Minimum operations to make two numbers equal

Given two numbers n and m, the task is to find the minimum number of operations required to make them equal if the following operations can be performed on them.

  • During the first operation, any of the two numbers can be increased by one.
  • During the second operation, any of the two numbers can be increased by two.
  • During the third operation, any of the two numbers can be increased by three and so on.

Examples:

Input : n = 1, m = 3
Output : 3
Explanation: 
Add 1 to n; n = 2
Add 2 to m; m = 5
Add 3 to n; n = 5
Both n and m are equal now
N of operations = 3

Input : n = 30, m = 20
Output : 4

Approach:
The approach used to solve the problem is the sum of N terms in an AP.
It is given by the formula



S(n) = (n*(n+1))/2

So, the task is to find the difference between those two numbers and see if the difference can be achieved by adding first n elements. Therefore,

S(n) = max(m,n) - min(m,n)

On substituting this value of sum in the first equation;
we get the number of elements n given by

n=(-1+sqrt(1+8*S(n)))/2

If this n is a perfect integer, then it is our final answer.
Else, we increment our target value to reach by 1 and continue.

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the minimum no of operations
int minOperations(int n, int m)
{
    int a = 0, k = 1;
  
    // find the maximum of two and store it in p
    int p = max(n, m);
  
    // increase it until it is achievable from
    // given n and m
    while (n != m) {
  
        // Here value added to n and m will be
        // S(n)=p-n+p-m;
        // check whether integer value of n exist
        // by the formula
        // n=(-1+sqrt(1+8*S(n)))/2 
        float s = (float)(p - n + p - m);
        float q = (-1 + sqrt(8 * s + 1)) / 2;
        if (q - floor(q) == 0) {
            a = q;
            n = m;
        }
  
        p = p + 1;
    }
    return a;
}
  
// Driver code
int main()
{
    int n = 1, m = 3;
  
    // Function calling
    cout << minOperations(n, m);
    return 0;
}

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Java

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// Java implementation of the above approach
import java.util.*;
  
class GFG
{
  
// Function to find the minimum no of operations
static int minOperations(int n, int m)
{
    int a = 0, k = 1;
  
    // find the maximum of two and store it in p
    int p = Math.max(n, m);
  
    // increase it until it is achievable from
    // given n and m
    while (n != m) 
    {
  
        // Here value added to n and m will be
        // S(n)=p-n+p-m;
        // check whether integer value of n exist
        // by the formula
        // n=(-1+Math.sqrt(1+8*S(n)))/2 
        float s = (float)(p - n + p - m);
        float q = (float) ((-1 + Math.sqrt(8 * s + 1)) / 2);
        if (q - Math.floor(q) == 0)
        {
            a = (int) q;
            n = m;
        }
  
        p = p + 1;
    }
    return a;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 1, m = 3;
  
    // Function calling
    System.out.print(minOperations(n, m));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation of 
# the above approach 
from math import sqrt, floor
  
# Function to find the minimum
# no. of operations 
def minOperations( n, m) :
  
    a = 0; k = 1
  
    # find the maximum of two and 
    # store it in p 
    p = max(n, m); 
  
    # increase it until it is achievable 
    # from given n and m 
    while (n != m) :
  
        # Here value added to n and m will be 
        # S(n)=p-n+p-m; 
        # check whether integer value of n  
        # exist by the formula 
        # n=(-1+sqrt(1+8*S(n)))/2 
        s = float(p - n + p - m); 
        q = (-1 + sqrt(8 * s + 1)) / 2
        if (q - floor(q) == 0) :
            a = q; 
            n = m; 
  
        p = p + 1
  
    return a; 
  
# Driver code 
if __name__ == "__main__"
  
    n = 1; m = 3
  
    # Function calling 
    print(minOperations(n, m)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the above approach
using System;
  
class GFG
{
      
    // Function to find the minimum no of operations
    static int minOperations(int n, int m)
    {
        int a = 0, k = 1;
      
        // find the maximum of two and store it in p
        int p = Math.Max(n, m);
      
        // increase it until it is achievable from
        // given n and m
        while (n != m) 
        {
      
            // Here value added to n and m will be
            // S(n)=p-n+p-m;
            // check whether integer value of n exist
            // by the formula
            // n=(-1+Math.sqrt(1+8*S(n)))/2 
            float s = (float)(p - n + p - m);
            float q = (float) ((-1 + Math.Sqrt(8 * s + 1)) / 2);
            if (q - Math.Floor(q) == 0)
            {
                a = (int) q;
                n = m;
            }
      
            p = p + 1;
        }
        return a;
    }
      
    // Driver code
    public static void Main()
    {
        int n = 1, m = 3;
      
        // Function calling
        Console.Write(minOperations(n, m));
    }
}
  
// This code is contributed by AnkitRai01

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Output:

3

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