Minimum steps to reach end from start by performing multiplication and mod operations with array elements

Given start, end and an array of N numbers. At each step, start is multiplied with any number in the array and then mod operation with 100000 is done to get the new start. The task is to find the minimum steps in which end can be achieved starting from start. 
Examples:

Input: start = 3 end = 30 a[] = {2, 5, 7} 
Output: 2 
Step 1: 3*2 = 6 % 100000 = 6 
Step 2: 6*5 = 30 % 100000 = 30 
Input: start = 7 end = 66175 a[] = {3, 4, 65} 
Output: 4 
Step 1: 7*3 = 21 % 100000 = 21 
Step 2: 21*3 = 6 % 100000 = 63 
Step 3: 63*65 = 4095 % 100000 = 4095 
Step 4: 4095*65 = 266175 % 100000 = 66175 

Approach: Since in the above problem the modulus given is 100000, therefore the maximum number of states will be 10⁵. All the states can be checked using simple BFS. Initialize an ans[] array with -1 which marks that the state has not been visited. ans[i] stores the number of steps taken to reach i from start. Initially push the start to the queue, then apply BFS. Pop the top element and check if it is equal to the end, if it is then print the ans[end]. If the element is not equal to the topmost element, then multiply top element with every element in the array and perform a mod operation. If the multiplied element state has not been visited previously, then push it into the queue. Initialize ans[pushed_element] by ans[top_element] + 1. Once all the states are visited, and the state cannot be reached by performing every possible multiplication, then print -1. 
Below is the implementation of the above approach: 

C++

// C++ program to find the minimum steps
// to reach end from start by performing
// multiplications and mod operations with array elements
#include <bits/stdc++.h>
using namespace std;

// Function that returns the minimum operations
int minimumMulitplications(int start, int end, int a[], int n)
{
    // array which stores the minimum steps
    // to reach i from start
    int ans[100001];

    // -1 indicated the state has not been visited
    memset(ans, -1, sizeof(ans));
    int mod = 100000;

    // queue to store all possible states
    queue<int> q;

    // initially push the start
    q.push(start % mod);

    // to reach start we require 0 steps
    ans[start] = 0;

    // till all states are visited
    while (!q.empty()) {

        // get the topmost element in the queue
        int top = q.front();

        // pop the topmost element
        q.pop();

        // if the topmost element is end
        if (top == end)
            return ans[end];

        // perform multiplication with all array elements
        for (int i = 0; i < n; i++) {
            int pushed = top * a[i];
            pushed = pushed % mod;

            // if not visited, then push it to queue
            if (ans[pushed] == -1) {
                ans[pushed] = ans[top] + 1;
                q.push(pushed);
            }
        }
    }
    return -1;
}

// Driver Code
int main()
{
    int start = 7, end = 66175;
    int a[] = { 3, 4, 65 };
    int n = sizeof(a) / sizeof(a[0]);

    // Calling function
    cout << minimumMulitplications(start, end, a, n);
    return 0;
}

Java

// Java program to find the minimum steps
// to reach end from start by performing
// multiplications and mod operations with array elements

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

class GFG {

// Function that returns the minimum operations
    static int minimumMulitplications(int start, int end, int a[], int n) {
        // array which stores the minimum steps
        // to reach i from start
        int ans[] = new int[100001];

        // -1 indicated the state has not been visited
        Arrays.fill(ans, -1);
        int mod = 100000;

        // queue to store all possible states
        Queue<Integer> q = new LinkedList<>();

        // initially push the start
        q.add(start % mod);

        // to reach start we require 0 steps
        ans[start] = 0;

        // till all states are visited
        while (!q.isEmpty()) {

            // get the topmost element in the queue
            int top = q.peek();

            // pop the topmost element
            q.remove();

            // if the topmost element is end
            if (top == end) {
                return ans[end];
            }

            // perform multiplication with all array elements
            for (int i = 0; i < n; i++) {
                int pushed = top * a[i];
                pushed = pushed % mod;

                // if not visited, then push it to queue
                if (ans[pushed] == -1) {
                    ans[pushed] = ans[top] + 1;
                    q.add(pushed);
                }
            }
        }
        return -1;
    }

// Driver Code
    public static void main(String args[]) {
        int start = 7, end = 66175;
        int a[] = {3, 4, 65};
        int n = a.length;

        // Calling function
        System.out.println(minimumMulitplications(start, end, a, n));

    }
}

// This code is contributed by PrinciRaj19992

Python3

# Python3 program to find the minimum steps
# to reach end from start by performing
# multiplications and mod operations with
# array elements
from collections import deque

# Function that returns the minimum operations
def minimumMulitplications(start, end, a, n):

    # array which stores the minimum
    # steps to reach i from start
    ans = [-1 for i in range(100001)]

    # -1 indicated the state has
    # not been visited
    mod = 100000

    q = deque()

    # queue to store all possible states
    # initially push the start
    q.append(start % mod)

    # to reach start we require 0 steps
    ans[start] = 0

    # till all states are visited
    while (len(q) > 0):

        # get the topmost element in the
        # queue, pop the topmost element
        top = q.popleft()

        # if the topmost element is end
        if (top == end):
            return ans[end]

        # perform multiplication with
        # all array elements
        for i in range(n):
            pushed = top * a[i]
            pushed = pushed % mod

            # if not visited, then push it to queue
            if (ans[pushed] == -1):
                ans[pushed] = ans[top] + 1
                q.append(pushed)

    return -1

# Driver Code
start = 7
end = 66175
a = [3, 4, 65]
n = len(a)

# Calling function
print(minimumMulitplications(start, end, a, n))

# This code is contributed by mohit kumar

C#

// C# program to find the minimum steps
// to reach end from start by performing
// multiplications and mod operations with array elements
using System;
using System.Collections.Generic;

class GFG
{

    // Function that returns the minimum operations
    static int minimumMulitplications(int start, int end,
                                            int []a, int n)
    {
        // array which stores the minimum steps
        // to reach i from start
        int []ans = new int[100001];

        // -1 indicated the state has not been visited
        for(int i = 0; i < ans.Length; i++)
            ans[i] = -1;
        int mod = 100000;

        // queue to store all possible states
        Queue<int> q = new Queue<int>();

        // initially push the start
        q.Enqueue(start % mod);

        // to reach start we require 0 steps
        ans[start] = 0;

        // till all states are visited
        while (q.Count != 0)
        {

            // get the topmost element in the queue
            int top = q.Peek();

            // pop the topmost element
            q.Dequeue();

            // if the topmost element is end
            if (top == end)
            {
                return ans[end];
            }

            // perform multiplication with all array elements
            for (int i = 0; i < n; i++)
            {
                int pushed = top * a[i];
                pushed = pushed % mod;

                // if not visited, then push it to queue
                if (ans[pushed] == -1)
                {
                    ans[pushed] = ans[top] + 1;
                    q.Enqueue(pushed);
                }
            }
        }
        return -1;
    }

    // Driver Code
    public static void Main(String []args)
    {
        int start = 7, end = 66175;
        int []a = {3, 4, 65};
        int n = a.Length;

        // Calling function
        Console.WriteLine(minimumMulitplications(start, end, a, n));

    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>
// Javascript program to find the minimum steps
// to reach end from start by performing
// multiplications and mod operations with array elements

    // Function that returns the minimum operations
    function minimumMulitplications(start,end,a,n)
    {
        // array which stores the minimum steps
        // to reach i from start
        let ans = new Array(100001);

        // -1 indicated the state has not been visited
        for(let i=0;i<ans.length;i++)
        {
            ans[i]=-1;
        }

        let mod = 100000;

        // queue to store all possible states
        let q = [];

        // initially push the start
        q.push(start % mod);

        // to reach start we require 0 steps
        ans[start] = 0;

        // till all states are visited
        while (q.length!=0) {

            // get the topmost element in the queue
            let top = q[0];

            // pop the topmost element
            q.shift();

            // if the topmost element is end
            if (top == end) {
                return ans[end];
            }

            // perform multiplication with all array elements
            for (let i = 0; i < n; i++) {
                let pushed = top * a[i];
                pushed = pushed % mod;

                // if not visited, then push it to queue
                if (ans[pushed] == -1) {
                    ans[pushed] = ans[top] + 1;
                    q.push(pushed);
                }
            }
        }
        return -1;
    }

    // Driver Code
    let start = 7, end = 66175;
    let a=[3, 4, 65];
    let n = a.length;

    // Calling function
    document.write(minimumMulitplications(start, end, a, n));


// This code is contributed by unknown2108
</script>

Output:

4

Time Complexity: O(n)

Auxiliary Space: O(n)

Another Approach: using Bidirectional Search

• Initialize two queues, one for start and one for end. Push start and end to their respective queues.
• Initialize two ans[] arrays for both start and end. ans[i] stores the number of steps taken to reach i from start or end, depending on which queue it belongs to.
• Initialize a visited[] array which marks if the state has been visited or not. visited[i] stores true if the state i has been visited.
• Initialize a commonNode variable to -1, indicating no common node has been found yet.
• While both queues are not empty, do the following:
  a. Choose the queue with the smaller size, and pop the front element. Let this element be x.
  b. If x is already visited, continue to the next element in the queue.
  c. For each element y in the array a[], calculate yx%mod. If this state is already visited by the other queue, then we have found a common node. Update commonNode to yx%mod and return ans[start] + ans[end].
  d. If this state has not been visited before, add it to the queue, mark it as visited, and update ans[] for this state.
• If no common node is found after visiting all possible states, return -1.
• Done

C++

// C++ program to find the minimum steps
// to reach end from start by performing
// multiplications and mod operations with array elements
#include <bits/stdc++.h>
using namespace std;

// Function that returns the minimum operations
int minimumMultiplications(int start, int end, int a[],
                           int n)
{
    // array which stores the minimum steps
    // to reach i from start or end
    int ans_start[100001], ans_end[100001];
    // visited array to keep track of visited states
    bool visited_start[100001], visited_end[100001];

    // -1 indicated the state has not been visited
    memset(ans_start, -1, sizeof(ans_start));
    memset(ans_end, -1, sizeof(ans_end));
    memset(visited_start, false, sizeof(visited_start));
    memset(visited_end, false, sizeof(visited_end));

    int mod = 100000;

    // queues to store all possible states
    queue<int> q_start, q_end;

    // initially push start and end
    q_start.push(start % mod);
    q_end.push(end % mod);

    // to reach start or end we require 0 steps
    ans_start[start] = 0;
    ans_end[end] = 0;

    // mark start and end as visited
    visited_start[start] = true;
    visited_end[end] = true;

    // variable to store common node
    int commonNode = -1;

    // till all states are visited or common node is found
    while (!q_start.empty() && !q_end.empty()) {

        // choose the queue with smaller size
        if (q_start.size() < q_end.size()) {
            int top = q_start.front();
            q_start.pop();

            // if the topmost element is end
            if (visited_end[top]) {
                commonNode = top;
                break;
            }

            // perform multiplication with all array
            // elements
            for (int i = 0; i < n; i++) {
                int pushed = top * a[i];
                pushed = pushed % mod;

                // if not visited, then push it to queue
                if (!visited_start[pushed]) {
                    ans_start[pushed] = ans_start[top] + 1;
                    q_start.push(pushed);
                    visited_start[pushed] = true;
                }
            }
        }
        else {
            int top = q_end.front();
            q_end.pop();

            // if the topmost element is start
            if (visited_start[top]) {
                commonNode = top;
                break;
            }

            // perform multiplication with all array
            // elements
            for (int i = 0; i < n; i++) {
                int pushed = top * a[i];
                pushed = pushed % mod;

                // if not visited, then push it to queue
                if (!visited_end[pushed]) {
                    ans_end[pushed] = ans_end[top] + 1;
                    q_end.push(pushed);
                    visited_end[pushed] = true;
                }
            }
        }
    }

    if (commonNode == -1)
        return -1;

    // return the sum of minimum steps from start and end to
    // common node
    return ans_start[commonNode] + ans_end[commonNode];
}

// Driver Code
int main()
{
    int start = 7, end = 66175;
    int a[] = { 3, 4, 65 };
    int n = sizeof(a) / sizeof(a[0]);
    // Calling function
    cout << minimumMultiplications(start, end, a, n);
    return 0;
}

Java

import java.util.*;

public class Main {
    // Function that returns the minimum operations
    static int minimumMultiplications(int start, int end, int a[], int n) {
        // array which stores the minimum steps
        // to reach i from start or end
        int ans_start[] = new int[100001];
        int ans_end[] = new int[100001];
        Arrays.fill(ans_start, -1);
        Arrays.fill(ans_end, -1);

        // visited array to keep track of visited states
        boolean visited_start[] = new boolean[100001];
        boolean visited_end[] = new boolean[100001];

        int mod = 100000;

        // queues to store all possible states
        Queue<Integer> q_start = new LinkedList<>();
        Queue<Integer> q_end = new LinkedList<>();

        // initially push start and end
        q_start.add(start % mod);
        q_end.add(end % mod);

        // to reach start or end we require 0 steps
        ans_start[start] = 0;
        ans_end[end] = 0;

        // mark start and end as visited
        visited_start[start] = true;
        visited_end[end] = true;

        // variable to store common node
        int commonNode = -1;

        // till all states are visited or common node is found
        while (!q_start.isEmpty() && !q_end.isEmpty()) {
            // choose the queue with smaller size
            if (q_start.size() < q_end.size()) {
                int top = q_start.poll();

                // if the topmost element is end
                if (visited_end[top]) {
                    commonNode = top;
                    break;
                }

                // perform multiplication with all array elements
                for (int i = 0; i < n; i++) {
                    int pushed = top * a[i];
                    pushed = pushed % mod;

                    // if not visited, then push it to queue
                    if (!visited_start[pushed]) {
                        ans_start[pushed] = ans_start[top] + 1;
                        q_start.add(pushed);
                        visited_start[pushed] = true;
                    }
                }
            } else {
                int top = q_end.poll();

                // if the topmost element is start
                if (visited_start[top]) {
                    commonNode = top;
                    break;
                }

                // perform multiplication with all array elements
                for (int i = 0; i < n; i++) {
                    int pushed = top * a[i];
                    pushed = pushed % mod;

                    // if not visited, then push it to queue
                    if (!visited_end[pushed]) {
                        ans_end[pushed] = ans_end[top] + 1;
                        q_end.add(pushed);
                        visited_end[pushed] = true;
                    }
                }
            }
        }

        if (commonNode == -1)
            return -1;

        // return the sum of minimum steps from start and end to common node
        return ans_start[commonNode] + ans_end[commonNode];
    }

    // Driver Code
    public static void main(String[] args) {
        int start = 7, end = 66175;
        int a[] = { 3, 4, 65 };
        int n = a.length;
        // Calling function
        System.out.println(minimumMultiplications(start, end, a, n));
    }
}

Python3

from queue import Queue

# Function that returns the minimum operations
def minimumMultiplications(start, end, a, n):
    # array which stores the minimum steps
    # to reach i from start or end
    ans_start, ans_end = [-1] * 100001, [-1] * 100001
    # visited array to keep track of visited states
    visited_start, visited_end = [False] * 100001, [False] * 100001

    mod = 100000

    # queues to store all possible states
    q_start, q_end = Queue(), Queue()

    # initially push start and end
    q_start.put(start % mod)
    q_end.put(end % mod)

    # to reach start or end we require 0 steps
    ans_start[start] = 0
    ans_end[end] = 0

    # mark start and end as visited
    visited_start[start] = True
    visited_end[end] = True

    # variable to store common node
    commonNode = -1

    # till all states are visited or common node is found
    while not q_start.empty() and not q_end.empty():

        # choose the queue with smaller size
        if q_start.qsize() < q_end.qsize():
            top = q_start.get()

            # if the topmost element is end
            if visited_end[top]:
                commonNode = top
                break

            # perform multiplication with all array elements
            for i in range(n):
                pushed = top * a[i]
                pushed = pushed % mod

                # if not visited, then push it to queue
                if not visited_start[pushed]:
                    ans_start[pushed] = ans_start[top] + 1
                    q_start.put(pushed)
                    visited_start[pushed] = True
        else:
            top = q_end.get()

            # if the topmost element is start
            if visited_start[top]:
                commonNode = top
                break

            # perform multiplication with all array elements
            for i in range(n):
                pushed = top * a[i]
                pushed = pushed % mod

                # if not visited, then push it to queue
                if not visited_end[pushed]:
                    ans_end[pushed] = ans_end[top] + 1
                    q_end.put(pushed)
                    visited_end[pushed] = True

    if commonNode == -1:
        return -1

    # return the sum of minimum steps from start and end to common node
    return ans_start[commonNode] + ans_end[commonNode]

# Driver Code
if __name__ == "__main__":
    start, end = 7, 66175
    a = [3, 4, 65]
    n = len(a)
    # Calling function
    print(minimumMultiplications(start, end, a, n))

Output

4

Time Complexity: O(n)

Auxiliary Space: O(n)