Longest subarray with odd product

Given an array arr[] consisting of N elements, the task is to find the length of longest subarray with odd product.

Examples:

Input: arr[] = {3, 5, 2, 1} 
Output:
Explanation: 
Subarrays with consecutive odd elements are {3, 5}, and {1}. 
Since, {3, 5} is the longer, the answer is 2.

Input: arr[] = {8, 5, 3, 1, 0} 
Output:
Explanation: 
Longest subarray with odd product is {5, 3, 1}. 

Approach: 
Following observations are required to solve the problem: 



The Product of two odd numbers generates an odd number. 
The Product of one odd and one even number generates an even number. 
The Product of two even numbers generates an even number. 

From the above observations, we can conclude that the longest subarray of consecutive odd elements in the array is the required answer. 
Follow the steps below to solve the problem: 

  • Traverse the array and check if the current element is even or odd.
  • If the current element is odd, set count to 1 and keep increasing count until an even element is encountered in the array.
  • Once an even element is encountered, compare count with ans and update ans storing maximum of the two.
  • Repeat the above steps for the remaining array.
  • Finally, print the value stored in ans.

Below is the implementation of the above approach:

C++

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// C++ Program to find the longest
// subarray with odd product
#include <bits/stdc++.h>
using namespace std;
  
// Function to return length of
// longest subarray with odd product
int Maxlen(int arr[], int n)
{
    int ans = 0;
    int count = 0;
    for (int i = 0; i < n; i++) {
  
        // If even element
        // is encountered
        if (arr[i] % 2 == 0)
            count = 0;
        else
            count++;
  
        // Update maximum
        ans = max(ans, count);
    }
    return ans;
}
  
// Driver Code
int main()
{
  
    // int arr[] = { 6, 3, 5, 1 };
    int arr[] = { 1, 7, 2 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << Maxlen(arr, n) << endl;
    return 0;
}

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Java

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// Java program to find the longest
// subarray with odd product
import java.util.*;
  
class GFG{
      
// Function to return length of
// longest subarray with odd product
static int Maxlen(int arr[], int n)
{
    int ans = 0;
    int count = 0;
    for(int i = 0; i < n; i++)
    {
          
        // If even element
        // is encountered
        if (arr[i] % 2 == 0)
            count = 0;
        else
            count++;
  
        // Update maximum
        ans = Math.max(ans, count);
    }
    return ans;
}
  
// Driver Code
public static void main(String s[])
{
    int arr[] = { 1, 7, 2 };
    int n = arr.length;
      
    System.out.println(Maxlen(arr, n));
}
  
// This code is contributed by rutvik_56

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Python3

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# Python3 program to find the longest 
# subarray with odd product
  
# Function to return length of 
# longest subarray with odd product 
def Maxlen(a, n):
      
    ans = 0
    count = 0
      
    for i in range(n): 
  
        # If even element 
        # is encountered 
        if a[i] % 2 == 0
            count = 0
        else
            count += 1
              
        # Update maximum 
        ans = max(ans, count)
          
    return ans
  
# Driver code
arr = [ 1, 7, 2
n = len(arr) 
  
print(Maxlen(arr, n))
  
# This code is contributed by amreshkumar3

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C#

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// C# program to find the longest 
// subarray with odd product 
using System; 
  
class GFG{ 
  
// Function to return length of 
// longest subarray with odd product 
static int Maxlen(int []arr, int n) 
    int ans = 0; 
    int count = 0; 
      
    for(int i = 0; i < n; i++) 
    
          
        // If even element 
        // is encountered 
        if (arr[i] % 2 == 0) 
            count = 0; 
        else
            count++; 
      
        // Update maximum 
        ans = Math.Max(ans, count); 
    
    return ans; 
  
// Driver Code 
public static void Main() 
    int []arr = { 1, 7, 2 }; 
    int n = arr.Length; 
      
    Console.WriteLine(Maxlen(arr, n)); 
  
// This code is contributed by amreshkumar3

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Output: 

2

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

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Improved By : rutvik_56, amreshkumar3