Count of setbits in bitwise OR of all K length substrings of given Binary String

Last Updated : 08 Mar, 2022

Given a binary string str of length N, the task is to find the number of setbits in the bitwise OR of all the K length substrings of string str.

Examples:

Input: N = 4, K = 3, str = "1111"
Output: 3
Explanation: All 3-sized substrings of S are:
"111" and "111". The OR of these strings is "111".
Therefore the number of 1 bits is 3.

Input: N = 4, K = 4, str = "0110"
Output: 2
Explanation: All 4-sized substrings of S are "0110".
The OR of these strings is "0110".
Therefore the number of 1 bits is 2.

Approach: The solution of the problem is based on the concept of Sliding window Technique. Follow the steps mentioned below:

Use sliding window technique and find all substrings of size K.
Store all substring of size K.(here vector of strings is used).
Do OR of all strings.
Count the number of setbits and return.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to make OR two string
string ORing(string a, string b)
{
    string ans = "";
    int n = a.size();
    for (int i = 0; i < n; i++) {
        if (a[i] == '1' || b[i] == '1')
            ans += "1";
        else
            ans += "0";
    }
    return ans;
}

// Function to check the setbits
// in OR of all K size substring
int solve(string str, int N, int K)
{
    // Making vector to store answer
    vector<string> v1;
    int windowsize = K;
    int i = 0;
    int j = 0;
    string temp = "";

    // Using sliding window technique
    while (j < N) {
        temp.push_back(str[j]);
        if (j - i + 1 < windowsize) {
            j++;
        }
        else {
            v1.push_back(temp);
            reverse(temp.begin(), temp.end());
            temp.pop_back();
            reverse(temp.begin(), temp.end());
            i++;
            j++;
        }
    }

    // OR of all strings which
    // are present in the vector
    string a = v1[0];
    for (int i = 1; i < v1.size(); i++) {
        a = ORing(a, v1[i]);
    }

    // Counting number of set bit
    int count = 0;
    for (int i = 0; i < a.size(); i++) {
        if (a[i] == '1') {
            count++;
        }
    }
    return count;
}

// Driver code
int main()
{
    int N = 4;
    int K = 3;
    string str = "1111";

    // Calling function
    cout << solve(str, N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG {

  // Function to make OR two String
  static String ORing(String a, String b) {
    String ans = "";
    int n = a.length();
    for (int i = 0; i < n; i++) {
      if (a.charAt(i) == '1' || b.charAt(i) == '1')
        ans += "1";
      else
        ans += "0";
    }
    return ans;
  }

  static String reverse(String input) {
    char[] a = input.toCharArray();
    int l, r = a.length - 1;
    for (l = 0; l < r; l++, r--) {
      char temp = a[l];
      a[l] = a[r];
      a[r] = temp;
    }
    return String.valueOf(a);
  }

  // Function to check the setbits
  // in OR of all K size subString
  static int solve(String str, int N, int K)
  {

    // Making vector to store answer
    Vector<String> v1 = new Vector<>();
    int windowsize = K;
    int i = 0;
    int j = 0;
    String temp = "";

    // Using sliding window technique
    while (j < N) {
      temp += (str.charAt(j));
      if (j - i + 1 < windowsize) {
        j++;
      } else {
        v1.add(temp);
        temp = reverse(temp);
        temp = temp.substring(0, temp.length() - 1);
        temp = reverse(temp);
        i++;
        j++;
      }
    }

    // OR of all Strings which
    // are present in the vector
    String a = v1.get(0);
    for (i = 1; i < v1.size(); i++) {
      a = ORing(a, v1.get(i));
    }

    // Counting number of set bit
    int count = 0;
    for (i = 0; i < a.length(); i++) {
      if (a.charAt(i) == '1') {
        count++;
      }
    }
    return count;
  }

  // Driver code
  public static void main(String[] args) {
    int N = 4;
    int K = 3;
    String str = "1111";

    // Calling function
    System.out.print(solve(str, N, K));
  }
}

// This code is contributed by Rajput-Ji

Python3

# Python code for the above approach

# Function to make OR two string
def ORing(a, b):
    ans = "";
    n = len(a)
    for i in range(n):
        if (a[i] == '1' or b[i] == '1'):
            ans += '1';
        else:
            ans += '0';
    
    return list(ans)

# Function to check the setbits
# in OR of all K size substring
def solve(str, N, K):
  
    # Making vector to store answer
    v1 = [];
    windowsize = K;
    i = 0;
    j = 0;
    temp = [];

    # Using sliding window technique
    while (j < N):
        temp.append(str[j]);
        if (j - i + 1 < windowsize):
            j += 1
        else:
            v1.append(''.join(temp));
            temp.pop(0)
            i += 1
            j += 1
        
    # OR of all strings which
    # are present in the vector
    a = v1[0];
    for i in range(1, len(v1)):
        a = ORing(a, v1[i]);
    
    # Counting number of set bit
    count = 0;
    for i in range(len(a)):
        if (a[i] == '1'):
            count = count + 1;
        
    return count;

# Driver code
N = 4;

K = 3;
str = "1111";

# Calling function
print(solve(str, N, K));

# This code is contributed by Saurabh Jaiswal

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG {

  // Function to make OR two String
  static String ORing(String a, String b) {
    String ans = "";
    int n = a.Length;
    for (int i = 0; i < n; i++) {
      if (a[i] == '1' || b[i] == '1')
        ans += "1";
      else
        ans += "0";
    }
    return ans;
  }

  static String reverse(String input) {
    char[] a = input.ToCharArray();
    int l, r = a.Length - 1;
    for (l = 0; l < r; l++, r--) {
      char temp = a[l];
      a[l] = a[r];
      a[r] = temp;
    }
    return String.Join("",a);
  }

  // Function to check the setbits
  // in OR of all K size subString
  static int solve(String str, int N, int K)
  {

    // Making vector to store answer
    List<String> v1 = new List<String>();
    int windowsize = K;
    int i = 0;
    int j = 0;
    String temp = "";

    // Using sliding window technique
    while (j < N) {
      temp += (str[j]);
      if (j - i + 1 < windowsize) {
        j++;
      } else {
        v1.Add(temp);
        temp = reverse(temp);
        temp = temp.Substring(0, temp.Length - 1);
        temp = reverse(temp);
        i++;
        j++;
      }
    }

    // OR of all Strings which
    // are present in the vector
    String a = v1[0];
    for (i = 1; i < v1.Count; i++) {
      a = ORing(a, v1[i]);
    }

    // Counting number of set bit
    int count = 0;
    for (i = 0; i < a.Length; i++) {
      if (a[i] == '1') {
        count++;
      }
    }
    return count;
  }

  // Driver code
  public static void Main(String[] args)
  {
    int N = 4;
    int K = 3;
    String str = "1111";

    // Calling function
    Console.Write(solve(str, N, K));
  }
}

// This code is contributed by 29AjayKumar

JavaScript

 <script>
        // JavaScript code for the above approach

        // Function to make OR two string
        function ORing(a, b) {
            let ans = "";
            let n = a.length;
            for (let i = 0; i < n; i++) {
                if (a[i] == '1' || b[i] == '1')
                    ans += '1';
                else
                    ans += '0';
            }
            return ans.split('');
        }

        // Function to check the setbits
        // in OR of all K size substring
        function solve(str, N, K) {
            // Making vector to store answer
            let v1 = [];
            let windowsize = K;
            let i = 0;
            let j = 0;
            let temp = [];

            // Using sliding window technique
            while (j < N) {
                temp.push(str[j]);
                if (j - i + 1 < windowsize) {
                    j++;
                }
                else {
                    v1.push(temp.join(''));
                    temp.shift()
                    i++;
                    j++;
                }
            }

            // OR of all strings which
            // are present in the vector
            let a = v1[0];
            for (let i = 1; i < v1.length; i++) {
                a = ORing(a, v1[i]);
            }

            // Counting number of set bit
            let count = 0;
            for (let i = 0; i < a.length; i++) {
                if (a[i] == '1') {
                    count = count + 1;
                }
            }
            return count;
        }

        // Driver code
        let N = 4;
        let K = 3;
        let str = "1111";

        // Calling function
        document.write(solve(str, N, K));

       // This code is contributed by Potta Lokesh
    </script>

Output

Time Complexity: O(N * N)
Auxiliary Space: O(N)

Count of setbits in bitwise OR of all K length substrings of given Binary String

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Count of setbits in bitwise OR of all K length substrings of given Binary String

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