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.

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

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