给定长度为N的数组arr [] ,任务是找到最小的索引对(i,j) ,以使子数组arr [i + 1,j – 1]中元素的乘积与任一子数组arr [0,i]或子数组arr [j,N]的乘积。如果不存在这样的对,则打印“ -1” 。
例子:
Input: arr[] = {2, 4, 1, 3, 7}
Output: (0, 2)
Explanation: The product of the subarray {arr[1], arr[1]} is 4. The product of right subarray = 1 * 3 * 7 = 21. Since 4 and 21 are co-primes, then print the range (0, 2) as the answer.
Input: arr[] = {21, 3, 11, 7, 18}
Output: (1, 3)
Explanation: The product of the subarray {arr[1], arr[2]} is 11. The product of right subarray is 7 * 18 = 126. Since 11 and 126 are co-primes, then print the range (1, 3) as the answer.
天真的方法:最简单的方法是遍历每对可能的索引(i,j) ,对于每对索引,找到子数组arr [0,i]和arr [j,N]的乘积,并检查其是否为co-是否使用子数组arr [i + 1,j – 1]的乘积来素数。如果发现是正确的,则打印那些索引对。如果不存在这样的对,则打印“ -1” 。
时间复杂度: O((N 3 )* log(M)),其中M是数组所有元素的乘积。
辅助空间: O(1)
高效方法:为了优化上述方法,其思想是使用辅助数组存储数组元素的后缀乘积。请按照以下步骤解决问题:
- 将后缀元素的乘积存储在rightProd []中,其中rightProd [i]存储来自arr [i],arr [N – 1]的元素乘积。
- 查找数组中所有元素的乘积作为totalProd。
- 使用变量i遍历给定数组并执行以下操作:
- 初始化一个变量,例如product 。
- 使用[i,N – 1]范围内的变量j遍历数组。
- 通过ARR [J]乘以产品的换代产品。
- 将leftProduct初始化为total / right_prod [i] 。
- 检查产品是否与leftProduct或rightProduct之一互底涂。如果发现是正确的,则打印该对(i – 1,j + 1)并退出循环。
- 在上述步骤之后,如果不存在这样的一对,则打印“ -1” 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
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 pr-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
(0, 2)
时间复杂度: O(N 2 * log(M)),其中M是数组中所有元素的乘积
辅助空间: O(N)