Minimum number of operation required to convert number x into y

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 <= 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.

Below is C++ implementation of above idea.

// 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;
}

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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