Making three numbers equal with the given operations

Given four positive integers A, B, C and K. The task is to check if it is possible to equalize the three integers A, B and C with the help of K and make K equal to 0. In one operation, you can subtract any value form K (if it remains greater than equal to 0 after subtraction) and add the new value to any of the three integers A, B or C.

Examples:

Input: A = 6, B = 3, C = 2, K = 7
Output: Yes
Operation 1: Add 3 to B and subtract 3 from K.
A = 6, B = 6, C = 2 and K = 4
Operation 2: Add 4 to C and subtract 4 from K.
A = 6, B = 6, C = 6 and K = 0



Input: A = 10, B = 20, C = 17, K = 15
Output: No

Approach: Check whether it is possible to equalize the three numbers by sorting the three numbers and subtracting the value of K by the sum of the difference of 3rd and 2nd element and the 3rd and 1st element. If K is still greater than 0 and can be divided among the three elements equally then only the three elements can be made equal and K can be made equal to 0.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if a, b and c can
// be made equal with the given operations
bool canBeEqual(int a, int b, int c, int k)
{
    int arr[3];
    arr[0] = a;
    arr[1] = b;
    arr[2] = c;
  
    // Sort the three numbers
    sort(arr, arr + 3);
  
    // Find the sum of difference of the 3rd and
    // 2nd element and the 3rd and 1st element
    int diff = 2 * arr[2] - arr[1] - arr[0];
  
    // Subtract the difference from k
    k = k - diff;
  
    // Check the required condition
    if (k < 0 || k % 3 != 0)
        return false;
  
    return true;
}
  
// Driver code
int main()
{
    int a1 = 6, b1 = 3, c1 = 2, k1 = 7;
  
    if (canBeEqual(a1, b1, c1, k1))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function that returns true if a, b and c can
// be made equal with the given operations
static boolean canBeEqual(int a, int b, int c, int k)
{
    int []arr = new int[3];
    arr[0] = a;
    arr[1] = b;
    arr[2] = c;
  
    // Sort the three numbers
    Arrays.sort(arr);
  
    // Find the sum of difference of the 3rd and
    // 2nd element and the 3rd and 1st element
    int diff = 2 * arr[2] - arr[1] - arr[0];
  
    // Subtract the difference from k
    k = k - diff;
  
    // Check the required condition
    if (k < 0 || k % 3 != 0)
        return false;
  
    return true;
}
  
// Driver code
public static void main(String[] args)
{
    int a1 = 6, b1 = 3, c1 = 2, k1 = 7;
  
    if (canBeEqual(a1, b1, c1, k1))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 implementation of the approach 
  
# Function that returns true if a, b and c can 
# be made equal with the given operations 
def canBeEqual(a, b, c, k) : 
  
    arr = [0] * 3
    arr[0] = a; 
    arr[1] = b; 
    arr[2] = c; 
  
    # Sort the three numbers 
    arr.sort()
  
    # Find the sum of difference of the 3rd and 
    # 2nd element and the 3rd and 1st element 
    diff = 2 * arr[2] - arr[1] - arr[0]; 
  
    # Subtract the difference from k 
    k = k - diff; 
  
    # Check the required condition 
    if (k < 0 or k % 3 != 0) :
        return False
  
    return True
  
# Driver code 
if __name__ == "__main__"
  
    a1 = 6; b1 = 3; c1 = 2; k1 = 7
  
    if (canBeEqual(a1, b1, c1, k1)) :
        print("Yes"); 
    else :
        print("No"); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function that returns true if a, b and c can
// be made equal with the given operations
static bool canBeEqual(int a, int b, int c, int k)
{
    int []arr = new int[3];
    arr[0] = a;
    arr[1] = b;
    arr[2] = c;
  
    // Sort the three numbers
    Array.Sort(arr);
  
    // Find the sum of difference of the 3rd and
    // 2nd element and the 3rd and 1st element
    int diff = 2 * arr[2] - arr[1] - arr[0];
  
    // Subtract the difference from k
    k = k - diff;
  
    // Check the required condition
    if (k < 0 || k % 3 != 0)
        return false;
  
    return true;
}
  
// Driver code
public static void Main(String[] args)
{
    int a1 = 6, b1 = 3, c1 = 2, k1 = 7;
  
    if (canBeEqual(a1, b1, c1, k1))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
  
// This code is contributed by 29AjayKumar

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Output:

Yes

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