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

• Difficulty Level : Medium
• Last Updated : 06 Oct, 2021

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

1. Multiply number by 2.
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
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 ``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 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 visited = ``new` `HashSet<>(``1000``);``        ``LinkedList queue = ``new` `LinkedList();` `        ``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`

## 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.``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG {``    ``public` `int` `val;``    ``public` `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)``    ``{` `        ``HashSet visited = ``new` `HashSet(1000);``        ``List queue = ``new` `List();` `        ``GFG node = ``new` `GFG(src, 0);` `        ``queue.Add(node);``        ``visited.Add(node);` `        ``while` `(queue.Count != 0) {``            ``GFG temp = queue;``            ``queue.RemoveAt(0);``            ``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.Add(nodeMul);``            ``}``            ``if` `(sub > 0 && sub < 1000) {``                ``GFG nodeSub = ``new` `GFG(sub, temp.steps + 1);``                ``queue.Add(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);``        ``Console.WriteLine(minOperations(x, y));``    ``}``}` `// This code is contributed by aashish1995`

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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Optimized solution

In the second approach, we will check the least most bit of the number and take a decision according to the value of that bit.

Instead of converting x into y, we will convert y into x and will reverse the operations which will take the same number of operations as converting x into y.

So, reversed operations for y will be:

1. Divide number by 2
2. Increment number by 1

## C++14

 `#include ``using` `namespace` `std;` `int` `min_operations(``int` `x, ``int` `y) {` `    ``// If both are equal then return 0``    ``if` `(x == y)``        ``return` `0;` `    ``// Check if conversion is possible or not``    ``if` `(x <= 0 && y > 0)``        ``return` `-1;` `    ``// If x > y then we can just increase y by 1``    ``// Therefore return the number of increments required``    ``if` `(x > y)``        ``return` `x - y;` `    ``// If last bit is odd``    ``// then increment y so that we can make it even``    ``if` `(y & 1)``        ``return` `1 + min_operations(x, y + 1);` `    ``// If y is even then divide it by 2 to make it closer to``    ``// x``    ``else``        ``return` `1 + min_operations(x, y / 2);``}` `// Driver code``signed` `main() {``    ``cout << min_operations(4, 7) << endl;``    ``return` `0;``}`

## Python3

 `def` `min_operations(x, y):``    ``# If both are equal then return 0``    ``if` `x ``=``=` `y:``        ``return` `0` `    ``# Check if conversion is possible or not``    ``if` `x <``=` `0` `and` `y > ``0``:``        ``return` `-``1` `    ``# If x > y then we can just increase y by 1``    ``# Therefore return the number of increments required``    ``if` `x > y:``        ``return` `a``-``b` `    ``# If last bit is odd``    ``# then increment y so that we can make it even``    ``if` `y & ``1` `=``=` `1``:``        ``return` `1``+``min_operations(x, y``+``1``)` `    ``# If y is even then divide it by 2 to make it closer to x``    ``else``:``        ``return` `1``+``min_operations(x, y``/``/``2``)`  `# Driver code``print``(min_operations(``4``, ``7``))`
Output
```2
```

The optimized solution is contributed by BurningTiles. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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