Average Value of Set Bit Count in Binary String After K Operations

In this problem, we need to find the average value of the count of set bits after performing K operations of all choices on the given string.

There can be a brute force approach to solve the problem, but we will use the probability principles to overcome the time complexity of the brute force approach.

Problem statement - We have given an integer N, array arr[] containing K positive integers, and a binary string of length N containing only set bits. We need to find the average value of the count of set bits after performing all possible choices of the K operations. IN ith operation, we can flip any arr[i] a bit of the given string.

Sample examples

Input - N = 2, arr[] = {1, 2}

Output- 1

Explanation- The initial binary string is 11.

• In the first move, we can flip the first character, and the string will be 01.

• In the second operation, we need to flip any two bits. So the string will be 10.

• The second choice can start by flipping the second character in the first move, and the string will be 10.

• In the second move of the current operation, we need to flip any 2 bits, and the string can be 01.

So, we have 2 choices, and the final strings can be 01 or 10.

Total choices = 2, total set bits in final string = 2, ans = 2/2 = 1.

Input - N = 3, arr[] = {2, 2}

Output- 1.6667

Explanation - We have an initial string is 111.

• In the first operation, we can flip any 2 characters. So, the string can be 001, 100, 010.

• In the second operation, we can flip the 2 bits in the resultant string we get from the first operation.

• When we flip any 2 bits of 001, we get 111, 010, and 100.

• When we flip any 2 bits of 100, we can get 010, 111, and 001.

• When we flip any 2 bits of 010, we can get 100, 001, and 111.

So, we got a total 9 different strings in the last operation.

Total set bits in 9 string = 15, total operations = 9, ans = 15/9 = 1.6667

Approach 1

Here we will use the probability principle to solve the problem. Let's assume that the average value of the set bit is p and the off bits is q after performing the i-1 operations. We need to count the average value of set bits and off bits in ith operation.

So, the updated value of p can be p + the average number of new set bits - the average number of new off bits.

Algorithm

• Initialize the P with N as we have N set bits initially and Q with 0 as we have 0 set bits initially.

• Traverse the array of operations.

• Initialize the prev_p and prev_q with P and Q values.

• Update the P value with prev_p - prev_p * arr[i] / N + prev_q * arr[i] / N, which adds average off bits flipped into set bits and removes average set bits flipped into the off bits

• Update the Q value.

• Return the P value.

Example

#include <bits/stdc++.h>
using namespace std;

double getAverageBits(int len, int K, int array[]) {
   // to store the average '1's in the binary string
   double P = len;
   // to store the average '0's in the binary string
   double Q = 0;
   // Traverse the array array[]
   for (int i = 0; i < K; i++) {
      // Initialize the prev_p and prev_q with P and Q, which we got from the previous iteration
      double prev_p = P, prev_q = Q;
      // Update the average '1's
      P = prev_p - prev_p * array[i] / len + prev_q * array[i] / len;
      // Update the average '0's
      Q = prev_q - prev_q * array[i] / len + prev_p * array[i] / len;
   }
   return P;
}
int main() {
   int N = 2;
   int array[] = {1};
   int K = sizeof(array) / sizeof(array[0]);
   cout << "The average number of set bits after performing the operations is " << getAverageBits(N, K, array);
   return 0;
}

Output

The average number of set bits after performing the operations is 1

Time complexity - O(K), where K is the length of the array.

Space complexity - O(1) as we don't use any extra space.

In this tutorial, we learned to find average set bits after performing all possible choices of K operation. In the single choice, we need to perform all operations given in the array.