Dado un arreglo arr[] de longitud N , la tarea es encontrar el par de índices más pequeño (i, j) tal que el producto de los elementos en el subarreglo arr[i + 1, j – 1] sea coprimo con el producto del subarreglo arr[0, i] o el del subarreglo arr[j, N] . Si no existe tal par, la impresión “-1” .
Ejemplos:
Entrada: arr[] = {2, 4, 1, 3, 7}
Salida: (0, 2)
Explicación: El producto del subarreglo {arr[1], arr[1]} es 4. El producto del subarreglo derecho = 1 * 3 * 7 = 21. Dado que 4 y 21 son coprimos, imprima el rango (0, 2) como respuesta.Entrada: arr[] = {21, 3, 11, 7, 18}
Salida: (1, 3)
Explicación: El producto del subarreglo {arr[1], arr[2]} es 11. El producto del subarreglo derecho es 7 * 18 = 126. Dado que 11 y 126 son coprimos, imprima el rango (1, 3) como respuesta.
Enfoque ingenuo: el enfoque más simple es iterar a través de cada par posible de índices (i, j) y para cada par, encontrar el producto del subarreglo arr[0, i] y arr[j, N] y verificar si es co- prime con el producto del subarreglo arr[i + 1, j – 1] o no. Si se encuentra que es cierto, imprima ese par de índices. Si no existe tal par, imprima “-1” .
Complejidad temporal: O((N 3 )*log(M)), donde M es el producto de todos los elementos del arreglo.
Espacio Auxiliar: O(1)
Enfoque eficiente: para optimizar el enfoque anterior, la idea es usar una array auxiliar para almacenar el producto de sufijos de los elementos de la array. Siga los pasos a continuación para resolver el problema:
- Almacene el producto de elementos de sufijo en rightProd[] , donde rightProd[i] almacena el producto de elementos de arr[i], arr[N – 1] .
- Encuentre el producto de todos los elementos en la array como totalProd.
- Recorra la array dada usando la variable i y realice las siguientes operaciones:
- Inicializa una variable, digamos product .
- Iterar a través de la array usando la variable j del rango [i, N – 1] .
- Actualice el producto multiplicando el producto por arr[j] .
- Inicialice leftProduct como total/right_prod[i] .
- Compruebe si el producto es coprimo con uno de los productos izquierdo o derecho o no. Si se determina que es cierto, imprima el par (i – 1, j + 1) y salga del ciclo .
- Después de los pasos anteriores, si no existe tal par, imprima «-1» .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate GCD
// of two integers
int gcd(int a, int b)
{
if (b == 0)
return a;
// Recursively calculate GCD
return gcd(b, a % b);
}
// Function to find the lexicographically
// smallest pair of indices whose product
// is co-prime with the product of the
// subarrays on its left and right
void findPair(int A[], int N)
{
// Stores the suffix product
// of array elements
int right_prod[N];
// Set 0/1 if pair satisfies the
// given condition or not
int flag = 0;
// Initialize array right_prod[]
right_prod[N - 1] = A[N - 1];
// Update the suffix product
for (int i = N - 2; i >= 0; i--)
right_prod[i] = right_prod[i + 1]
* A[i];
// Stores product of all elements
int total_prod = right_prod[0];
// Stores the product of subarray
// in between the pair of indices
int product;
// Iterate through every pair of
// indices (i, j)
for (int i = 1; i < N - 1; i++) {
product = 1;
for (int j = i; j < N - 1; j++) {
// Store product of A[i, j]
product *= A[j];
// Check if product is co-prime
// to product of either the left
// or right subarrays
if (gcd(product,
right_prod[j + 1])
== 1
|| gcd(product,
total_prod
/ right_prod[i])
== 1) {
flag = 1;
cout << "(" << i - 1
<< ", " << j + 1
<< ")";
break;
}
}
if (flag == 1)
break;
}
// If no such pair is found,
// then print -1
if (flag == 0)
cout << -1;
}
// Driver Code
int main()
{
int arr[] = { 2, 4, 1, 3, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
findPair(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to calculate GCD
// of two integers
static int gcd(int a, int b)
{
if (b == 0)
return a;
// Recursively calculate GCD
return gcd(b, a % b);
}
// Function to find the lexicographically
// smallest pair of indices whose product
// is co-prime with the product of the
// subarrays on its left and right
static void findPair(int A[], int N)
{
// Stores the suffix product
// of array elements
int right_prod[] = new int[N];
// Set 0/1 if pair satisfies the
// given condition or not
int flag = 0;
// Initialize array right_prod[]
right_prod[N - 1] = A[N - 1];
// Update the suffix product
for(int i = N - 2; i >= 0; i--)
right_prod[i] = right_prod[i + 1] * A[i];
// Stores product of all elements
int total_prod = right_prod[0];
// Stores the product of subarray
// in between the pair of indices
int product;
// Iterate through every pair of
// indices (i, j)
for(int i = 1; i < N - 1; i++)
{
product = 1;
for(int j = i; j < N - 1; j++)
{
// Store product of A[i, j]
product *= A[j];
// Check if product is co-prime
// to product of either the left
// or right subarrays
if (gcd(product, right_prod[j + 1]) == 1 ||
gcd(product, total_prod /
right_prod[i]) == 1)
{
flag = 1;
System.out.println("(" + (i - 1) + ", " +
(j + 1) + ")");
break;
}
}
if (flag == 1)
break;
}
// If no such pair is found,
// then print -1
if (flag == 0)
System.out.print(-1);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 4, 1, 3, 7 };
int N = arr.length;
// Function Call
findPair(arr, N);
}
}
// This code is contributed by chitranayal
Python3
# Python3 program for the above approach
# Function to calculate GCD
# of two integers
def gcd(a, b):
if (b == 0):
return a
# Recursively calculate GCD
return gcd(b, a % b)
# Function to find the lexicographically
# smallest pair of indices whose product
# is co-prime with the product of the
# subarrays on its left and right
def findPair(A, N):
# Stores the suffix product
# of array elements
right_prod = [0] * N
# Set 0/1 if pair satisfies the
# given condition or not
flag = 0
# Initialize array right_prod
right_prod[N - 1] = A[N - 1]
# Update the suffix product
for i in range(N - 2, 0, -1):
right_prod[i] = right_prod[i + 1] * A[i]
# Stores product of all elements
total_prod = right_prod[0]
# Stores the product of subarray
# in between the pair of indices
product = 1
# Iterate through every pair of
# indices (i, j)
for i in range(1, N - 1):
product = 1
for j in range(i, N - 1):
# Store product of A[i, j]
product *= A[j]
# Check if product is co-prime
# to product of either the left
# or right subarrays
if (gcd(product, right_prod[j + 1]) == 1 or
gcd(product, total_prod /
right_prod[i]) == 1):
flag = 1
print("(" , (i - 1) , ", " ,
(j + 1) ,")")
break
if (flag == 1):
break
# If no such pair is found,
# then print -1
if (flag == 0):
print(-1)
# Driver Code
if __name__ == '__main__':
arr = [ 2, 4, 1, 3, 7 ]
N = len(arr)
# Function Call
findPair(arr, N)
# This code is contributed by Amit Katiyar
C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate GCD
// of two integers
static int gcd(int a, int b)
{
if (b == 0)
return a;
// Recursively calculate GCD
return gcd(b, a % b);
}
// Function to find the lexicographically
// smallest pair of indices whose product
// is co-prime with the product of the
// subarrays on its left and right
static void findPair(int []A, int N)
{
// Stores the suffix product
// of array elements
int []right_prod = new int[N];
// Set 0/1 if pair satisfies the
// given condition or not
int flag = 0;
// Initialize array right_prod[]
right_prod[N - 1] = A[N - 1];
// Update the suffix product
for(int i = N - 2; i >= 0; i--)
right_prod[i] = right_prod[i + 1] * A[i];
// Stores product of all elements
int total_prod = right_prod[0];
// Stores the product of subarray
// in between the pair of indices
int product;
// Iterate through every pair of
// indices (i, j)
for(int i = 1; i < N - 1; i++)
{
product = 1;
for(int j = i; j < N - 1; j++)
{
// Store product of A[i, j]
product *= A[j];
// Check if product is co-prime
// to product of either the left
// or right subarrays
if (gcd(product, right_prod[j + 1]) == 1 ||
gcd(product, total_prod /
right_prod[i]) == 1)
{
flag = 1;
Console.WriteLine("(" + (i - 1) + ", " +
(j + 1) + ")");
break;
}
}
if (flag == 1)
break;
}
// If no such pair is found,
// then print -1
if (flag == 0)
Console.Write(-1);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 4, 1, 3, 7 };
int N = arr.Length;
// Function Call
findPair(arr, N);
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript program for the above approach
// Function to calculate GCD
// of two integers
function gcd(a, b)
{
if (b == 0)
return a;
// Recursively calculate GCD
return gcd(b, a % b);
}
// Function to find the lexicographically
// smallest pair of indices whose product
// is co-prime with the product of the
// subarrays on its left and right
function findPair(A, N)
{
// Stores the suffix product
// of array elements
let right_prod = [];
// Set 0/1 if pair satisfies the
// given condition or not
let flag = 0;
// Initialize array right_prod[]
right_prod[N - 1] = A[N - 1];
// Update the suffix product
for(let i = N - 2; i >= 0; i--)
right_prod[i] = right_prod[i + 1] * A[i];
// Stores product of all elements
let total_prod = right_prod[0];
// Stores the product of subarray
// in between the pair of indices
let product;
// Iterate through every pair of
// indices (i, j)
for(let i = 1; i < N - 1; i++)
{
product = 1;
for(let j = i; j < N - 1; j++)
{
// Store product of A[i, j]
product *= A[j];
// Check if product is co-prime
// to product of either the left
// or right subarrays
if (gcd(product, right_prod[j + 1]) == 1 ||
gcd(product, total_prod /
right_prod[i]) == 1)
{
flag = 1;
document.write("(" + (i - 1) + ", " +
(j + 1) + ")");
break;
}
}
if (flag == 1)
break;
}
// If no such pair is found,
// then print -1
if (flag == 0)
document.write(-1);
}
// Driver code
let arr = [ 2, 4, 1, 3, 7 ];
let N = arr.length;
// Function Call
findPair(arr, N);
// This code is contributed by code_hunt
</script>
Producción:
(0, 2)
Complejidad de tiempo: O(N 2 *log(M)), donde M es el producto de todos los elementos del arreglo
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por ManikantaBandla y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA