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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:  

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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. 



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

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 

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 <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;
}

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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