# Minimum operations to make two numbers equal

• Last Updated : 22 Apr, 2021

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:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

```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++

 `// C++ implementation of the above approach``#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``
Output:
`3`

My Personal Notes arrow_drop_up