Colocación de Sudo[1.3] | Destino final

Dada una array de enteros y un número K con valores inicial y final. Su tarea es encontrar la cantidad mínima de pasos necesarios para obtener el valor final a partir del valor inicial utilizando los elementos de la array. Solo puede agregar (operación de agregar% 1000) en valores para obtener el valor final. En cada paso, puede agregar cualquiera de los elementos de la array con la operación de módulo.

Ejemplos:  

Entrada: inicial = 1, final = 6, a[] = {1, 2, 3, 4} 
Salida: 2 
Paso 1: (1 + 1 ) % 1000 = 2. 
Paso 2: (2 + 4) % 1000 = 6 (que es el valor final requerido).

Entrada: inicio = 998 fin = 2 a[] = {2, 1, 3} 
Salida: 2 
Paso 1: (998 + 2) % 1000 = 0. 
Paso 2: (0 + 2) % 1000 = 2. 
O 
Paso 1: (998 + 1) % 1000 = 999. 
Paso 2: (999 + 3) % 1000 = 2

Enfoque: Dado que en el problema anterior el módulo dado es 1000, el número máximo de estados será 10 ³ . Todos los estados se pueden verificar usando BFS simple. Inicialice una array ans[] con -1 que marca que el estado no ha sido visitado. ans[i] almacena el número de pasos dados para llegar a i desde el inicio. Inicialmente empuje el inicio a la cola, luego aplique BFS. Haga estallar el elemento superior y verifique si es igual al final, si es así, imprima el ans[end]. Si el elemento no es igual al elemento superior, agregue el elemento superior con todos los elementos de la array y realice una operación de modificación con 1000. Si el estado del elemento agregado no se ha visitado anteriormente, insértelo en la cola. Inicialice ans[pushed_element] por ans[top_element] + 1. Una vez que se visitan todos los estados, y no se puede alcanzar el estado realizando todas las multiplicaciones posibles, imprima -1. 

A continuación se muestra la implementación del enfoque anterior:  

// C++ program to find the minimum steps
// to reach end from start by performing
// additions and mod operations with array elements
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns the minimum operations
int minimumAdditions(int start, int end, int a[], int n)
{
    // array which stores the minimum steps
    // to reach i from start
    int ans[1001];
 
    // -1 indicated the state has not been visited
    memset(ans, -1, sizeof(ans));
    int mod = 1000;
 
    // 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 addition 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 = 998, end = 2;
    int a[] = { 2, 1, 3 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // Calling function
    cout << minimumAdditions(start, end, a, n);
    return 0;
}

// Java program to find the minimum steps
// to reach end from start by performing
// additions and mod operations with array elements
import java.util.*;
 
class GFG
{
 
    // Function that returns
    // the minimum operations
    static int minimumAdditions(int start,
                    int end, int a[], int n)
    {
        // array which stores the minimum steps
        // to reach i from start
        int ans[] = new int[1001];
 
        // -1 indicated the state has not been visited
        Arrays.fill(ans, -1);
        int mod = 1000;
 
        // queue to store all possible states
        Queue<Integer> q = new java.util.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.poll();
 
            // if the topmost element is end
            if (top == end)
            {
                return ans[end];
            }
 
            // perform addition 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 = 998, end = 2;
        int a[] = {2, 1, 3};
        int n = a.length;
 
        // Calling function
        System.out.println(minimumAdditions(start, end, a, n));
    }
}
 
/* This code contributed by PrinciRaj1992 */

# Python3 program to find the minimum steps
# to reach end from start by performing
# additions and mod operations with array
# elements
from collections import deque
from typing import List
 
# Function that returns the minimum operations
def minimumAdditions(start: int, end: int,
                     a: List[int], n: int) -> int:
                          
    # Array which stores the minimum steps
    # to reach i from start
    # -1 indicated the state has not been visited
    ans = [-1] * 1001
 
    mod = 1000
 
    # Queue to store all possible states
    q = deque()
 
    # Initially push the start
    q.append(start % mod)
 
    # To reach start we require 0 steps
    ans[start] = 0
 
    # Till all states are visited
    while q:
         
        # Get the topmost element in the queue
        top = q[0]
 
        # Pop the topmost element
        q.popleft()
 
        # If the topmost element is end
        if (top == end):
            return ans[end]
 
        # Perform addition 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
if __name__ == "__main__":
 
    start = 998
    end = 2
    a = [ 2, 1, 3 ]
    n = len(a)
 
    # Calling function
    print(minimumAdditions(start, end, a, n))
 
# This code is contributed by sanjeev2552

// C# program to find the minimum steps
// to reach end from start by performing
// additions and mod operations with array elements
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function that returns
    // the minimum operations
    static int minimumAdditions(int start,
                    int end, int []a, int n)
    {
        // array which stores the minimum steps
        // to reach i from start
        int []ans = new int[1001];
         
        // -1 indicated the state has not been visited
        for(int i = 0; i < 1001; i++)
        {
            ans[i] = -1;
        }
        int mod = 1000;
 
        // 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 addition 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 = 998, end = 2;
        int []a = {2, 1, 3};
        int n = a.Length;
 
        // Calling function
        Console.WriteLine(minimumAdditions(start, end, a, n));
    }
}
 
// This code has been contributed by 29AjayKumar

<script>
// Javascript program to find the minimum steps
// to reach end from start by performing
// additions and mod operations with array elements
 
// Function that returns
    // the minimum operations
function minimumAdditions(start,end,a,n)
{
    // array which stores the minimum steps
        // to reach i from start
        let ans = new Array(1001);
  
        // -1 indicated the state has not been visited
        for(let i=0;i<1001;i++)
            ans[i]=-1;
        let mod = 1000;
  
        // 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 addition 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 = 998, end = 2;
let a=[2, 1, 3];
let n = a.length;
  
// Calling function
document.write(minimumAdditions(start, end, a, n));
 
// This code is contributed by rag2127
</script>

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Complejidad de tiempo : O(N)