Recuento de arrays no decrecientes arr3[] tales que arr1[i] <= arr3[i] <= arr2[i]

Dados dos arreglos arr1[] y arr2[] que tienen N enteros en orden no decreciente, la tarea es encontrar el conteo de arreglos no decrecientes arr3[] de longitud N tal que arr1[i] <= arr3[i] < = arr2[i] para todos los valores de i en el rango [0, N) .

Ejemplos :

Entrada : arr1[] = {1, 1}, arr2[] = {2, 3}
Salida : 5
Explicación: Los 5 arreglos posibles que siguen las condiciones requeridas son {1, 1}, {1, 2}, {1 , 3}, {2, 2}, {2, 3}

Entrada : rangos[] = {{-12, 15}, {3, 9}, {-5, -2}, {20, 25}, {16, 20}}
Salida : 247

Enfoque: El problema dado se puede resolver usando Programación Dinámica . Considere una array 2D dp[][] tal que dp[i][j] representa el recuento de arrays de longitud i tal que el i ^-ésimo elemento es j . Inicialice todos los elementos de la array dp como 0 y dp[0][0] como 1 . Tras la observación, la relación DP del problema anterior se puede establecer de la siguiente manera:

pd[i][j] = $\sum_{x = arr1[i]}^{arr2[i]}dp[i-1][x] + dp[i][j-1]$

Por lo tanto, usando la relación anterior, calcule el valor de dp[i][j] para cada i en el rango [0, N] y para cada j en el rango [0, M] donde M representa el número entero máximo tanto en el arrays dadas arr1[] y arr2[] . Por lo tanto, el valor almacenado en dp[N][M] es la respuesta requerida.

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the count of
// valid sorted arrays
int arrCount(int arr1[], int arr2[], int N)
{
 
    // Maximum possible value
    // of arr1 and arr2
    int M = 1000;
 
    // Stores the dp states
    vector<vector<int> > dp(
        N + 1,
        vector<int>(M + 1, 0));
 
    // Initial condition
    dp[0][0] = 1;
 
    // Loop to iterate over range [0, N]
    for (int i = 0; i <= N; i++) {
 
        // Loop to iterate over
        // the range [0, M]
        for (int j = 0; j < M; j++) {
            dp[i][j + 1] += dp[i][j];
        }
 
        // If current index is not
        // the final index
        if (i != N) {
 
            // Loop to iterate in the
            // range [arr1[i], arr2[i]]
            for (int j = arr1[i]; j <= arr2[i]; j++)
                dp[i + 1][j] += dp[i][j];
        }
    }
 
    // Return Answer
    return dp[N][M];
}
 
// Driver Code
int main()
{
    int arr1[] = { 1, 1 };
    int arr2[] = { 2, 3 };
    int N = sizeof(arr1) / sizeof(int);
 
    cout << arrCount(arr1, arr2, N);
 
    return 0;
}

Java

// Java Program of the above approach
import java.util.*;
 
public class GFG{
     
// Function to find the count of
// valid sorted arrays
static int arrCount(int[] arr1, int[] arr2, int N)
{
     
    // Maximum possible value
    // of arr1 and arr2
    int M = 1000;
 
    // Stores the dp states
    int[][] dp = new int[N + 1][M + 1];
 
    // Initial condition
    dp[0][0] = 1;
 
    // Loop to iterate over range [0, N]
    for(int i = 0; i <= N; i++)
    {
         
        // Loop to iterate over
        // the range [0, M]
        for(int j = 0; j < M; j++)
        {
            dp[i][j + 1] += dp[i][j];
        }
 
        // If current index is not
        // the final index
        if (i != N)
        {
             
            // Loop to iterate in the
            // range [arr1[i], arr2[i]]
            for(int j = arr1[i]; j <= arr2[i]; j++)
                dp[i + 1][j] += dp[i][j];
        }
    }
 
    // Return Answer
    return dp[N][M];
}
 
// Driver Code
public static void main(String args[])
{
    int[] arr1 = { 1, 1 };
    int[] arr2 = { 2, 3 };
    int N = arr1.length;
 
    System.out.println(arrCount(arr1, arr2, N));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python Program to implement
# the above approach
 
# Function to find the count of
# valid sorted arrays
def arrCount(arr1, arr2, N):
 
    # Maximum possible value
    # of arr1 and arr2
    M = 1000
 
    # Stores the dp states
    dp = [0] * (N + 1)
    for i in range(len(dp)):
        dp[i] = [0] * (M + 1)
 
    # Initial condition
    dp[0][0] = 1
 
    # Loop to iterate over range [0, N]
    for i in range(N + 1):
 
        # Loop to iterate over
        # the range [0, M]
        for j in range(M):
            dp[i][j + 1] += dp[i][j]
 
        # If current index is not
        # the final index
        if (i != N):
 
            # Loop to iterate in the
            # range [arr1[i], arr2[i]]
            for j in range(arr1[i], arr2[i] + 1):
                dp[i + 1][j] += dp[i][j]
 
    # Return Answer
    return dp[N][M]
 
# Driver Code
arr1 = [1, 1]
arr2 = [2, 3]
N = len(arr1)
 
print(arrCount(arr1, arr2, N))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# Program of the above approach
using System;
 
class GFG{
     
// Function to find the count of
// valid sorted arrays
static int arrCount(int[] arr1, int[] arr2, int N)
{
     
    // Maximum possible value
    // of arr1 and arr2
    int M = 1000;
 
    // Stores the dp states
    int[,] dp = new int[N + 1, M + 1];
 
    // Initial condition
    dp[0, 0] = 1;
 
    // Loop to iterate over range [0, N]
    for(int i = 0; i <= N; i++)
    {
         
        // Loop to iterate over
        // the range [0, M]
        for(int j = 0; j < M; j++)
        {
            dp[i, j + 1] += dp[i, j];
        }
 
        // If current index is not
        // the final index
        if (i != N)
        {
             
            // Loop to iterate in the
            // range [arr1[i], arr2[i]]
            for(int j = arr1[i]; j <= arr2[i]; j++)
                dp[i + 1, j] += dp[i, j];
        }
    }
 
    // Return Answer
    return dp[N, M];
}
 
// Driver Code
public static void Main()
{
    int[] arr1 = { 1, 1 };
    int[] arr2 = { 2, 3 };
    int N = arr1.Length;
 
    Console.WriteLine(arrCount(arr1, arr2, N));
}
}
 
// This code is contributed by ukasp

Javascript

<script>
    // JavaScript Program to implement
    // the above approach
 
    // Function to find the count of
    // valid sorted arrays
    function arrCount(arr1, arr2, N) {
 
        // Maximum possible value
        // of arr1 and arr2
        let M = 1000;
 
        // Stores the dp states
        let dp = new Array(N + 1);
        for (let i = 0; i < dp.length; i++) {
            dp[i] = new Array(M + 1).fill(0);
        }
 
        // Initial condition
        dp[0][0] = 1;
 
        // Loop to iterate over range [0, N]
        for (let i = 0; i <= N; i++) {
 
            // Loop to iterate over
            // the range [0, M]
            for (let j = 0; j < M; j++) {
                dp[i][j + 1] += dp[i][j];
            }
 
            // If current index is not
            // the final index
            if (i != N) {
 
                // Loop to iterate in the
                // range [arr1[i], arr2[i]]
                for (let j = arr1[i]; j <= arr2[i]; j++)
                    dp[i + 1][j] += dp[i][j];
            }
        }
 
        // Return Answer
        return dp[N][M];
    }
 
    // Driver Code
 
    let arr1 = [1, 1];
    let arr2 = [2, 3];
    let N = arr1.length;
 
    document.write(arrCount(arr1, arr2, N));
 
// This code is contributed by Potta Lokesh
</script>

Producción

Complejidad de tiempo : O(N * M), donde M representa el valor máximo de los enteros en la array arr1[] y arr2[].
Espacio Auxiliar: O(N * M)

Publicación traducida automáticamente

Artículo escrito por apratyush41 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

C++

Java

Python3

C#

Javascript

Deja una respuesta Cancelar la respuesta