Se requieren incrementos mínimos de prefijo para hacer que todos los elementos de una array sean múltiplos de otra array

Dados dos arreglos A[] y B[] de tamaño N , la tarea es encontrar el recuento mínimo de operaciones requeridas para hacer A[i] múltiplos de B[i] incrementando los subarreglos de prefijos en 1 .

Ejemplos:

Entrada: A[ ] = {3, 2, 9}, B[ ] = {5, 7, 4}, N = 3
Salida: 7
Explicación:
Incrementar {A[0]} dos veces modifica A[] a { 5 , 2, 9}
Incrementar todos los elementos del subarreglo {A[0], A[1]} modifica dos veces A[] a { 7, 4, 9}
Incrementar todos los elementos del subarreglo {A[0], A[1 ], A[2]} tres veces modifica A[] a { 10, 7, 12 } 
Por lo tanto, el total de operaciones requeridas = 2 + 2 + 3 = 7.

Entrada: A[ ] = {3, 4, 5, 2, 5, 5, 9}, B[ ] = {1, 1, 9, 6, 3, 8, 7}, N = 7
Salida: 22

Planteamiento: El problema se puede resolver usando la técnica Greedy . Para minimizar el número de operaciones, la idea es encontrar el elemento mayor o igual más pequeño más cercano de A[i] que sea un múltiplo de B[i] :

1. Atraviesa la array el
2. la la
3. las operacioneslas
4. hacia

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
int MinimumMoves(int A[], int B[], int N)
{
 
    // Stores  minimum count of operations
    // required to make A[i] multiple of B[i]
    // by  incrementing prefix subarray
    int totalOperations = 0;
 
    // Stores the carry
    int carry = 0;
 
    // Stores minimum difference of
    // correspoinding element in
    // prefix subarray
    int K = 0;
 
    // Traverse the array
    for (int i = N - 1; i >= 0; i--) {
 
        // Stores the closest greater or equal number
        // to A[i] which is a multiple of B[i]
        int nearestMultiple = ceil((double)(A[i] + carry)
                                   / (double)(B[i]))
                              * B[i];
 
        // Stores minimum difference
        K = nearestMultiple - (A[i] + carry);
 
        // Update totalOperations
        totalOperations += K;
 
        // Update carry
        carry += K;
    }
 
    return totalOperations;
}
 
// Driver Code
int main()
{
 
    // Input arrays A[] and B[]
    int A[] = { 3, 4, 5, 2, 5, 5, 9 };
    int B[] = { 1, 1, 9, 6, 3, 8, 7 };
 
    // Length of arrays
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << MinimumMoves(A, B, N) << endl;
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find minimum count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
static int MinimumMoves(int A[], int B[], int N)
{
 
    // Stores  minimum count of operations
    // required to make A[i] multiple of B[i]
    // by  incrementing prefix subarray
    int totalOperations = 0;
 
    // Stores the carry
    int carry = 0;
 
    // Stores minimum difference of
    // correspoinding element in
    // prefix subarray
    int K = 0;
 
    // Traverse the array
    for (int i = N - 1; i >= 0; i--)
    {
 
        // Stores the closest greater or equal number
        // to A[i] which is a multiple of B[i]
        int nearestMultiple = (int) (Math.ceil((double)(A[i] + carry)
                                   / (double)(B[i]))
                              * B[i]);
 
        // Stores minimum difference
        K = nearestMultiple - (A[i] + carry);
 
        // Update totalOperations
        totalOperations += K;
 
        // Update carry
        carry += K;
    }
    return totalOperations;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Input arrays A[] and B[]
    int A[] = { 3, 4, 5, 2, 5, 5, 9 };
    int B[] = { 1, 1, 9, 6, 3, 8, 7 };
 
    // Length of arrays
    int N = A.length;
    System.out.print(MinimumMoves(A, B, N) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

# Python3 program for the above approach
from math import ceil,floor
 
# Function to find minimum count of operations
# required to make A[i] multiple of B[i] by
# incrementing prefix subarray
def MinimumMoves(A, B, N):
 
    # Stores  minimum count of operations
    # required to make A[i] multiple of B[i]
    # by  incrementing prefix subarray
    totalOperations = 0
 
    # Stores the carry
    carry = 0
 
    # Stores minimum difference of
    # correspoinding element in
    # prefix subarray
    K = 0
 
    # Traverse the array
    for i in range(N - 1, -1, -1):
 
        # Stores the closest greater or equal number
        # to A[i] which is a multiple of B[i]
        nearestMultiple = ceil((A[i] + carry)/ B[i])* B[i]
 
        # Stores minimum difference
        K = nearestMultiple - (A[i] + carry)
 
        # Update totalOperations
        totalOperations += K
 
        # Update carry
        carry += K
    return totalOperations
 
# Driver Code
if __name__ == '__main__':
 
    # Input arrays A[] and B[]
    A = [3, 4, 5, 2, 5, 5, 9]
    B = [1, 1, 9, 6, 3, 8, 7]
 
    # Length of arrays
    N = len(A)
    print (MinimumMoves(A, B, N))
 
# This code is contributed by mohit kumar 29.

// C# program for the above approach
using System;
class GFG{
 
// Function to find minimum count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
static int MinimumMoves(int[] A, int[] B, int N)
{
 
    // Stores  minimum count of operations
    // required to make A[i] multiple of B[i]
    // by  incrementing prefix subarray
    int totalOperations = 0;
 
    // Stores the carry
    int carry = 0;
 
    // Stores minimum difference of
    // correspoinding element in
    // prefix subarray
    int K = 0;
 
    // Traverse the array
    for (int i = N - 1; i >= 0; i--)
    {
 
        // Stores the closest greater or equal number
        // to A[i] which is a multiple of B[i]
        int nearestMultiple = (int) (Math.Ceiling((double)(A[i] + carry)
                                   / (double)(B[i]))
                              * B[i]);
 
        // Stores minimum difference
        K = nearestMultiple - (A[i] + carry);
 
        // Update totalOperations
        totalOperations += K;
 
        // Update carry
        carry += K;
    }
    return totalOperations;
}
 
// Driver Code
public static void Main(string[] args)
{
    // Input arrays A[] and B[]
    int[] A = { 3, 4, 5, 2, 5, 5, 9 };
    int[] B = { 1, 1, 9, 6, 3, 8, 7 };
 
    // Length of arrays
    int N = A.Length;
    Console.Write(MinimumMoves(A, B, N) +"\n");
}
}
 
// This code is contributed by sanjoy_62.

<script>
// javascript program of the above approach
 
// Function to find minimum count of operations
// required to make A[i] multiple of B[i] by
// incrementing prefix subarray
function MinimumMoves(A, B, N)
{
 
    // Stores  minimum count of operations
    // required to make A[i] multiple of B[i]
    // by  incrementing prefix subarray
    let totalOperations = 0;
 
    // Stores the carry
    let carry = 0;
 
    // Stores minimum difference of
    // correspoinding element in
    // prefix subarray
    let K = 0;
 
    // Traverse the array
    for (let i = N - 1; i >= 0; i--)
    {
 
        // Stores the closest greater or equal number
        // to A[i] which is a multiple of B[i]
        let nearestMultiple =  (Math.ceil((A[i] + carry)
                                   / (B[i]))
                              * B[i]);
 
        // Stores minimum difference
        K = nearestMultiple - (A[i] + carry);
 
        // Update totalOperations
        totalOperations += K;
 
        // Update carry
        carry += K;
    }
    return totalOperations;
}
 
    // Driver Code
     
    // Input arrays A[] and B[]
    let A = [ 3, 4, 5, 2, 5, 5, 9 ];
    let B = [ 1, 1, 9, 6, 3, 8, 7 ];
 
    // Length of arrays
    let N = A.length;
    document.write(MinimumMoves(A, B, N) + "<br/>");
 
</script>

22

Complejidad temporal: O(N)
Espacio auxiliar: O(1)