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Making three numbers equal with the given operations

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  • Last Updated : 21 Jun, 2022
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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 from 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++




// 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;
}


Java




// 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


Python3




# 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


C#




// 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


Javascript




<script>
 
// Javascript implementation of the approach
 
// Function that returns true if a, b and c can
// be made equal with the given operations
function canBeEqual(a, b, c, k)
{
    var arr = Array(3);
    arr[0] = a;
    arr[1] = b;
    arr[2] = c;
 
    // Sort the three numbers
    arr.sort((a,b)=> a-b);
 
    // Find the sum of difference of the 3rd and
    // 2nd element and the 3rd and 1st element
    var 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
var a1 = 6, b1 = 3, c1 = 2, k1 = 7;
if (canBeEqual(a1, b1, c1, k1))
    document.write( "Yes");
else
    document.write( "No");
 
// This code is contributed by rrrtnx.
</script>


Output: 

Yes

 

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


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