Given binary string str, the task is to find the count of K length subarrays containing only 1s.
Examples
Input: str = “0101000”, K=1
Output: 2
Explanation: 0101000 -> There are 2 subarrays of length 1 containing only 1s.Input: str = “11111001”, K=3
Output: 3
Approach: The given problem can also be solved by using the Sliding Window technique. Create a window of size K initially with the count of 1s from range 0 to K-1. Then traverse the string from index 1 to N-1 and subtract the value of i-1 and add the value of i+K to the current count. Here if the current count is equal to K, increment the possible count of subarrays.
Below is the implementation of the above approach.
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the count of all possible // k length subarrays int get(string s, int k)
{ int n = s.length();
int cntOf1s = 0;
for ( int i = 0; i < k; i++)
if (s[i] == '1' )
cntOf1s++;
int ans = cntOf1s == k ? 1 : 0;
for ( int i = 1; i < n; i++) {
cntOf1s = cntOf1s - (s[i - 1] - '0' )
+ (s[i + k - 1] - '0' );
if (cntOf1s == k)
ans++;
}
return ans;
} // Driver code int main()
{ string str = "0110101110" ;
int K = 2;
cout << get(str, K) << endl;
return 0;
} |
// Java code to implement above approach import java.util.*;
public class GFG {
// Function to find the count of all possible
// k length subarrays
static int get(String s, int k)
{
int n = s.length();
int cntOf1s = 0 ;
for ( int i = 0 ; i < k; i++) {
if (s.charAt(i) == '1' ) {
cntOf1s++;
}
}
int ans = cntOf1s == k ? 1 : 0 ;
for ( int i = 1 ; i < n; i++) {
if (i + k - 1 < n) {
cntOf1s = cntOf1s - (s.charAt(i - 1 ) - '0' )
+ (s.charAt(i + k - 1 ) - '0' );
}
if (cntOf1s == k)
ans++;
}
return ans;
}
// Driver code
public static void main(String args[])
{
String str = "0110101110" ;
int K = 2 ;
System.out.println(get(str, K));
}
} // This code is contributed by Samim Hossain Mondal. |
# Python code to implement above approach # Function to find the count of all possible # k length subarrays def get(s, k):
n = len (s);
cntOf1s = 0 ;
for i in range ( 0 ,k):
if (s[i] = = '1' ):
cntOf1s + = 1 ;
ans = i if (cntOf1s = = k) else 0 ;
for i in range ( 1 , n):
if (i + k - 1 < n):
cntOf1s = cntOf1s - ( ord (s[i - 1 ]) - ord ( '0' )) + ( ord (s[i + k - 1 ]) - ord ( '0' ));
if (cntOf1s = = k):
ans + = 1 ;
return ans;
# Driver code if __name__ = = '__main__' :
str = "0110101110" ;
K = 2 ;
print (get( str , K));
# This code is contributed by 29AjayKumar |
// C# code to implement above approach using System;
public class GFG {
// Function to find the count of all possible
// k length subarrays
static int get ( string s, int k)
{
int n = s.Length;
int cntOf1s = 0;
for ( int i = 0; i < k; i++) {
if (s[i] == '1' ) {
cntOf1s++;
}
}
int ans = cntOf1s == k ? 1 : 0;
for ( int i = 1; i < n; i++) {
if (i + k - 1 < n) {
cntOf1s = cntOf1s - (s[i - 1] - '0' )
+ (s[i + k - 1] - '0' );
}
if (cntOf1s == k)
ans++;
}
return ans;
}
// Driver code
public static void Main()
{
string str = "0110101110" ;
int K = 2;
Console.WriteLine( get (str, K));
}
} // This code is contributed by ukasp. |
<script> // JavaScript code for the above approach
// Function to find the count of all possible
// k length subarrays
function get(s, k)
{
let n = s.length;
let cntOf1s = 0;
for (let i = 0; i < k; i++)
if (s[i] == '1' )
cntOf1s++;
let ans = cntOf1s == k ? 1 : 0;
for (let i = 1; i < n; i++)
{
cntOf1s = cntOf1s - (s[i - 1] - '0' )
+ (s[i + k - 1] - '0' );
if (cntOf1s == k)
ans++;
}
return ans;
}
// Driver code
let str = "0110101110" ;
let K = 2;
document.write(get(str, K) + '<br>' )
// This code is contributed by Potta Lokesh
</script>
|
3
Time Complexity: O(N), Where N is the length of the string.
Auxiliary Space: O(1).