# Length of maximum product subarray

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 
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 ` `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; ` `         `  `        ``// 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 zeroindex = ``new` `Vector();  ` `     `  `    ``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``;  ` `    ``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;  ` `         `  `        ``// 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 `

Output:

```3
```

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

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