</script>


Output

A = 3, B = 2, C = 0

Time Complexity: O(N2)
Auxiliary Space: O(1)



Last Updated : 27 Apr, 2021
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