# Minimum number of operation required to convert number x into y

Given a initial number x and two operations which are given below:

- Multiply number by 2.
- Subtract 1 from the number.

The task is to find out minimum number of operation required to convert number x into y using only above two operations. We can apply these operations any number of times.

Constraints:

1 <= x, y <= 10000

**Example:**

Input : x = 4, y = 7 Output : 2 We can transform x into y using following two operations. 1. 4*2 = 8 2. 8-1 = 7 Input : x = 2, y = 5 Output : 4 We can transform x into y using following four operations. 1. 2*2 = 4 2. 4-1 = 3 3. 3*2 = 6 4. 6-1 = 5 Answer = 4 Note that other sequences of two operations would take more operations.

The idea is to use BFS for this. We run a BFS and create nodes by multiplying with 2 and subtracting by 1, thus we can obtain all possible numbers reachable from starting number.

Important Points :

1) When we subtract 1 from a number and if it becomes < 0 i.e. Negative then there is no reason to create next node from it (As per input constraints, numbers x and y are positive).

2) Also, if we have already created a number then there is no reason to create it again. i.e. we maintain a visited array.

## C++

`// C++ program to find minimum number of steps needed ` `// to covert a number x into y with two operations ` `// allowed : (1) multiplication with 2 (2) subtraction ` `// with 1. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// A node of BFS traversal ` `struct` `node ` `{ ` ` ` `int` `val; ` ` ` `int` `level; ` `}; ` ` ` `// Returns minimum number of operations ` `// needed to covert x into y using BFS ` `int` `minOperations(` `int` `x, ` `int` `y) ` `{ ` ` ` `// To keep track of visited numbers ` ` ` `// in BFS. ` ` ` `set<` `int` `> visit; ` ` ` ` ` `// Create a queue and enqueue x into it. ` ` ` `queue<node> q; ` ` ` `node n = {x, 0}; ` ` ` `q.push(n); ` ` ` ` ` ` ` `// Do BFS starting from x ` ` ` `while` `(!q.empty()) ` ` ` `{ ` ` ` `// Remove an item from queue ` ` ` `node t = q.front(); ` ` ` `q.pop(); ` ` ` ` ` `// If the removed item is target ` ` ` `// number y, return its level ` ` ` `if` `(t.val == y) ` ` ` `return` `t.level; ` ` ` ` ` `// Mark dequeued number as visited ` ` ` `visit.insert(t.val); ` ` ` ` ` `// If we can reach y in one more step ` ` ` `if` `(t.val*2 == y || t.val-1 == y) ` ` ` `return` `t.level+1; ` ` ` ` ` `// Insert children of t if not visited ` ` ` `// already ` ` ` `if` `(visit.find(t.val*2) == visit.end()) ` ` ` `{ ` ` ` `n.val = t.val*2; ` ` ` `n.level = t.level+1; ` ` ` `q.push(n); ` ` ` `} ` ` ` `if` `(t.val-1>=0 && visit.find(t.val-1) == visit.end()) ` ` ` `{ ` ` ` `n.val = t.val-1; ` ` ` `n.level = t.level+1; ` ` ` `q.push(n); ` ` ` `} ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `x = 4, y = 7; ` ` ` `cout << minOperations(x, y); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find minimum ` `// number of steps needed to ` `// covert a number x into y ` `// with two operations allowed : ` `// (1) multiplication with 2 ` `// (2) subtraction with 1. ` ` ` `import` `java.util.HashSet; ` `import` `java.util.LinkedList; ` `import` `java.util.Set; ` ` ` `class` `GFG ` `{ ` ` ` `int` `val; ` ` ` `int` `steps; ` ` ` ` ` `public` `GFG(` `int` `val, ` `int` `steps) ` ` ` `{ ` ` ` `this` `.val = val; ` ` ` `this` `.steps = steps; ` ` ` `} ` `} ` ` ` `public` `class` `GeeksForGeeks ` `{ ` ` ` `private` `static` `int` `minOperations(` `int` `src, ` ` ` `int` `target) ` ` ` `{ ` ` ` ` ` `Set<GFG> visited = ` `new` `HashSet<>(` `1000` `); ` ` ` `LinkedList<GFG> queue = ` `new` `LinkedList<GFG>(); ` ` ` ` ` `GFG node = ` `new` `GFG(src, ` `0` `); ` ` ` ` ` `queue.offer(node); ` ` ` `visited.add(node); ` ` ` ` ` `while` `(!queue.isEmpty()) ` ` ` `{ ` ` ` `GFG temp = queue.poll(); ` ` ` `visited.add(temp); ` ` ` ` ` `if` `(temp.val == target) ` ` ` `{ ` ` ` `return` `temp.steps; ` ` ` `} ` ` ` ` ` `int` `mul = temp.val * ` `2` `; ` ` ` `int` `sub = temp.val - ` `1` `; ` ` ` ` ` `// given constraints ` ` ` `if` `(mul > ` `0` `&& mul < ` `1000` `) ` ` ` `{ ` ` ` `GFG nodeMul = ` `new` `GFG(mul, temp.steps + ` `1` `); ` ` ` `queue.offer(nodeMul); ` ` ` `} ` ` ` `if` `(sub > ` `0` `&& sub < ` `1000` `) ` ` ` `{ ` ` ` `GFG nodeSub = ` `new` `GFG(sub, temp.steps + ` `1` `); ` ` ` `queue.offer(nodeSub); ` ` ` `} ` ` ` `} ` ` ` `return` `-` `1` `; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `// int x = 2, y = 5; ` ` ` `int` `x = ` `4` `, y = ` `7` `; ` ` ` `GFG src = ` `new` `GFG(x, y); ` ` ` `System.out.println(minOperations(x, y)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Rahul ` |

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**Output :**

2

This article is contributed by **Vipin Khushu**. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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