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 <= 1000**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 convert 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 convert 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;` `}` |

## Java

`// Java program to find minimum ` `// number of steps needed to` `// convert 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` |

## Python3

`# Python3 program to find minimum number of ` `# steps needed to convert a number x into y ` `# with two operations allowed : ` `# (1) multiplication with 2 ` `# (2) subtraction with 1. ` `import` `queue` `# A node of BFS traversal ` `class` `node:` ` ` `def` `__init__(` `self` `, val, level):` ` ` `self` `.val ` `=` `val ` ` ` `self` `.level ` `=` `level` `# Returns minimum number of operations ` `# needed to convert x into y using BFS ` `def` `minOperations(x, y):` ` ` ` ` `# To keep track of visited numbers ` ` ` `# in BFS. ` ` ` `visit ` `=` `set` `() ` ` ` `# Create a queue and enqueue x into it. ` ` ` `q ` `=` `queue.Queue()` ` ` `n ` `=` `node(x, ` `0` `) ` ` ` `q.put(n) ` ` ` `# Do BFS starting from x ` ` ` `while` `(` `not` `q.empty()):` ` ` ` ` `# Remove an item from queue ` ` ` `t ` `=` `q.get() ` ` ` `# If the removed item is target ` ` ` `# number y, return its level ` ` ` `if` `(t.val ` `=` `=` `y):` ` ` `return` `t.level ` ` ` `# Mark dequeued number as visited ` ` ` `visit.add(t.val) ` ` ` `# If we can reach y in one more step ` ` ` `if` `(t.val ` `*` `2` `=` `=` `y ` `or` `t.val ` `-` `1` `=` `=` `y): ` ` ` `return` `t.level` `+` `1` ` ` `# Insert children of t if not visited ` ` ` `# already ` ` ` `if` `(t.val ` `*` `2` `not` `in` `visit):` ` ` `n.val ` `=` `t.val ` `*` `2` ` ` `n.level ` `=` `t.level ` `+` `1` ` ` `q.put(n)` ` ` `if` `(t.val ` `-` `1` `>` `=` `0` `and` `t.val ` `-` `1` `not` `in` `visit):` ` ` `n.val ` `=` `t.val ` `-` `1` ` ` `n.level ` `=` `t.level ` `+` `1` ` ` `q.put(n)` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `x ` `=` `4` ` ` `y ` `=` `7` ` ` `print` `(minOperations(x, y))` `# This code is contributed by PranchalK` |

**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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