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 <= 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. 

Implementation:

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<Integer> visited = new HashSet<>(1000);
        LinkedList<GFG> queue = new LinkedList<GFG>();

        GFG node = new GFG(src, 0);

        queue.offer(node);

        while (!queue.isEmpty()) {
            GFG temp = queue.poll();
            if(visited.contains(temp.val)) {
              continue;
            }
            visited.add(temp.val);

            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<GFG> visited = new HashSet<GFG>(1000);
        List<GFG> queue = new List<GFG>();

        GFG node = new GFG(src, 0);

        queue.Add(node);
        visited.Add(node);

        while (queue.Count != 0) {
            GFG temp = queue[0];
            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

Javascript

// JavaScript 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.

// A node of BFS traversal
class node {
constructor(val, level) {
this.val = val;
this.level = level;
}
}

// Returns minimum number of operations
// needed to convert x into y using BFS
function minOperations(x, y) {

// To keep track of visited numbers
// in BFS.
const visit = new Set();

// Create a queue and enqueue x into it.
const q = [];
const n = new node(x, 0);
q.push(n);

// Do BFS starting from x
while (q.length > 0) {


// Remove an item from queue
const t = q.shift();

// 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 || t.val - 1 == y) {
  return t.level + 1;
}

// Insert children of t if not visited
// already
if (!visit.has(t.val * 2)) {
  n.val = t.val * 2;
  n.level = t.level + 1;
  q.push(Object.assign({}, n));
}
if (t.val - 1 >= 0 && !visit.has(t.val - 1)) {
  n.val = t.val - 1;
  n.level = t.level + 1;
  q.push(Object.assign({}, n));
}
}
}

// Driver code
const x = 4;
const y = 7;
console.log(minOperations(x, y));

// This code is contributed by lokeshpotta20.

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

Implementation:

C++14

#include <iostream>
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;
}

C

#include <stdio.h>

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()
{
    printf("%d", min_operations(4, 7));
    return 0;
}

// This code is contributed by Rohit Pradhan

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG
{
    static int minOperations(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 % 2 != 0)
            return 1 + minOperations(x, y + 1);

        // If y is even then divide it by 2 to make it
        // closer to x
        else
            return 1 + minOperations(x, y / 2);
    }

    public static void main(String[] args)
    {
        System.out.println(minOperations(4, 7));
    }
}

// This code is contributed by Shobhit Yadav

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))

C#

using System;
class GFG {

  static 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 % 2 == 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
  public static int Main()
  {
    Console.WriteLine(min_operations(4, 7));
    return 0;
  }
}

// This code is contributed by Taranpreet

Javascript

<script>

function 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 && 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
document.write(min_operations(4, 7));

// This code is contributed by Taranpreet
</script>

Output

2

Time complexity:O(Y-X), where X, Y is the given number in the problem. 

Space complexity: O(1), since no extra space used.

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.