# 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 = 3Output :3Explanation: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 = 3Input :n = 30, m = 20Output :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 <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

`// 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

`# 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#

`// 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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