Minimum number of moves to make M and N equal by repeatedly adding any divisor of number to itself except 1 and the number

Given two numbers N and M, the task is to find the minimum number of moves to change N to M or -1 if it’s impossible. In one move, add to the current number any of its divisors not equal to 1 and the number itself.

Examples:

Input: N = 4, M = 24
Output: 5
Explanation: the number 24 can be reached starting from 4 using 5 operations: 4->6->8->12->18->24.

Input: N = 4, M = 576
Output: 14

Approach: Build a graph where vertices are numbers from N to M and there is an edge from vertex v1 to vertex v2 if v2 can be obtained from v1 using exactly one operation from the problem statement. To solve the problem, find the shortest path from vertex N to vertex M in this graph. This can be done using the breadth first search algorithm. Follow the steps below to solve the problem: 

• Initialize a boolean array visited[] to store whether a particular number is counted or not.
• Fill all the indices of the bool array visited[] with the value false and set the value of visited[N] to true.
• Initialize a queue of pairs q to store the number visited and the number of operations done.
• Push the pair {N, 0} into the queue of pairs q.
• Iterate in a range till the queue of pairs q is not empty.
  • Initialize the variables aux as the first value of the front pair of the queue q and cont as the second value of the front pair of the queue q.
  • Pop the front pair from the queue of pairs q.
  • If aux is equal to M, then, return the value of cont.
  • Iterate in a range [2, aux^1/2] and perform the following steps.
    • If i is a factor of aux, then take the following steps.
      • If aux+i is less than equal to M and visited[i+aux] is false, then set the value of visited[aux+i] to true and push the pair {aux+i, cont+1} into the queue of pairs q.
      • If aux+aux/i is less than equal to M and visited[aux/i+aux] is false, then set the value of visited[aux+aux/i] to true and push the pair {aux+aux/i, cont+1} into the queue of pairs q.
• Return -1 as it is impossible to make N equal to M.

Below is the implementation of the above approach.

C++

// C++ program for the above approach.
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum number
// of moves to make N and M equal.
int countOperations(int N, int M)
{
    // Array to maintain the numbers
    // included.
    bool visited[100001];
    fill(visited, visited + 100001, false);

    // pair of vertex, count
    queue<pair<int, int> > Q;
    Q.push(make_pair(N, 0));
    visited[N] = true;

    // run bfs from N
    while (!Q.empty()) {
        int aux = Q.front().first;
        int cont = Q.front().second;
        Q.pop();

        // if we reached goal
        if (aux == M)
            return cont;

        // Iterate in the range
        for (int i = 2; i * i <= aux; i++)
            // If i is a factor of aux
            if (aux % i == 0) {
                // If i is less than M-aux and
                // is not included earlier.
                if (aux + i <= M && !visited[aux + i]) {
                    Q.push(make_pair(aux + i, cont + 1));
                    visited[aux + i] = true;
                }
                // If aux/i is less than M-aux and
                // is not included earlier.
                if (aux + aux / i <= M
                    && !visited[aux + aux / i]) {
                    Q.push(
                        make_pair(aux + aux / i, cont + 1));
                    visited[aux + aux / i] = true;
                }
            }
    }

    // Not possible
    return -1;
}

// Driver Code
int main()
{

    int N = 4, M = 24;

    cout << countOperations(N, M);

    return 0;
}

Java

// Java program for above approach
import java.util.*;

class GFG{

static class pair<T, V>
{
    T first;
    V second;
}

static pair<Integer, Integer> make_pair(int f, int s)
{
    pair<Integer, Integer> p = new pair<>();
    p.first = f; p.second = s;
    return p;
}

// Function to find the minimum number
// of moves to make N and M equal.
static int countOperations(int N, int M)
{

    // Array to maintain the numbers
    // included.
    boolean[] visited = new boolean[100001];
    Arrays.fill(visited, false);

    // Pair of vertex, count
    Queue<pair<Integer, Integer>> Q = new LinkedList<>();
    Q.add(make_pair(N, 0));
    visited[N] = true;

    // Run bfs from N
    while (!Q.isEmpty())
    {
        int aux = Q.peek().first;
        int cont = Q.peek().second;
        Q.remove();

        // If we reached goal
        if (aux == M)
            return cont;

        // Iterate in the range
        for(int i = 2; i * i <= aux; i++)

            // If i is a factor of aux
            if (aux % i == 0)
            {

                // If i is less than M-aux and
                // is not included earlier.
                if (aux + i <= M && !visited[aux + i])
                {
                    Q.add(make_pair(aux + i, cont + 1));
                    visited[aux + i] = true;
                }

                // If aux/i is less than M-aux and
                // is not included earlier.
                if (aux + aux / i <= M &&
                    !visited[aux + aux / i])
                {
                    Q.add(make_pair(aux + aux / i,
                                   cont + 1));
                    visited[aux + aux / i] = true;
                }
            }
    }

    // Not possible
    return -1;
}

// Driver Code
public static void main(String[] args)
{
    int N = 4, M = 24;

    System.out.println(countOperations(N, M));
}
}

// This code is contributed by hritikrommie

Python3

# Python3 program for the above approach.

# Function to find the minimum number
# of moves to make N and M equal.
def countOperations(N, M):

    # Array to maintain the numbers
    # included.
    visited = [False] * (100001)

    # Pair of vertex, count
    Q = []
    Q.append([N, 0])
    visited[N] = True

    # Run bfs from N
    while (len(Q) > 0):
        aux = Q[0][0]
        cont = Q[0][1]
        Q.pop(0)

        # If we reached goal
        if (aux == M):
            return cont

        # Iterate in the range
        i = 2

        while i * i <= aux:

            # If i is a factor of aux
            if (aux % i == 0):

                # If i is less than M-aux and
                # is not included earlier.
                if (aux + i <= M and not visited[aux + i]):
                    Q.append([aux + i, cont + 1])
                    visited[aux + i] = True

                # If aux/i is less than M-aux and
                # is not included earlier.
                if (aux + int(aux / i) <= M and not
                    visited[aux + int(aux / i)]):
                    Q.append([aux + int(aux / i), cont + 1])
                    visited[aux + int(aux / i)] = True

            i += 1

    # Not possible
    return -1

# Driver code
N, M = 4, 24

print(countOperations(N, M))

# This code is contributed by mukesh07

C#

// C# program for the above approach.
using System;
using System.Collections;
class GFG
{

    // Function to find the minimum number
    // of moves to make N and M equal.
    static int countOperations(int N, int M)
    {
        // Array to maintain the numbers
        // included.
        bool[] visited = new bool[100001];

        // pair of vertex, count
        Queue Q = new Queue();
        Q.Enqueue(new Tuple<int, int>(N, 0));
        visited[N] = true;
        // run bfs from N
        while (Q.Count > 0) {
            int aux = ((Tuple<int,int>)(Q.Peek())).Item1;
            int cont = ((Tuple<int,int>)(Q.Peek())).Item2;
            Q.Dequeue();

            // if we reached goal
            if (aux == M)
                return cont;

            // Iterate in the range
            for (int i = 2; i * i <= aux; i++)

                // If i is a factor of aux
                if (aux % i == 0)
                {

                    // If i is less than M-aux and
                    // is not included earlier.
                    if (aux + i <= M && !visited[aux + i]) {
                        Q.Enqueue(new Tuple<int, int>(aux + i, cont + 1));
                        visited[aux + i] = true;
                    }

                    // If aux/i is less than M-aux and
                    // is not included earlier.
                    if (aux + aux / i <= M
                        && !visited[aux + aux / i]) {
                        Q.Enqueue(new Tuple<int, int>(aux + aux / i, cont + 1));
                        visited[aux + aux / i] = true;
                    }
                }
        }

        // Not possible
        return -1;
    }

  static void Main ()
  {
    int N = 4, M = 24;

    Console.WriteLine(countOperations(N, M));
  }
}

// This code is contributed by suresh07.

Javascript

<script>
    // Javascript program for the above approach.

    // Function to find the minimum number
    // of moves to make N and M equal.
    function countOperations(N, M)
    {

        // Array to maintain the numbers
        // included.
        let visited = new Array(100001);

        // pair of vertex, count
        let Q = [];
        Q.push([N, 0]);
        visited[N] = true;

        // run bfs from N
        while (Q.length > 0) {
            let aux = Q[0][0];
            let cont = Q[0][1];
            Q.shift();

            // if we reached goal
            if (aux == M)
                return cont;

            // Iterate in the range
            for (let i = 2; i * i <= aux; i++)

                // If i is a factor of aux
                if (aux % i == 0)
                {

                    // If i is less than M-aux and
                    // is not included earlier.
                    if (aux + i <= M && !visited[aux + i]) {
                        Q.push([aux + i, cont + 1]);
                        visited[aux + i] = true;
                    }

                    // If aux/i is less than M-aux and
                    // is not included earlier.
                    if (aux + parseInt(aux / i, 10) <= M
                        && !visited[aux + parseInt(aux / i, 10)]) {
                        Q.push([aux + parseInt(aux / i, 10), cont + 1]);
                        visited[aux + parseInt(aux / i, 10)] = true;
                    }
                }
        }

        // Not possible
        return -1;
    }

    let N = 4, M = 24;

    document.write(countOperations(N, M));

// This code is contributed by rameshtravel07.
</script>

Output

5

Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(N)

Another Approach: Use a more efficient way to generate the factors of a number. Instead of iterating over all numbers in the range [2, aux1/2] to check if they divide the number, we can use a more optimized approach to generate factors.

We can observe that every factor of a number n lies in the range [2, sqrt(n)]. Therefore, we only need to iterate over this range and check if the current number is a factor of n. If it is, we can add both the number and n/number to the queue of pairs.

1. Define the problem:
   Given two integers N and M, we need to find the minimum number of moves required to change N to M. In one move, we can add to the current number any of its divisors not equal to 1 and the number itself.
2. Build a graph:
   Build a graph where the vertices are the numbers from N to M and there is an edge from vertex v1 to vertex v2 if v2 can be obtained from v1 using exactly one operation from the problem statement.
3. Find the shortest path:
   To solve the problem, we need to find the shortest path from vertex N to vertex M in this graph. This can be done using the breadth-first search (BFS) algorithm.
4. Initialize the boolean array:
   Initialize a boolean array visited[] to store whether a particular number is counted or not. Fill all the indices of the bool array visited[] with the value false and set the value of visited[N] to true.
5. Initialize the queue:
   Initialize a queue of pairs q to store the number visited and the number of operations done. Push the pair {N, 0} into the queue of pairs q.
6. Iterate through the queue:
   Iterate in a range till the queue of pairs q is not empty.
7. Dequeue and set variables:
   Initialize the variables aux as the first value of the front pair of the queue q and cont as the second value of the front pair of the queue q. Pop the front pair from the queue of pairs q.
8. Check if we have reached the goal:
   If aux is equal to M, then return the value of cont, which is the minimum number of moves required to change N to M.
9. Iterate through the range:
   Iterate in a range [2, aux/2] and perform the following steps.
10. Check if i is a factor of aux:
    If i is a factor of aux, then take the following steps.
11. Check if aux+i is less than equal to M and not visited:
    If aux+i is less than equal to M and visited[aux+i] is false, then set the value of visited[aux+i] to true and push the pair {aux+i, cont+1} into the queue of pairs q.
12. Check if aux+aux/i is less than equal to M and not visited:
    If aux+aux/i is less than equal to M and visited[aux+aux/i] is false, then set the value of visited[aux+aux/i] to true and push the pair {aux+aux/i, cont+1} into the queue of pairs q.
13. Return -1 if impossible:
    If we cannot make N equal to M, return -1.
14. Output the result:
    The output of the program is the minimum number of moves required to change N to M.

C++

// C++ program for the above approach.
#include <bits/stdc++.h>
using namespace std;
int countOperations(int N, int M)
{
    // Array to maintain the numbers
    // included.
    bool visited[100001];
    fill(visited, visited + 100001, false);

    // pair of vertex, count
    queue<pair<int, int> > Q;
    Q.push(make_pair(N, 0));
    visited[N] = true;

    // run bfs from N
    while (!Q.empty()) {
        int aux = Q.front().first;
        int cont = Q.front().second;
        Q.pop();

        // if we reached goal
        if (aux == M)
            return cont;

        // Iterate over factors
        for (int i = 2; i * i <= aux; i++) {
            if (aux % i == 0) {
                int factor1 = i;
                int factor2 = aux / i;
                if (aux + factor1 <= M && !visited[aux + factor1]) {
                    Q.push(make_pair(aux + factor1, cont + 1));
                    visited[aux + factor1] = true;
                }
                if (factor1 != factor2 && aux + factor2 <= M && !visited[aux + factor2]) {
                    Q.push(make_pair(aux + factor2, cont + 1));
                    visited[aux + factor2] = true;
                }
            }
        }
    }

    // Not possible
    return -1;
}
// Driver Code
int main()
{

    int N = 4, M = 24;

    cout << countOperations(N, M);

    return 0;
}

Java

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

public class MinMovesToChangeNToM {

    // Function to find the minimum number of moves to change N to M
    // using only allowed operations as described in the problem statement
    public static int countOperations(int N, int M) {
        boolean[] visited = new boolean[M + 1];
        Arrays.fill(visited, false);

        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{N, 0});
        visited[N] = true;

        while (!queue.isEmpty()) {
            int[] pair = queue.poll();
            int aux = pair[0];
            int cont = pair[1];

            if (aux == M) {
                return cont;
            }

            for (int i = 2; i * i <= aux; i++) {
                if (aux % i == 0) {
                    int sum = aux + i;
                    if (sum <= M && !visited[sum]) {
                        queue.offer(new int[]{sum, cont + 1});
                        visited[sum] = true;
                    }
                    sum = aux + aux / i;
                    if (sum <= M && !visited[sum]) {
                        queue.offer(new int[]{sum, cont + 1});
                        visited[sum] = true;
                    }
                }
            }
        }

        // Not possible
        return -1;
    }

    public static void main(String[] args) {
        int N = 4, M = 24;
        System.out.println(countOperations(N, M));
    }
}

C#

using System;
using System.Collections.Generic;

public class Solution
{
    public static int CountOperations(int N, int M)
    {
        // Array to maintain the numbers included.
        bool[] visited = new bool[100001];
        Array.Fill(visited, false);

        // Queue of vertex-count pairs.
        Queue<Tuple<int, int>> Q = new Queue<Tuple<int, int>>();
        Q.Enqueue(new Tuple<int, int>(N, 0));
        visited[N] = true;

        // Run BFS from N.
        while (Q.Count > 0)
        {
            Tuple<int, int> front = Q.Dequeue();
            int aux = front.Item1;
            int cont = front.Item2;

            // If we reached the goal.
            if (aux == M)
            {
                return cont;
            }

            // Iterate over divisors.
            for (int i = 2; i * i <= aux; i++)
            {
                // If i is a factor of aux.
                if (aux % i == 0)
                {
                    // If i is less than M-aux and is not included earlier.
                    if (aux + i <= M && !visited[aux + i])
                    {
                        Q.Enqueue(new Tuple<int, int>(aux + i, cont + 1));
                        visited[aux + i] = true;
                    }

                    // If aux/i is less than M-aux and is not included earlier.
                    if (aux + aux / i <= M && !visited[aux + aux / i])
                    {
                        Q.Enqueue(new Tuple<int, int>(aux + aux / i, cont + 1));
                        visited[aux + aux / i] = true;
                    }
                }
            }
        }

        // Not possible.
        return -1;
    }

    public static void Main()
    {
        int N = 4, M = 24;

        Console.WriteLine(CountOperations(N, M));
    }
}

Python3

def count_operations(N: int, M: int) -> int:
    # Array to maintain the numbers included.
    visited = [False] * (M + 1)

# pair of vertex, count
    Q = [(N, 0)]
    visited[N] = True

# run bfs from N
    while Q:
        aux, cont = Q.pop(0)

# if we reached goal
        if aux == M:
            return cont

# Iterate in the range
        for i in range(2, int(aux ** 0.5) + 1):
            # If i is a factor of aux
            if aux % i == 0:
                # If i is less than M-aux and is not included earlier.
                if aux + i <= M and not visited[aux + i]:
                    Q.append((aux + i, cont + 1))
                    visited[aux + i] = True
                # If aux/i is less than M-aux and is not included earlier.
                if aux + aux // i <= M and not visited[aux + aux // i]:
                    Q.append((aux + aux // i, cont + 1))
                    visited[aux + aux // i] = True

# Not possible
    return -1

# Example usage
N = 4
M = 24
print(count_operations(N, M)) # Output: 5

Javascript

function countOperations(N, M) {
  // Array to maintain the numbers included
  const visited = new Array(M + 1).fill(false);

  // Queue of pairs [vertex, count]
  const queue = [[N, 0]];
  visited[N] = true;

  // Run BFS from N
  while (queue.length) {
    const [aux, cont] = queue.shift();

    // If we reached goal
    if (aux === M) {
      return cont;
    }

    // Iterate in the range
    for (let i = 2; i * i <= aux; i++) {
      // If i is a factor of aux
      if (aux % i === 0) {
        // If i is less than M-aux and is not included earlier
        if (aux + i <= M && !visited[aux + i]) {
          queue.push([aux + i, cont + 1]);
          visited[aux + i] = true;
        }
        // If aux/i is less than M-aux and is not included earlier
        if (aux + aux / i <= M && !visited[aux + aux / i]) {
          queue.push([aux + aux / i, cont + 1]);
          visited[aux + aux / i] = true;
        }
      }
    }
  }

  // Not possible
  return -1;
}

// Example usage:
console.log(countOperations(4, 24)); // Output: 5
console.log(countOperations(4, 576)); // Output: 14

Output

5

Time Complaxity: O(N*logN)

Auxiliary Space: O(N)