Smallest pair of indices with product of subarray co-prime with product of the subarray on the left or right

Given an array arr[] of length N, the task is to find the smallest pair of indices (i, j) such that the product of elements in the subarray arr[i + 1, j – 1] is co-prime with either the product of the subarray arr[0, i] or that of the subarray arr[j, N]. If no such pair exists, the print “-1”.

Examples:

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.

Naive Approach: The simplest approach is to iterate through every possible pair of indices (i, j) and for each pair, find the product of subarray arr[0, i] and arr[j, N] and check if it is co-prime with the product of the subarray arr[i + 1, j – 1] or not. If found to be true, then print those pair of indices. If no such pair exists, then print “-1”. 

Time Complexity: O((N³)*log(M)), where M is the product of all elements of the array.
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use an auxiliary array to store the suffix product of the array elements. Follow the steps below to solve the problem:

• Store the product of suffix elements in rightProd[], where rightProd[i] stores the product of elements from arr[i], arr[N – 1].
• Find the product of all elements in the array as totalProd.
• Traverse the given array using the variable i and perform the following operations:
  • Initialize a variable, say product.
  • Iterate through the array using variable j from the range [i, N – 1].
  • Update product by multiplying product by arr[j].
  • Initialize leftProduct as total/right_prod[i].
  • Check if the product is co-prime with one of the leftProduct or rightProduct or not. If found to be true, then print the pair (i – 1, j + 1) and break out of the loop.
• After the above steps, if no such pair exists then print “-1”.

Below is the implementation of the above approach:

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;
}

C

// C program for the above approach
#include <stdio.h>

// 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;
                printf("(%d, %d)", i - 1, j + 1);
                break;
            }
        }
        if (flag == 1)
            break;
    }

    // If no such pair is found, then print -1
    if (flag == 0)
        printf("-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;
}

// This code is contributed by Aditya Kumar (adityakumar129)

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>

Output:

(0, 2)

Time Complexity: O(N²*log(M)), where M is the product of all elements in the array
Auxiliary Space: O(N)