Given two integers X and Y. X and Y represent any two values among (A + B), (A + C) and (B + C). The task is to find A, B and C such that A + B + C is minimum possible.
Examples:
Input: X = 3, Y = 4
Output: 2 1 2
A = 2, B = 1, C = 2.
Then A + B = 3 and A + C = 4.
A + B + C = 5 which is minimum possible.
Input: X = 123, Y = 13
Output: 1 12 111
Approach: Let X = A + B and Y = B + C. If X > Y let’s swap them. Note that A + B + C = A + B + (Y – B) = A + Y. That’s why it’s optimal to minimize the value of A. So the value of A can always be 1. Then B = X – A and C = Y – B.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to find A, B and C void MinimumValue( int x, int y)
{ // Keep minimum number in x
if (x > y)
swap(x, y);
// Find the numbers
int a = 1;
int b = x - 1;
int c = y - b;
cout << a << " " << b << " " << c;
} // Driver code int main()
{ int x = 123, y = 13;
// Function call
MinimumValue(x, y);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG
{ // Function to find A, B and C static void MinimumValue( int x, int y)
{ // Keep minimum number in x
if (x > y)
{
int temp = x;
x = y;
y = temp;
}
// Find the numbers
int a = 1 ;
int b = x - 1 ;
int c = y - b;
System.out.print( a + " " + b + " " + c);
} // Driver code public static void main (String[] args)
{ int x = 123 , y = 13 ;
// Function call
MinimumValue(x, y);
} } // This code is contributed by anuj_67.. |
# Python3 implementation of the approach # Function to find A, B and C def MinimumValue(x, y):
# Keep minimum number in x
if (x > y):
x, y = y, x
# Find the numbers
a = 1
b = x - 1
c = y - b
print (a, b, c)
# Driver code x = 123
y = 13
# Function call MinimumValue(x, y) # This code is contributed by Mohit Kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to find A, B and C static void MinimumValue( int x, int y)
{ // Keep minimum number in x
if (x > y)
{
int temp = x;
x = y;
y = temp;
}
// Find the numbers
int a = 1;
int b = x - 1;
int c = y - b;
Console.WriteLine( a + " " + b + " " + c);
} // Driver code public static void Main ()
{ int x = 123, y = 13;
// Function call
MinimumValue(x, y);
} } // This code is contributed by anuj_67.. |
<script> // javascript implementation of the approach // Function to find A, B and C function MinimumValue(x, y)
{ // Keep minimum number in x
if (x > y)
{
var temp = x;
x = y;
y = temp;
}
// Find the numbers
var a = 1;
var b = x - 1;
var c = y - b;
document.write( a + " " + b + " " + c);
} // Driver code var x = 123, y = 13;
// Function call
MinimumValue(x, y);
// This code is contributed by Amit Katiyar </script> |
1 12 111
Time Complexity: O(1)
Auxiliary Space: O(1)