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Split array into three continuous subarrays with negative, 0 and positive product respectively

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  • Last Updated : 22 Jun, 2021
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Given an array arr[] size N such that each array element is either -1, 0, or 1, the task is to check if is it possible to split the array into 3 contiguous subarrays such that the product of the first, second and third subarrays is negative, 0 and positive respectively.

Examples:

Input: arr[] = {-1, 0, 1}, N = 3
Output: -1
0
1
Explanation: Split the array into subarrays {-1}, {0}, {1}.

Input: arr[] = {1, -1, 1, -1}, N = 4
Output: Not Possible

 

Approach: The problem can be solved by greedily selecting the elements up to first ‘-1‘ having no 0′s as the first subarray and then selecting the elements from the last until the product becomes 1 as the third subarray and the remaining elements as the second subarray. Follow the steps below to solve the problem:

  • Initialize two variables, say l and r as -1, to store the index of the last element of the first subarray and the first element of the third subarray.
  • Traverse the array arr[] and if l is -1 and arr[i] is 0, then print “Not possible”. Otherwise, if arr[i] is -1, then assign i to l and break out of the loop.
  • Traverse the array arr[] in reverse order and if the product in the range [i, N – 1] is greater than 0, then assign i to r and break out of the loop.
  • Check if the value l or r is -1 or l ≥ r or count of 0s in the range [l, r] is 0. If any of the condition is true, print “Not Possible”.
  • Print the first subarray over the range [0, l], the second subarray over the range [l+1, r-1], and then the last subarray over the range [r, N-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 split an array into 3 contiguous
// subarrays satisfying the given condition
void PrintAllArrays(int arr[], int N)
{
    // Store the index of rightmost
    // element of the first and leftmost
    // element of the third subarray
    int l = -1, r = -1;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // If l is equal to -1
        if (l == -1) {
 
            // If arr[i] is -1
            if (arr[i] == -1) {
 
                l = i;
                break;
            }
 
            // If arr[i] is 0
            if (arr[i] == 0) {
 
                cout << "Not Possible" << endl;
                return;
            }
        }
    }
 
    // Stores the product
    // of the last subarray
    int p = 1;
 
    // Traverse the array in reverse
    for (int i = N - 1; i >= 0; i--) {
 
        // Update the product
        p *= arr[i];
 
        // If p is greater than 0,
        // assign i to r and break
        if (p > 0) {
            r = i;
            break;
        }
    }
 
    // If l or r is -1 or l to r, there
    // P are no 0's, print "Not possible"
    if (l == -1 || r == -1 || l >= r
        || count(arr + l, arr + r + 1, 0) == 0) {
 
        cout << "Not Possible" << endl;
        return;
    }
 
    // Print the three subarrays
    for (int i = 0; i <= l; i++)
        cout << arr[i] << " ";
 
    cout << "\n";
 
    for (int i = l + 1; i < r; i++)
        cout << arr[i] << " ";
    cout << "\n";
 
    for (int i = r; i < N; i++)
        cout << arr[i] << " ";
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { -1, 0, 1 };
    int N = sizeof(arr) / sizeof(int);
 
    PrintAllArrays(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to split an array into 3 contiguous
// subarrays satisfying the given condition
static void PrintAllArrays(int arr[], int N)
{
     
    // Store the index of rightmost
    // element of the first and leftmost
    // element of the third subarray
    int l = -1, r = -1;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // If l is equal to -1
        if (l == -1)
        {
             
            // If arr[i] is -1
            if (arr[i] == -1)
            {
                l = i;
                break;
            }
 
            // If arr[i] is 0
            if (arr[i] == 0)
            {
                System.out.println("Not Possible");
                return;
            }
        }
    }
 
    // Stores the product
    // of the last subarray
    int p = 1;
 
    // Traverse the array in reverse
    for(int i = N - 1; i >= 0; i--)
    {
         
        // Update the product
        p *= arr[i];
 
        // If p is greater than 0,
        // assign i to r and break
        if (p > 0)
        {
            r = i;
            break;
        }
    }
 
    // If l or r is -1 or l to r, there
    // P are no 0's, print "Not possible"
    if (l == -1 || r == -1 || l >= r ||
       count(arr, l, r + 1, 0) == 0)
    {
        System.out.println("Not Possible");
        return;
    }
 
    // Print the three subarrays
    for(int i = 0; i <= l; i++)
        System.out.print(arr[i] + " ");
         
    System.out.println();
 
    for(int i = l + 1; i < r; i++)
        System.out.print(arr[i] + " ");
         
    System.out.println();
 
    for(int i = r; i < N; i++)
        System.out.print(arr[i] + " ");
}
 
// Function to return the number of occurrence of
// the element 'val' in the range [start, end).
static int count(int arr[], int start,
                 int end, int val)
{
    int count = 0;
    for(int i = start; i < end; i++)
    {
        if (arr[i] == val)
        {
            count++;
        }
    }
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int arr[] = { -1, 0, 1 };
    int N = arr.length;
 
    PrintAllArrays(arr, N);
}
}
 
// This code is contributed by Kingash

Python3




# Python3 program for the above approach
 
# Function to split an array into 3 contiguous
# subarrays satisfying the given condition
def PrintAllArrays(arr, N):
     
    # Store the index of rightmost
    # element of the first and leftmost
    # element of the third subarray
    l = -1
    r = -1
 
    # Traverse the array arr[]
    for i in range(N):
         
        # If l is equal to -1
        if (l == -1):
             
            # If arr[i] is -1
            if (arr[i] == -1):
                l = i
                break
             
            # If arr[i] is 0
            if (arr[i] == 0):
                print("Not Possible")
                return
             
    # Stores the product
    # of the last subarray
    p = 1
 
    # Traverse the array in reverse
    for i in range(N - 1, -1, -1):
         
        # Update the product
        p *= arr[i]
 
        # If p is greater than 0,
        # assign i to r and break
        if (p > 0):
            r = i
            break
         
    # If l or r is -1 or l to r, there
    # P are no 0's, pr"Not possible"
    if (l == -1 or r == -1 or l >= r or
       count(arr, l, r + 1, 0) == 0):
        print("Not Possible")
        return
 
    # Printthe three subarrays
    for i in range(0, l + 1, 1):
        print(arr[i], end = " ")
         
    print()
 
    for i in range(l + 1, r, 1):
        print(arr[i], end = " ")
         
    print()
 
    for i in range(r, N, 1):
        print(arr[i], end = " ")
 
# Function to return the number of occurrence of
# the element 'val' in the range [start, end).
def count(arr, start, end, val):
     
    count = 0
    for i in range(start, end, 1):
        if (arr[i] == val):
            count += 1
         
    return count
 
# Driver Code
 
# Given Input
arr = [ -1, 0, 1 ]
N = len(arr)
 
PrintAllArrays(arr, N)
  
# This code is contriobuted by code_hunt

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to split an array into 3 contiguous
// subarrays satisfying the given condition
static void PrintAllArrays(int[] arr, int N)
{
     
    // Store the index of rightmost
    // element of the first and leftmost
    // element of the third subarray
    int l = -1, r = -1;
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // If l is equal to -1
        if (l == -1)
        {
             
            // If arr[i] is -1
            if (arr[i] == -1)
            {
                l = i;
                break;
            }
 
            // If arr[i] is 0
            if (arr[i] == 0)
            {
                Console.WriteLine("Not Possible");
                return;
            }
        }
    }
 
    // Stores the product
    // of the last subarray
    int p = 1;
 
    // Traverse the array in reverse
    for(int i = N - 1; i >= 0; i--)
    {
         
        // Update the product
        p *= arr[i];
 
        // If p is greater than 0,
        // assign i to r and break
        if (p > 0)
        {
            r = i;
            break;
        }
    }
 
    // If l or r is -1 or l to r, there
    // P are no 0's, print "Not possible"
    if (l == -1 || r == -1 || l >= r ||
       count(arr, l, r + 1, 0) == 0)
    {
        Console.WriteLine("Not Possible");
        return;
    }
 
    // Print the three subarrays
    for(int i = 0; i <= l; i++)
        Console.Write(arr[i] + " ");
         
    Console.WriteLine();
 
    for(int i = l + 1; i < r; i++)
        Console.Write(arr[i] + " ");
         
    Console.WriteLine();
 
    for(int i = r; i < N; i++)
        Console.Write(arr[i] + " ");
}
 
// Function to return the number of occurrence of
// the element 'val' in the range [start, end).
static int count(int[] arr, int start,
                 int end, int val)
{
    int count = 0;
    for(int i = start; i < end; i++)
    {
        if (arr[i] == val)
        {
            count++;
        }
    }
    return count;
}
 
// Driver code
public static void Main(String []args)
{
     
    // Given Input
    int[] arr = { -1, 0, 1 };
    int N = arr.Length;
     
    PrintAllArrays(arr, N);
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
 
// JavaScript Program to implement
// the above approach
 
// Function to split an array into 3 contiguous
// subarrays satisfying the given condition
function PrintAllArrays(arr, N)
{
     
    // Store the index of rightmost
    // element of the first and leftmost
    // element of the third subarray
    let l = -1, r = -1;
 
    // Traverse the array arr[]
    for(let i = 0; i < N; i++)
    {
         
        // If l is equal to -1
        if (l == -1)
        {
             
            // If arr[i] is -1
            if (arr[i] == -1)
            {
                l = i;
                break;
            }
 
            // If arr[i] is 0
            if (arr[i] == 0)
            {
                document.write("Not Possible" + "<br/>");
                return;
            }
        }
    }
 
    // Stores the product
    // of the last subarray
    let p = 1;
 
    // Traverse the array in reverse
    for(let i = N - 1; i >= 0; i--)
    {
         
        // Update the product
        p *= arr[i];
 
        // If p is greater than 0,
        // assign i to r and break
        if (p > 0)
        {
            r = i;
            break;
        }
    }
 
    // If l or r is -1 or l to r, there
    // P are no 0's, print "Not possible"
    if (l == -1 || r == -1 || l >= r ||
       count(arr, l, r + 1, 0) == 0)
    {
        document.write("Not Possible" + "<br/>");
        return;
    }
 
    // Print the three subarrays
    for(let i = 0; i <= l; i++)
        document.write(arr[i] + " ");
         
    document.write("<br/>");
 
    for(let i = l + 1; i < r; i++)
        document.write(arr[i] + " ");
         
    document.write("<br/>");
 
    for(let i = r; i < N; i++)
        document.write(arr[i] + " ");
}
 
// Function to return the number of occurrence of
// the element 'val' in the range [start, end).
function count(arr, start, end, val)
{
    let count = 0;
    for(let i = start; i < end; i++)
    {
        if (arr[i] == val)
        {
            count++;
        }
    }
    return count;
}
 
// Driver Code
 
     // Given Input
    let arr = [ -1, 0, 1 ];
    let N = arr.length;
 
    PrintAllArrays(arr, N);
 
</script>

Output: 

-1 
0 
1

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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