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Number of subarrays with m odd numbers

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  • Difficulty Level : Hard
  • Last Updated : 12 Sep, 2022
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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++




// CPP program to count the
// Number of subarrays with
// m odd numbers
#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;
 
    // 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;
}


Java




// 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.


Python3




# 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


C#




// 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.


PHP




<?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.
?>


Javascript




<script>
// javascript 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)
    {
  
        var count = 0;
  
        // traverse for all possible
        // subarrays
        for (var i = 0; i < n; i++)
        {
            var odd = 0;
            for (var 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
        var a = [ 2, 2, 5, 6, 9, 2, 11 ];
        var n = a.length;
        var m = 2;
  
        document.write(countSubarrays(a, n, m));
 
// This code is contributed by bunnyram19.
 
</script>


Output

8

Time Complexity: O(n2
Auxiliary Space : O(1)

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++




// 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 + 1] = { 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;
}


Java




// 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 + 1];
        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;
 
        // Function call
        System.out.println(countSubarrays(a, n, m));
    }
}
 
// This code is contributed by akash1295.


Python3




# 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] * (n+1)
    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


C#




// 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 + 1];
        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.


PHP




<?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+1] = 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
?>


Javascript




<script>
    // Javascript 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)
    {
        let count = 0;
        let prefix = new Array(n + 1);
        prefix.fill(0);
        let odd = 0;
  
        // traverse in the array
        for (let 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;
    }
     
    let a = [ 2, 2, 5, 6, 9, 2, 11 ];
    let n = a.length;
    let m = 2;
 
    document.write(countSubarrays(a, n, m));
 
// This code is contributed by divyeshrabadiya07.
</script>


Output

8

Time Complexity: O(n)
Auxiliary Space: O(n)

Alternative approach: An alternative approach is to replace all the odd numbers with 1 and all the even numbers with zero and then calculate the number of subarrays with sum equal to m.

An efficient solution for this is while traversing the array, storing the sum so far in currsum. Also, maintain the count of different values of currsum in a map. If the value of currsum is equal to the desired sum at any instance, increment the count of subarrays by one.
The value of currsum exceeds the desired sum by currsum – sum. If this value is removed from currsum, the desired sum can be obtained. From the map, find the number of subarrays previously found having sum equal to currsum-sum. Excluding all those subarrays from the current subarray gives new subarrays having the desired sum.
So increase the count by the number of such subarrays. Note that when currsum is equal to the desired sum, check the number of subarrays previously having a sum equal to 0. Excluding those subarrays from the current subarray gives new subarrays having the desired sum. Increase the count by the number of subarrays having the sum of 0 in that case.
Below is the implementation of the approach:

C++




// CPP program to count the Number
// of subarrays with m odd numbers
// Alternative O(N) approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find number of subarrays
// with sum exactly equal to k.
int countSubarrays(int arr[], int n, int m)
{
    // STL map to store number of subarrays starting from
    // index zero having particular value of sum.
    unordered_map<int, int> prevSum;
 
    int res = 0;
 
    // Variable to store sum
    int currSum = 0;
 
    for (int i = 0; i < n; i++) {
 
        currSum += arr[i];
 
        // If currsum is equal to m
        if (currSum == m)
            res++;
 
        // currsum exceeds m find number of
        // subarrays having this sum and exclude
        // those subarrays from currsum by
        // increasing count by same amount.
        if (prevSum.find(currSum - m) != prevSum.end())
            res += (prevSum[currSum - m]);
 
        // Increment currSum count
        prevSum[currSum]++;
    }
 
    return res;
}
 
// Driver Code
int main()
{
    int a[] = { 2, 2, 5, 6, 9, 2, 11 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = 2;
   
    // replace all the odd numbers with 1
    // and all the even numbers with 0
    for (int i = 0; i < n; i++) {
        if (a[i] % 2 == 0)
            a[i] = 0;
        else
            a[i] = 1;
    }
   
      // Function Call
    cout << countSubarrays(a, n, m);
 
    return 0;
}


Output

8

Time Complexity: O(n)
Auxiliary Space: O(n)


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