Número mínimo de operaciones requeridas para convertir el número x en y

Dado un número inicial x y dos operaciones que se dan a continuación: 

1. Multiplica el número por 2.
2. Resta 1 del número.

La tarea es averiguar el número mínimo de operaciones requeridas para convertir el número x en y usando solo las dos operaciones anteriores. Podemos aplicar estas operaciones cualquier número de veces.
Restricciones: 
1 <= x, y <= 1000

Ejemplo:  

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.

La idea es usar BFS para esto. Ejecutamos un BFS y creamos Nodes multiplicando por 2 y restando por 1, por lo que podemos obtener todos los números posibles alcanzables desde el número inicial. 
Puntos importantes : 

1. Cuando restamos 1 de un número y se convierte en <0, es decir, Negativo, entonces no hay razón para crear el siguiente Node a partir de él (según las restricciones de entrada, los números x e y son positivos). 
2. Además, si ya hemos creado un número, no hay motivo para volver a crearlo. es decir, mantenemos una array visitada. 

Implementación:

// 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 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 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# 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

2

Este artículo es una contribución de Vipin Khushu . Si le gusta GeeksforGeeks y le gustaría contribuir, también puede escribir un artículo y enviarlo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.

Solución optimizada:

En el segundo enfoque, verificaremos el bit más pequeño del número y tomaremos una decisión de acuerdo con el valor de ese bit.

En lugar de convertir x en y, convertiremos y en x e invertiremos las operaciones, lo que requerirá el mismo número de operaciones que para convertir x en y.

Entonces, las operaciones inversas para y serán:

1. Dividir número por 2
2. Incrementar el número en 1

Implementación:

#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;
}

#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

/*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

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))

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

<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>

2

La solución optimizada es aportada por BurningTiles. Si le gusta GeeksforGeeks y le gustaría contribuir, también puede escribir un artículo y enviarlo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.