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:
// 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 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 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# 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 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. ?> |
<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> |
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:
// 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 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 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# 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 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 ?> |
<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> |
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:
// 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;
} |
// Java program to count the Number // of subarrays with m odd numbers // Alternative O(N) approach import java.util.*;
public class GFG {
// Function to find number of subarrays
// with sum exactly equal to k.
static int countSubarrays( int arr[], int n, int m)
{
// STL map to store number of subarrays starting
// from index zero having particular value of sum.
HashMap<Integer, Integer> prevSum = new HashMap<>();
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.containsKey(currSum - m))
res += (prevSum.get(currSum - m));
// Increment currSum count
prevSum.put(currSum,
prevSum.getOrDefault(currSum, 0 )
+ 1 );
}
return res;
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 2 , 2 , 5 , 6 , 9 , 2 , 11 };
int n = a.length;
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
System.out.println(countSubarrays(a, n, m));
}
} // This code is contributed by Karandeep1234 |
# Python program to count the Number # of subarrays with m odd numbers # Alternative O(N) approach # Function to find number of subarrays # with sum exactly equal to k. def countSubarrays(arr, n, m):
# map to store number of subarrays starting from
# index zero having particular value of sum.
prevSum = {}
# Variable to store sum
res = currSum = 0
for i in range (n):
currSum + = arr[i]
# If currsum is equal to m
if currSum = = m: res + = 1
# currsum exceeds m find number of
# subarrays having this sum and exclude
# those subarrays from currsum by
# increasing count by same amount.
if currSum - m in prevSum: res + = prevSum[currSum - m]
# Increment currSum count
if currSum in prevSum: prevSum[currSum] + = 1
else : prevSum[currSum] = 1
return res
# Driver Code a = [ 2 , 2 , 5 , 6 , 9 , 2 , 11 ]
n = len (a)
m = 2
# replace all the odd numbers with 1 # and all the even numbers with 0 for i in range (n):
if (a[i] % 2 = = 0 ): a[i] = 0
else : a[i] = 1
# Function Call print (countSubarrays(a, n, m))
# This code is contributed by hardikkushwaha. |
// C# program to count the Number // of subarrays with m odd numbers // Alternative O(N) approach using System;
using System.Collections;
using System.Collections.Generic;
public class GFG {
// Function to find number of subarrays
// with sum exactly equal to k.
static int countSubarrays( int [] arr, int n, int m)
{
// STL map to store number of subarrays starting
// from index zero having particular value of sum.
Dictionary< int , int > prevSum
= new Dictionary< int , int >();
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.ContainsKey(currSum - m))
res += (prevSum[currSum - m]);
// Increment currSum count
if (prevSum.ContainsKey(currSum))
prevSum[currSum] = prevSum[currSum] + 1;
else
prevSum.Add(currSum, 1);
}
return res;
}
// Driver Code
public static void Main( string [] args)
{
int [] a = { 2, 2, 5, 6, 9, 2, 11 };
int n = a.Length;
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
Console.WriteLine(countSubarrays(a, n, m));
}
} // This code is contributed by Karandeep1234 |
// Function to find number of subarrays // with sum exactly equal to k. function countSubarrays(arr, n, m) {
// Map to store number of subarrays starting from
// index zero having particular value of sum.
const prevSum = new Map();
let res = 0;
// Variable to store sum
let currSum = 0;
for (let 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.has(currSum - m)) res += prevSum.get(currSum - m);
// Increment currSum count
prevSum.set(currSum, (prevSum.get(currSum) || 0) + 1);
}
return res;
} // Driver Code function main() {
const a = [2, 2, 5, 6, 9, 2, 11];
const n = a.length;
const m = 2;
// Replace all the odd numbers with 1
// and all the even numbers with 0
for (let i = 0; i < n; i++) {
if (a[i] % 2 === 0) a[i] = 0;
else a[i] = 1;
}
// Function Call
console.log(countSubarrays(a, n, m));
} main(); // This code is contributed by shvrekhan. |
8
Time Complexity: O(n)
Auxiliary Space: O(n)
Another Approach:
Exactly(k) = atMost(k) – atMost(k-1)
While traversing, calculate the count of odd numbers. If count of odd numbers became greater than m, then increment the index i and check whether the arr[i] is an odd element or not. If arr[i] is odd then decrement the count of odd till the count of odd becomes equal to m. Store the length of subarray in ans.
Do the above process for subarrays with at most k odd elements and for the subarrays with at most k-1 odd elements.
If we subtract the subarrays with at most k-1 odd elements occur from the subarrays with at most k odd elements occur, we get exactly the subarrays with k odd elements.
#include <bits/stdc++.h> using namespace std;
int atMost( int arr[], int n, int m){
int i=0, ans=0, odd=0;
for ( int j=0;j<n;j++){
//count odd elements
if (arr[j]%2==1){
odd++;
}
/* if count of odd elements is greater than m,
then increment the i index and check whether
arr[i] is odd or not*/
while (i<= j && odd>m){
// if arr[i] is odd, then decrement the odd.
if (arr[i]%2==1){
odd--;
}
// increment the index i
i++;
}
ans+= j-i+1;
}
return ans;
} int countSubarrays( int arr[], int n, int m){
// subtract the subarrays with at most k-1 odd elements occur from
// the subarrays with at most k odd elements occur, we get exactly
// subarray with k odd elements.
return atMost(arr, n, m) - atMost(arr, n, m-1);
} int main() {
int arr[] = { 2, 2, 5, 6, 9, 2, 11 };
int n = sizeof (arr) / sizeof (arr[0]);
int m = 2;
cout << countSubarrays(arr, n, m);
return 0;
} // this code is contributed by 525tamannacse11 // and minor correction by Rahul Verma |
import java.util.*;
public class Main {
public static int atMost( int [] arr, int n, int m) {
int i = 0 , ans = 0 , odd = 0 ;
for ( int j = 0 ; j < n; j++) {
// count odd elements
if (arr[j] % 2 == 1 ) {
odd++;
}
/* if count of odd elements is greater than m,
then increment the i index and check whether
arr[i] is odd or not*/
while (i <= j && odd > m) {
// if arr[i] is odd, then decrement the odd.
if (arr[i] % 2 == 1 ) {
odd--;
}
// increment the index i
i++;
}
ans += j - i + 1 ;
}
return ans;
}
public static int countSubarrays( int [] arr, int n, int m) {
// subtract the subarrays with at most k-1 odd elements occur from
// the subarrays with at most k odd elements occur, we get exactly
// subarray with k odd elements.
return atMost(arr, n, m) - atMost(arr, n, m- 1 );
}
public static void main(String[] args) {
int [] arr = { 2 , 2 , 5 , 6 , 9 , 2 , 11 };
int n = arr.length;
int m = 2 ;
System.out.println(countSubarrays(arr, n, m));
}
} |
def atMost(arr, n, m):
i = 0
ans = 0
odd = 0
for j in range (n): #count odd elements
if arr[j] % 2 = = 1 :
odd + = 1
# if count of odd elements is greater than m,
# then increment the i index and check whether
# arr[i] is odd or not
while i < = j and odd > m:
# if arr[i] is odd, then decrement the odd.
if arr[i] % 2 = = 1 :
odd - = 1
# increment the index i
i + = 1
ans + = j - i + 1
return ans
def countSubarrays(arr, n, m):
# subtract the subarrays with at most k-1 odd elements occur from
# the subarrays with at most k odd elements occur, we get exactly
# subarray with k odd elements.
return atMost(arr, n, m) - atMost(arr, n, m - 1 )
arr = [ 2 , 2 , 5 , 6 , 9 , 2 , 11 ]
n = len (arr)
m = 2
print (countSubarrays(arr, n, m))
|
using System;
public class Program {
public static int AtMost( int [] arr, int n, int m) {
int i = 0, ans = 0, odd = 0;
for ( int j = 0; j < n; j++)
{
// count odd elements
if (arr[j] % 2 == 1) {
odd++;
}
// if count of odd elements is greater than m,
// then increment the i index and check whether
// arr[i] is odd or not
while (i<= j && odd > m)
{
// if arr[i] is odd, then decrement odd
if (arr[i] % 2 == 1) {
odd--;
}
// increment the index i
i++;
}
ans += j - i + 1;
}
return ans;
}
public static int CountSubarrays( int [] arr, int n, int m)
{
// subtract the subarrays with at most k-1 odd elements occur from
// the subarrays with at most k odd elements occur, we get exactly
// subarray with k odd elements.
return AtMost(arr, n, m) - AtMost(arr, n, m - 1);
}
public static void Main() {
int [] arr = { 2, 2, 5, 6, 9, 2, 11 };
int n = arr.Length;
int m = 2;
Console.WriteLine(CountSubarrays(arr, n, m));
}
} // This code is contributed by Prajwal Kandekar // and minor correction by Rahul Verma |
function atMost(arr, n, m) {
let i = 0;
let ans = 0;
let odd = 0;
for (let j = 0; j < n; j++) { //count odd elements
if (arr[j] % 2 === 1) {
odd++;
}
/* if count of odd elements is greater than m,
then increment the i index and check whether
arr[i] is odd or not*/
while (i<= j && odd > m) {
// if arr[i] is odd, then decrement the odd.
if (arr[i] % 2 === 1) {
odd--;
}
// increment the index i
i++;
}
ans += j - i + 1;
}
return ans;
} function countSubarrays(arr, n, m) {
// subtract the subarrays with at most k-1 odd elements occur from
// the subarrays with at most k odd elements occur, we get exactly
// subarray with k odd elements.
return atMost(arr, n, m) - atMost(arr, n, m - 1);
} const arr = [2, 2, 5, 6, 9, 2, 11]; const n = arr.length; const m = 2; console.log(countSubarrays(arr, n, m)); |
8
Time Complexity: O(n)
Space Complexity: O(1)