Given a binary string S of size N and an integer K. The task is to find the maximum number of set bit appears in a substring of size K.
Examples:
Input: S = “100111010”, K = 3
Output: 3
Explanation:
The substring “111” contains 3 set bits.Input:S = “0000000”, K = 4
Output: 0
Explanation: S doesn’t have any set bits in it, so ans is 0.
Naive Approach:
- Generate all substring of size K.
- Find maximum of count of set bits in all substrings.
Time Complexity: O( N2).
Auxiliary Space: O(1).
Efficient Approach: The problem can be solved using Sliding window technique.
- Take maxcount variable to store maximum count of set bit and Count variable to store count set bit of current window.
- Traverse string from 1 to K and calculate the count of set bits and store as maxcount.
- Traverse string from K + 1 to length of the string.
- At every iteration, decrease count if (K – i)th bit is set. Increase count if ith bit is set. Compare and update maxcount.
- After complete array traversal, finally return maxcount.
Below is the implementation of the above approach:
C++
// C++ program to find the maximum // set bits in a substring of size K #include<bits/stdc++.h> using namespace std; // Function that find Maximum number // of set bit appears in a substring // of size K. int maxSetBitCount(string s, int k) { int maxCount = 0, n = s.length(); int count = 0; // Traverse string 1 to k for ( int i = 0; i < k; i++) { // Increment count if // character is set bit if (s[i] == '1' ) count++; } maxCount = count; // Traverse string k+1 // to length of string for ( int i = k; i < n; i++) { // Remove the contribution of the // (i - k)th character which is no // longer in the window if (s[i - k] == '1' ) count--; // Add the contribution of // the current character if (s[i] == '1' ) count++; // Update maxCount at for // each window of size k maxCount = max(maxCount, count); } // Return maxCount return maxCount; } // Driver code int main() { string s = "100111010" ; int k = 3; cout << (maxSetBitCount(s, k)); return 0; } // This code is contributed by Rajput-Ji |
Java
// Java program to find the maximum // set bits in a substring of size K import java.util.*; class GFG { // Function that find Maximum number // of set bit appears in a substring // of size K. static int maxSetBitCount(String s, int k) { int maxCount = 0 , n = s.length(); int count = 0 ; // Traverse string 1 to k for ( int i = 0 ; i < k; i++) { // Increment count if // character is set bit if (s.charAt(i) == '1' ) count++; } maxCount = count; // Traverse string k+1 // to length of string for ( int i = k; i < n; i++) { // remove the contribution of the // (i - k)th character which is no // longer in the window if (s.charAt(i - k) == '1' ) count--; // add the contribution of // the current character if (s.charAt(i) == '1' ) count++; // update maxCount at for // each window of size k maxCount = Math.max(maxCount, count); } // return maxCount return maxCount; } // Driver Program public static void main(String[] args) { String s = "100111010" ; int k = 3 ; System.out.println(maxSetBitCount(s, k)); } } |
Python3
# Python3 program to find the maximum # set bits in a substring of size K # Function that find Maximum number # of set bit appears in a substring # of size K. def maxSetBitCount(s, k): maxCount = 0 n = len (s) count = 0 # Traverse string 1 to k for i in range (k): # Increment count if # character is set bit if (s[i] = = '1' ): count + = 1 maxCount = count # Traverse string k+1 # to length of string for i in range (k, n): # Remove the contribution of the # (i - k)th character which is no # longer in the window if (s[i - k] = = '1' ): count - = 1 # Add the contribution of # the current character if (s[i] = = '1' ): count + = 1 # Update maxCount at for # each window of size k maxCount = max (maxCount, count) # Return maxCount return maxCount # Driver code if __name__ = = '__main__' : s = "100111010" k = 3 print (maxSetBitCount(s, k)) # This code is contributed by mohit kumar 29 |
C#
// C# program to find the maximum // set bits in a substring of size K using System; class GFG { // Function that find Maximum number // of set bit appears in a substring // of size K. static int maxSetBitCount( string s, int k) { int maxCount = 0, n = s.Length; int count = 0; // Traverse string 1 to k for ( int i = 0; i < k; i++) { // Increment count if // character is set bit if (s[i] == '1' ) count++; } maxCount = count; // Traverse string k+1 // to length of string for ( int i = k; i < n; i++) { // remove the contribution of the // (i - k)th character which is no // longer in the window if (s[i - k] == '1' ) count--; // add the contribution of // the current character if (s[i] == '1' ) count++; // update maxCount at for // each window of size k maxCount = Math.Max(maxCount, count); } // return maxCount return maxCount; } // Driver Program public static void Main() { string s = "100111010" ; int k = 3; Console.Write(maxSetBitCount(s, k)); } } // This code is contributed by Code_Mech |
Javascript
<script> // Javascript program to find the maximum // set bits in a substring of size K // Function that find Maximum number // of set bit appears in a substring // of size K. function maxSetBitCount(s, k) { var maxCount = 0, n = s.length; var count = 0; // Traverse string 1 to k for ( var i = 0; i < k; i++) { // Increment count if // character is set bit if (s[i] == '1' ) count++; } maxCount = count; // Traverse string k+1 // to length of string for ( var i = k; i < n; i++) { // Remove the contribution of the // (i - k)th character which is no // longer in the window if (s[i - k] == '1' ) count--; // Add the contribution of // the current character if (s[i] == '1' ) count++; // Update maxCount at for // each window of size k maxCount = Math.max(maxCount, count); } // Return maxCount return maxCount; } // Driver code var s = "100111010" ; var k = 3; document.write(maxSetBitCount(s, k)); // This code is contributed by famously. </script> |
3
Time Complexity: O(N).
Auxiliary Space: O(1).
Brute Force in python:
Approach:
One simple approach to solve the problem is to generate all possible substrings of length K from the binary string S and count the number of set bits in each substring. Finally, return the maximum count of set bits among all the substrings.
- Define a function count_set_bits(s) that takes a binary string s and returns the count of set bits (i.e., number of ‘1’s) in it.
- Define a function max_set_bits(s, k) that takes a binary string s and an integer k and returns the maximum count of set bits in a substring of length k of s.
- Get the length n of the binary string s.
- Initialize a variable ans to 0, which will hold the maximum count of set bits among all the substrings.
- Iterate over all possible substrings of length k in the binary string s. We do this by iterating over all starting positions i from 0 to n-k and taking the substring s[i:i+k].
- For each substring, compute the count of set bits by calling the function count_set_bits(s[i:i+k]).
- Update the ans variable to hold the maximum count of set bits among all the substrings seen so far.
- Return the ans variable as the output.
Example usage: Call the max_set_bits() function with the binary string “100111010” and k=3 as inputs and print the output, which should be 3. Similarly, call the function with the binary string “0000000” and k=4 as inputs and print the output, which should be 0.
C++
#include <iostream> using namespace std; int countSetBits(string s) { int count = 0; for ( char c : s) { if (c == '1' ) { count++; } } return count; } int maxSetBits(string s, int k) { int n = s.length(); int ans = 0; for ( int i = 0; i <= n - k; i++) { string subStr = s.substr(i, k); int setBitsCount = countSetBits(subStr); ans = max(ans, setBitsCount); } return ans; } int main() { // Example usage cout << maxSetBits( "100111010" , 3) << endl; cout << maxSetBits( "0000000" , 4) << endl; return 0; } |
Java
public class Main { // Function to count the number of '1's in a given // string. public static int countSetBits(String s) { int count = 0 ; for ( char c : s.toCharArray()) { if (c == '1' ) { count++; } } return count; } // Function to find the maximum count of '1's in a // substring of length 'k'. public static int maxSetBits(String s, int k) { int n = s.length(); int ans = 0 ; // Iterate through the string to find substrings of // length 'k'. for ( int i = 0 ; i <= n - k; i++) { String subStr = s.substring(i, i + k); int setBitsCount = countSetBits(subStr); ans = Math.max(ans, setBitsCount); } return ans; } public static void main(String[] args) { // Example usage and testing of the maxSetBits // function. System.out.println( "Maximum set bits in '100111010' with k=3: " + maxSetBits( "100111010" , 3 )); System.out.println( "Maximum set bits in '0000000' with k=4: " + maxSetBits( "0000000" , 4 )); } } |
Python3
def count_set_bits(s): return s.count( '1' ) def max_set_bits(s, k): n = len (s) ans = 0 for i in range (n - k + 1 ): ans = max (ans, count_set_bits(s[i:i + k])) return ans # Example usage print (max_set_bits( "100111010" , 3 )) # Output: 3 print (max_set_bits( "0000000" , 4 )) # Output: 0 |
C#
using System; class Program { // Function to count the number of set bits in a string static int CountSetBits( string s) { int count = 0; foreach ( char c in s) { if (c == '1' ) { count++; } } return count; } // Function to find the maximum number of set bits in a // substring of length k static int MaxSetBits( string s, int k) { int n = s.Length; int ans = 0; for ( int i = 0; i <= n - k; i++) { string subStr = s.Substring(i, k); int setBitsCount = CountSetBits(subStr); ans = Math.Max(ans, setBitsCount); } return ans; } static void Main() { // Example usage Console.WriteLine(MaxSetBits( "100111010" , 3)); Console.WriteLine(MaxSetBits( "0000000" , 4)); } } |
Javascript
function countSetBits(s) { let count = 0; for (let i = 0; i < s.length; i++) { if (s[i] === '1' ) { count++; } } return count; } function maxSetBits(s, k) { const n = s.length; let ans = 0; for (let i = 0; i <= n - k; i++) { const subStr = s.substring(i, i + k); const setBitsCount = countSetBits(subStr); ans = Math.max(ans, setBitsCount); } return ans; } // Example usage console.log(maxSetBits( "100111010" , 3)); console.log(maxSetBits( "0000000" , 4)); |
3 0
Time Complexity: O(N * K^2), where N is the length of the binary string S.
Space Complexity: O(1)
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