Number of subarrays with m odd numbers

Given an array of n elements and an integer m, we need to write a program to find the number of contiguous subarrays in the array which contains exactly m odd numbers.

Examples :

Input : arr = {2, 5, 6, 9},  m = 2 
Output : 2
Explanation: subarrays are [2, 5, 6, 9] 
and [5, 6, 9]

Input : arr = {2, 2, 5, 6, 9, 2, 11},  m = 2
Output : 8
Explanation: subarrays are [2, 2, 5, 6, 9], 
[2, 5, 6, 9], [5, 6, 9], [2, 2, 5, 6, 9, 2], 
[2, 5, 6, 9, 2], [5, 6, 9, 2], [6, 9, 2, 11] 
and [9, 2, 11]

Naive Approach: The naive approach is to generate all possible subarrays and simultaneously checking for the subarrays with m odd numbers.



Below is the implementation of the above approach:

C++

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// CPP program to count the
// Number of subarrays with 
// m odd numbers
#include <bits/stdc++.h>
using namespace std;
  
// fucntion that returns 
// the count of subarrays
// with m odd numbers
int countSubarrays(int a[], 
                   int n, int m)
{
    int count = 0;
  
    // traverse for all 
    // possible subarrays
    for (int i = 0; i < n; i++) 
    {
        int odd = 0;
        for (int j = i; j < n; j++)
        {
            if (a[j] % 2)
                odd++;
  
            // if count of odd numbers in
            // subarray is m
            if (odd == m)
                count++;
        }
    }
    return count;
}
  
// Driver Code
int main()
{
    int a[] = { 2, 2, 5, 6, 9, 2, 11 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = 2;
  
    cout << countSubarrays(a, n, m);
    return 0;
}

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Java

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// Java program to count the number of 
// subarrays with m odd numbers
import java.util.*;
  
class GFG {
  
    // function that returns the count of
    // subarrays with m odd numbers   
       static int countSubarrays(int a[],
                              int n, int m)
    {
          
        int count = 0;
  
        // traverse for all possible
        // subarrays
        for (int i = 0; i < n; i++) 
        {
            int odd = 0;
            for (int j = i; j < n; j++) 
            {
                if (a[j] % 2 ==0)
                    odd++;
      
                // if count of odd numbers
                // in subarray is m
                if (odd == m)
                    count++;
            }
        }
          
        return count;
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int a[] = { 2, 2, 5, 6, 9, 2, 11 };
        int n = a.length;
        int m = 2;
          
        System.out.println(countSubarrays(a, n, m));
    }
}
  
// This code is contributed by akash1295.

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Python3

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# Python3 program to count the
# Number of subarrays with 
# m odd numbers
  
# function that returns the count 
# of subarrays with m odd numbers
def countSubarrays(a, n, m):
    count = 0;
  
    # traverse for all
    # possible subarrays
    for i in range(n): 
        odd = 0;
        for j in range(i, n): 
            if (a[j] % 2):
                odd += 1;
  
            # if count of odd numbers 
            # in subarray is m
            if (odd == m):
                count += 1;
    return count;
  
# Driver Code
a = [ 2, 2, 5, 6, 9, 2, 11 ];
n = len(a);
m = 2;
  
print(countSubarrays(a, n, m));
  
# This code is contributed by mits

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C#

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// C# program to count the number of 
// subarrays with m odd numbers
using System;
  
class GFG {
  
    // function that returns the count of
    // subarrays with m odd numbers 
    static int countSubarrays(int []a,
                            int n, int m)
    {
          
        int count = 0;
  
        // traverse for all possible
        // subarrays
        for (int i = 0; i < n; i++) 
        {
            int odd = 0;
            for (int j = i; j < n; j++) 
            {
                if (a[j] % 2 ==0)
                    odd++;
      
                // if count of odd numbers
                // in subarray is m
                if (odd == m)
                    count++;
            }
        }
          
        return count;
    }
      
    // Driver code
    public static void Main ()
    {
        int []a = { 2, 2, 5, 6, 9, 2, 11 };
        int n = a.Length;
        int m = 2;
          
        Console.WriteLine(
                  countSubarrays(a, n, m));
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP program to count the
// Number of subarrays with 
// m odd numbers
  
  
// function that returns the count 
// of subarrays with m odd numbers
function countSubarrays( $a, $n, $m)
{
    $count = 0;
  
    // traverse for all
    // possible subarrays
    for ( $i = 0; $i < $n; $i++) 
    {
        $odd = 0;
        for ( $j = $i; $j < $n; $j++) 
        {
            if ($a[$j] % 2)
                $odd++;
  
            // if count of odd numbers in
            // subarray is m
            if ($odd == $m)
                $count++;
        }
    }
    return $count;
}
  
// Driver Code
$a = array( 2, 2, 5, 6, 9, 2, 11 );
$n = count($a);
$m = 2;
  
echo countSubarrays($a, $n, $m);
  
// This code is contributed by anuj_67.
?>

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Output :

8

Time Complexity: O(n2)

Efficient Approach: An efficient approach is to while traversing, compute the prefix[] array. Prefix[i] stores the number of prefixes which has ‘i’ odd numbers in it. We increase the count of odd numbers if the array element is an odd one. When the count of odd numbers exceeds or is equal to m, add the number of prefixes which has “(odd-m)” numbers to the answer. At every step odd>=m, we calculate the number of subarrays formed till a particular index with the help of prefix array. prefix[odd-m] provides us with the number of prefixes which has “odd-m” odd numbers, which is added to the count to get the number of subarrays till the index.

Below is the implementation of the above approach:

C++

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// CPP program to count the Number 
// of subarrays with m odd numbers
// O(N) approach
#include <bits/stdc++.h>
using namespace std;
  
// function that returns the count
// of subarrays with m odd numbers
int countSubarrays(int a[], int n, int m)
{
    int count = 0;
    int prefix[n] = { 0 };
    int odd = 0;
  
    // traverse in the array
    for (int i = 0; i < n; i++) 
    {
  
        prefix[odd]++;
  
        // if array element is odd
        if (a[i] & 1)
            odd++;
  
        // when number of odd elements>=M
        if (odd >= m)
            count += prefix[odd - m];
    }
  
    return count;
}
  
// Driver Code
int main()
{
    int a[] = { 2, 2, 5, 6, 9, 2, 11 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = 2;
      
    cout << countSubarrays(a, n, m);
      
    return 0;
}

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Java

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// Java program to count the 
// number of subarrays with
// m odd numbers
import java.util.*;
  
class GFG
{
  
    // function that returns the count of
    // subarrays with m odd numbers 
    public static int countSubarrays(int a[],
                                     int n, int m)
    {
        int count = 0;
        int prefix[] = new int[n];
        int odd = 0;
      
        // traverse in the array
        for (int i = 0; i < n; i++)
        {
            prefix[odd]++;
      
            // if array element is odd
            if ((a[i] & 1) == 1)
                odd++;
      
            // when number of odd 
            // elements >= M
            if (odd >= m)
                count += prefix[odd - m];
        }
      
        return count;
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int a[] = { 2, 2, 5, 6, 9, 2, 11 };
        int n = a.length;
        int m = 2;
          
        System.out.println(countSubarrays(a, n, m));
    }
}
  
// This code is contributed by akash1295.

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Python3

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# Python3 program to count the Number 
# of subarrays with m odd numbers 
# O(N) approach 
  
# function that returns the count 
# of subarrays with m odd numbers 
def countSubarrays(a, n, m): 
    count = 0
    prefix = [0] *
    odd = 0
  
    # traverse in the array 
    for i in range(n):
        prefix[odd] += 1
  
        # if array element is odd 
        if (a[i] & 1): 
            odd += 1
  
        # when number of odd elements>=M 
        if (odd >= m): 
            count += prefix[odd - m] 
  
    return count 
      
# Driver Code 
a = [2, 2, 5, 6, 9, 2, 11]
n = len(a) 
m = 2
  
print(countSubarrays(a, n, m))
  
# This code is contributed 29Ajaykumar

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C#

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// C# program to count the number of
// subarrays with m odd numbers
using System;
  
class GFG {
  
    // function that returns the count of
    // subarrays with m odd numbers 
    public static int countSubarrays(int []a,
                                int n, int m)
    {
        int count = 0;
        int []prefix = new int[n];
        int odd = 0;
      
        // traverse in the array
        for (int i = 0; i < n; i++)
        {
            prefix[odd]++;
      
            // if array element is odd
            if ((a[i] & 1) == 1)
                odd++;
      
            // when number of odd 
            // elements >= M
            if (odd >= m)
                count += prefix[odd - m];
        }
      
        return count;
    }
      
    // Driver code
    public static void Main ()
    {
        int []a = { 2, 2, 5, 6, 9, 2, 11 };
        int n = a.Length;
        int m = 2;
          
        Console.WriteLine(
                  countSubarrays(a, n, m));
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP program to count the Number 
// of subarrays with m odd numbers 
// O(N) approach 
  
// function that returns the count 
// of subarrays with m odd numbers 
function countSubarrays(&$a, $n, $m
    $count = 0; 
    $prefix[$n] = array(); 
    $odd = 0; 
  
    // traverse in the array 
    for ($i = 0; $i < $n; $i++) 
    
  
        $prefix[$odd]++; 
  
        // if array element is odd 
        if ($a[$i] & 1) 
            $odd++; 
  
        // when number of odd elements>=M 
        if ($odd >= $m
            $count += $prefix[$odd - $m]; 
    
  
    return $count
  
// Driver Code 
$a = array(2, 2, 5, 6, 9, 2, 11 ); 
$n = sizeof($a); 
$m = 2; 
  
echo countSubarrays($a, $n, $m); 
  
// This code is contributed 
// by Shivi_Aggarwal
?>

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Output :

8

Time Complexity: O(n)



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