# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

Output:

```3
```

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