Number of subarrays with m odd numbers

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

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 ` `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

 ``

Javascript

 ``

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 ` `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

 `=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

 ``

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 ` `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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