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.

1 <= x, y <= 10000


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.
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};

    // Do BFS starting from x
    while (!q.empty())
        // Remove an item from queue
        node t = q.front();

        // If the removed item is target
        // number y, return its level
        if (t.val == y)
            return t.level;

        // Mark dequeued number as visited

        // 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;
        if (t.val-1>=0 && visit.find(t.val-1) == visit.end())
            n.val = t.val-1;
            n.level = t.level+1;

// Driver code
int main()
    int x = 4, y = 7;
    cout << minOperations(x, y);
    return 0;

Output :


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 See your article appearing on the GeeksforGeeks main page and help other Geeks.

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