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Length of maximum product subarray

  • Last Updated : 12 May, 2021

Given an integer array arr[] of size N, the task is to find the maximum length subarray whose products of element is non zero. . 
Examples:

Input: arr[] = [1, 1, 0, 2, 1, 0, 1, 6, 1] 
Output:
Explanation 
Possible subarray whose product are non zero are [1, 1], [2, 1] and [1, 6, 1] 
So maximum possible length is 3.
Input: arr[] = [0, 1, 2, 1, 3, 0, 0, 1] 
Output:
Explanation 
Possible subarray whose product are non zero are [1, 2, 1, 3] and [1] 
So maximum possible length is 4 .

Approach:

  1. Save all the indices of zero from input array.
  2. Longest subarray must lie inside below three ranges:
    • Start from zero index and end at first zero index – 1.
    • Lies between two zero index.
    • Start at last zero index + 1 and end at N-1.
  3. Finally find the maximum length from all cases.

Here is implementation of above approach :

C++




// C++ program to find maximum
// length subarray having non
// zero product
#include<bits/stdc++.h>
using namespace std;
 
// Function that returns the
// maximum length subarray
// having non zero product
void Maxlength(int arr[],int N)
{
    vector<int> zeroindex;
    int maxlen;
     
    // zeroindex list to store indexex
    // of zero
    for(int i = 0; i < N; i++)
    {
        if(arr[i] == 0)
            zeroindex.push_back(i);
    }
             
    if(zeroindex.size() == 0)
    {
         
        // If zeroindex list is empty
        // then Maxlength is as
        // size of array
        maxlen = N;
    }
     
    // If zeroindex list is not empty
    else
    {
         
        // for example list 1 1 0 0 1
        // is on indexex 0 1 2 3 4
         
        // first zero is on index 2
        // that means two numbers positive,
        // before index 2 so as
        // their product is positive to
        maxlen = zeroindex[0];
         
        // Checking for other indexex
        for(int i = 0;
                i < zeroindex.size() - 1; i++)
        {
             
            // If the difference is greater
            // than maxlen then maxlen
            // is updated
            if(zeroindex[i + 1]-
               zeroindex[i] - 1 > maxlen)
            {
                maxlen = zeroindex[i + 1] -
                         zeroindex[i] - 1;
            }
        }
         
        // To check the length of remaining
        // array after last zeroindex
        if(N - zeroindex[zeroindex.size() - 1] -
           1 > maxlen)
        {
            maxlen = N - zeroindex[
                         zeroindex.size() - 1] - 1;
        }
    }
    cout << maxlen << endl;
}
 
// Driver Code
int main()
{
    int N = 9;
    int arr[] = {7, 1, 0, 1, 2, 0, 9, 2, 1};
     
    Maxlength(arr, N);
}
     
// This code is contributed by Surendra_Gangwar

Java




// Java program to find maximum
// length subarray having non
// zero product
import java.util.*;
 
class GFG{
 
// Function that returns the
// maximum length subarray
// having non zero product
static void Maxlength(int arr[],int N)
{
    Vector<Integer> zeroindex = new Vector<Integer>();
     
    int maxlen;
     
    // zeroindex list to store indexex
    // of zero
    for(int i = 0; i < N; i++)
    {
        if (arr[i] == 0)
            zeroindex.add (i);
    }
             
    if (zeroindex.size() == 0)
    {
         
        // If zeroindex list is empty
        // then Maxlength is as
        // size of array
        maxlen = N;
    }
     
    // If zeroindex list is not empty
    else
    {
         
        // for example list 1 1 0 0 1
        // is on indexex 0 1 2 3 4
         
        // first zero is on index 2
        // that means two numbers positive,
        // before index 2 so as
        // their product is positive to
        maxlen = (int)zeroindex.get(0);
         
        // Checking for other indexex
        for(int i = 0;
                i < zeroindex.size() - 1; i++)
        {
             
            // If the difference is greater
            // than maxlen then maxlen
            // is updated
            if ((int)zeroindex.get(i + 1) -
                (int)zeroindex.get(i) - 1 > maxlen)
            {
                maxlen = (int)zeroindex.get(i + 1) -
                         (int)zeroindex.get(i) - 1;
            }
        }
         
        // To check the length of remaining
        // array after last zeroindex
        if (N - (int)zeroindex.get(
                     zeroindex.size() - 1) -
                                    1 > maxlen)
        {
            maxlen = N - (int)zeroindex.get(
                              zeroindex.size() - 1) - 1;
        }
    }
    System.out.println(maxlen);
}
 
// Driver code
public static void main(String args[])
{
    int N = 9;
    int arr[] = { 7, 1, 0, 1, 2, 0, 9, 2, 1 };
     
    Maxlength(arr, N);
}
}
 
// This code is contributed by amreshkumar3

Python3




# Python3 program to find
# maximum length subarray
# having non zero product
 
# function that returns the
# maximum length subarray
# having non zero product
def Maxlength(arr, N):
     
    zeroindex =[]
     
    # zeroindex list to store indexex
    # of zero
    for i in range(N):
        if(arr[i] == 0):
            zeroindex.append(i)
             
     
    if(len(zeroindex) == 0):
        # if zeroindex list is empty
        # then Maxlength is as
        # size of array
        maxlen = N
    # if zeroindex list is not empty
    else:
        # for example list 1 1 0 0 1
        # is on indexex 0 1 2 3 4
         
        # first zero is on index 2
        # that means two numbers positive,
        # before index 2 so as
        # their product is positive to
         
        maxlen = zeroindex[0]
         
        # checking for other indexex
        for i in range(0, len(zeroindex)-1):
             
            # if the difference is greater
            # than maxlen then maxlen
            # is updated
            if(zeroindex[i + 1]\
               - zeroindex[i] - 1\
               > maxlen):
                maxlen = zeroindex[i + 1]\
                         - zeroindex[i] - 1
             
         
        # to check the length of remaining
        # array after last zeroindex
        if(N - zeroindex[len(zeroindex) - 1]\
                                 - 1 > maxlen):
            maxlen = N\
             - zeroindex[len(zeroindex) - 1] - 1
     
    print(maxlen)
 
 
# Driver Code
if __name__ == "__main__":
    N = 9
    arr = [7, 1, 0, 1, 2, 0, 9, 2, 1]
    Maxlength(arr, N)

C#




// C# program to find maximum
// length subarray having non
// zero product
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function that returns the
// maximum length subarray
// having non zero product
static void Maxlength(int []arr,int N)
{
    int[] zeroindex = new int[20000];
    int maxlen;
     
    // zeroindex list to store indexex
    // of zero
    int size = 0;
    for(int i = 0; i < N; i++)
    {
        if (arr[i] == 0)
            zeroindex[size++] = i;
    }
             
    if (size == 0)
    {
         
        // If zeroindex list is empty
        // then Maxlength is as
        // size of array
        maxlen = N;
    }
     
    // If zeroindex list is not empty
    else
    {
         
        // for example list 1 1 0 0 1
        // is on indexex 0 1 2 3 4
         
        // first zero is on index 2
        // that means two numbers positive,
        // before index 2 so as
        // their product is positive to
        maxlen = zeroindex[0];
         
        // Checking for other indexex
        for(int i = 0; i < size; i++)
        {
             
            // If the difference is greater
            // than maxlen then maxlen
            // is updated
            if (zeroindex[i + 1]-
                zeroindex[i] - 1 > maxlen)
            {
                maxlen = zeroindex[i + 1] -
                         zeroindex[i] - 1;
            }
        }
         
        // To check the length of remaining
        // array after last zeroindex
        if (N - zeroindex[size - 1] - 1 > maxlen)
        {
            maxlen = N - zeroindex[size - 1] - 1;
        }
    }
    Console.WriteLine(maxlen);
}
 
// Driver code
public static void Main()
{
    int N = 9;
    int []arr = { 7, 1, 0, 1, 2, 0, 9, 2, 1 };
     
    Maxlength(arr, N);
}
}
 
// This code is contributed by amreshkumar3

Javascript




<script>
// Javascript program to find maximum
// length subarray having non
// zero product
 
// Function that returns the
// maximum length subarray
// having non zero product
function Maxlength(arr, N)
{
    let zeroindex = Array.from({length: 20000}, (_, i) => 0);
    let maxlen;
       
    // zeroindex list to store indexex
    // of zero
    let size = 0;
    for(let i = 0; i < N; i++)
    {
        if (arr[i] == 0)
            zeroindex[size++] = i;
    }
               
    if (size == 0)
    {
           
        // If zeroindex list is empty
        // then Maxlength is as
        // size of array
        maxlen = N;
    }
       
    // If zeroindex list is not empty
    else
    {
           
        // for example list 1 1 0 0 1
        // is on indexex 0 1 2 3 4
           
        // first zero is on index 2
        // that means two numbers positive,
        // before index 2 so as
        // their product is positive to
        maxlen = zeroindex[0];
           
        // Checking for other indexex
        for(let i = 0; i < size; i++)
        {
               
            // If the difference is greater
            // than maxlen then maxlen
            // is updated
            if (zeroindex[i + 1]-
                zeroindex[i] - 1 > maxlen)
            {
                maxlen = zeroindex[i + 1] -
                         zeroindex[i] - 1;
            }
        }
           
        // To check the length of remaining
        // array after last zeroindex
        if (N - zeroindex[size - 1] - 1 > maxlen)
        {
            maxlen = N - zeroindex[size - 1] - 1;
        }
    }
    document.write(maxlen);
}
 
  // Driver Code
     
    let N = 9;
    let arr = [ 7, 1, 0, 1, 2, 0, 9, 2, 1 ];
       
    Maxlength(arr, N);
               
</script>
Output: 



3

Time complexity: O (N) 
Auxiliary Space: O (N)
 

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