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Maximize “10” Subsequences by replacing at most one 0 with 1

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  • Difficulty Level : Medium
  • Last Updated : 30 Nov, 2022
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Given a binary string S consisting of ‘0‘ and ‘1‘ only, the task is to find the maximum possible number of “10” subsequences by replacing at most one 0 with 1.

Examples:

Input: 1110
Output: 3
Explanation: Here initial answer = 3, since pairs: {0, 3}, {1, 3} and {2, 3} initially form the given subsequence. So no need to perform any operations since after performing operation answer would decrease. Hence answer is 3

Input: 101100
Output: 8
Explanation: Here initial answer = 7, since pairs {0, 1}, {0, 4}, {0, 5}, {2, 4}, {2, 5}, {3, 5} and {3, 4} formed the subsequence. If we perform operation on index 1 we get new string as 111100 which gives 8 possible subsequence.

Approach: To solve the problem follow the below idea :

Since it’s allowed to convert a value from ‘0’ to ‘1’, we can think greedily to get our answer.

If we change one ‘0’ to ‘1’, It results in the addition of extra subsequences formed due to changing of ‘0’ to ‘1’. However, it also leads to a loss in a certain amount of subsequences that ‘0’ was initially forming with its previous ‘1’s. Hence we need to select a position such that greedily Net profit = gain(from extra good pairs generated) – loss (originally good pairs destroyed) is maximized.

So how do we calculate it efficiently:

For this purpose, we need to create a suffix array that stores the number of ‘0’ to its right in the string so that we can find the profit after converting that ‘0’ to ‘1’. Also, we need to create a prefix array that stores the number of ‘1’ to its left such that we can find loss after converting that ‘0’ to ‘1’.

Follow the steps to solve the problem:

  • Create an array of size N + 1 for storing how many zeros appear after a character.
    • Check the last char first, then for the remaining chars by using a loop running from (i = n – 2 to 0).
  • Again, do the same thing for storing the appearance of ones in the prefix of any character. 
  • Calculate the initial answer by adding 0s in the suffix for all indices.
  • Run a loop from i = 0 to N
    • Calculate the maximum profit among all the indices.
  • Add the initial answer and maximum profit, which is our final answer, and return that value.

Below is the implementation of the approach.

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// good pairs that satisfy the conditions
int maximum_subseq(string s)
{
    int n = s.length();
 
    // To store how many zero appear
    // in front of that to form good pair
    vector<int> number_of_zero_suffix(n + 1);
 
    // Checking if its value is 0
    number_of_zero_suffix[n - 1] = (s[n - 1] == '0');
 
    for (int i = n - 2; i >= 0; i--) {
        number_of_zero_suffix[i]
            = number_of_zero_suffix[i + 1] + (s[i] == '0');
    }
 
    // Prefix array to count the number of
    // '1' which appear before to it which
    // would decrease the number of good pairs
    vector<int> number_of_one_prefix(n);
 
    number_of_one_prefix[0] = (s[0] == '1');
 
    for (int i = 1; i < n; i++) {
        number_of_one_prefix[i]
            = number_of_one_prefix[i - 1] + (s[i] == '1');
    }
 
    int initial_answer = 0;
 
    for (int i = 0; i < n; i++) {
        if (s[i] == '1')
 
            // Counting initial answer in
            // the original string
            initial_answer += number_of_zero_suffix[i + 1];
    }
 
    int maxi = 0;
 
    int profit = 0, loss = 0;
    for (int i = 0; i < n; i++) {
        if (s[i] == '0') {
            if (i == n - 1)
                profit = 0;
            else
                profit = number_of_zero_suffix[i + 1];
            if (i == 0)
                loss = 0;
            else
                loss = number_of_one_prefix[i - 1];
 
            // Calculating net profit
            int net = profit - loss;
 
            // Storing maximum value of net
            // profit after performing
            // the operation.
            maxi = max(maxi, net);
        }
    }
 
    // Calculating final answer.
    int answer = initial_answer + maxi;
 
    return answer;
}
 
// Driver code
int main()
{
    // First test case
    string S1 = "1110";
    cout << maximum_subseq(S1) << "\n";
 
    // Second test case
    string S2 = "10110";
    cout << maximum_subseq(S2) << "\n";
 
    // Third test case
    string S3 = "101100";
    cout << maximum_subseq(S3) << "\n";
 
    return 0;
}

Java




// Java code to implement the approach
 
import java.io.*;
 
class GFG {
 
    // Function to count the number of
    // good pairs that satisfy the conditions
    static int maximum_subseq(String s)
    {
        int n = s.length();
 
        // To store how many zero appear
        // in front of that to form good pair
        int[] number_of_zero_suffix = new int[n + 1];
 
        // Checking if its value is 0
        number_of_zero_suffix[n - 1]
            = ((s.charAt(n - 1) == '0') ? 1 : 0);
 
        for (int i = n - 2; i >= 0; i--) {
            number_of_zero_suffix[i]
                = number_of_zero_suffix[i + 1]
                  + ((s.charAt(i) == '0') ? 1 : 0);
        }
 
        // Prefix array to count the number of
        // '1' which appear before to it which
        // would decrease the number of good pairs
        int[] number_of_one_prefix = new int[n];
 
        number_of_one_prefix[0]
            = (s.charAt(0) == '1' ? 1 : 0);
 
        for (int i = 1; i < n; i++) {
            number_of_one_prefix[i]
                = number_of_one_prefix[i - 1]
                  + ((s.charAt(i) == '1' ? 1 : 0));
        }
 
        int initial_answer = 0;
 
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') {
                // Counting initial answer in
                // the original string
                initial_answer
                    += number_of_zero_suffix[i + 1];
            }
        }
 
        int maxi = 0;
 
        int profit = 0, loss = 0;
 
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '0') {
                if (i == n - 1) {
                    profit = 0;
                }
                else {
                    profit = number_of_zero_suffix[i + 1];
                }
                if (i == 0) {
                    loss = 0;
                }
                else {
                    loss = number_of_one_prefix[i - 1];
                }
 
                // Calculating net profit
                int net = profit - loss;
 
                // Storing maximum value of net
                // profit after performing
                // the operation.
                maxi = Math.max(maxi, net);
            }
        }
 
        // Calculating final answer.
        int answer = initial_answer + maxi;
 
        return answer;
    }
 
    public static void main(String[] args)
    {
 
        // First test case
        String S1 = "1110";
        System.out.println(maximum_subseq(S1));
 
        // Second test case
        String S2 = "10110";
        System.out.println(maximum_subseq(S2));
 
        // Third test case
        String S3 = "101100";
        System.out.println(maximum_subseq(S3));
    }
}
 
// This code is contributed by lokeshmvs21.

C#




// C# implementation
using System;
public class GFG {
 
  // Function to count the number of
  // good pairs that satisfy the conditions
  public static int maximum_subseq(string s)
  {
    int n = s.Length;
 
    // To store how many zero appear
    // in front of that to form good pair
    int[] number_of_zero_suffix = new int[n + 1];
 
    // Checking if its value is 0
 
    if (s[n - 1] == '0') {
      number_of_zero_suffix[n - 1] = 1;
    }
    else
      number_of_zero_suffix[n - 1] = 0;
 
    for (int i = n - 2; i >= 0; i--) {
      if (s[i] == '0') {
        number_of_zero_suffix[i]
          = number_of_zero_suffix[i + 1] + 1;
      }
      else {
        number_of_zero_suffix[i]
          = number_of_zero_suffix[i + 1] + 0;
      }
    }
 
    // Prefix array to count the number of
    // '1' which appear before to it which
    // would decrease the number of good pairs
    int[] number_of_one_prefix = new int[n];
 
    if (s[0] == '1')
      number_of_one_prefix[0] = 1;
    else
      number_of_one_prefix[0] = 0;
 
    for (int i = 1; i < n; i++) {
      if (s[i] == '1') {
        number_of_one_prefix[i]
          = number_of_one_prefix[i - 1] + 1;
      }
      else {
        number_of_one_prefix[i]
          = number_of_one_prefix[i - 1] + 0;
      }
    }
 
    int initial_answer = 0;
 
    for (int i = 0; i < n; i++) {
      if (s[i] == '1')
        // Counting initial answer in
        // the original string
        initial_answer
        += number_of_zero_suffix[i + 1];
    }
 
    int maxi = 0;
 
    int profit = 0, loss = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == '0') {
        if (i == n - 1)
          profit = 0;
        else
          profit = number_of_zero_suffix[i + 1];
        if (i == 0)
          loss = 0;
        else
          loss = number_of_one_prefix[i - 1];
 
        // Calculating net profit
        int net = profit - loss;
 
        // Storing maximum value of net
        // profit after performing
        // the operation.
        maxi = Math.Max(maxi, net);
      }
    }
 
    // Calculating final answer.
    int answer = initial_answer + maxi;
 
    return answer;
  }
 
  static public void Main()
  {
    // First test case
    string S1 = "1110";
    Console.WriteLine(maximum_subseq(S1));
 
    // Second test case
    string S2 = "10110";
    Console.WriteLine(maximum_subseq(S2));
 
    // Third test case
    string S3 = "101100";
    Console.WriteLine(maximum_subseq(S3));
  }
}
// This code is contributed by ksam24000

Javascript




// Function to count the number of
// good pairs that satisfy the conditions
function maximum_subseq(s)
{
    let n = s.length;
 
    // To store how many zero appear
    // in front of that to form good pair
    let number_of_zero_suffix = [];
    for(let i=0;i<(n + 1);i++)
    {
        number_of_zero_suffix.push(0);
    }
 
    // Checking if its value is 0
    number_of_zero_suffix[n - 1] = (s[n - 1] == '0');
 
    for (let i = n - 2; i >= 0; i--) {
        number_of_zero_suffix[i]
            = number_of_zero_suffix[i + 1] + (s[i] == '0');
    }
 
    // Prefix array to count the number of
    // '1' which appear before to it which
    // would decrease the number of good pairs
    let number_of_one_prefix = [];
    for(let i=0;i<n;i++)
    {
        number_of_one_prefix.push(0);
    };
 
    number_of_one_prefix[0] = (s[0] == '1');
 
    for (let i = 1; i < n; i++) {
        number_of_one_prefix[i]
            = number_of_one_prefix[i - 1] + (s[i] == '1');
    }
 
    let initial_answer = 0;
 
    for (let i = 0; i < n; i++) {
        if (s[i] == '1')
 
            // Counting initial answer in
            // the original string
            initial_answer += number_of_zero_suffix[i + 1];
    }
 
    let maxi = 0;
 
    let profit = 0, loss = 0;
    for (let i = 0; i < n; i++) {
        if (s[i] == '0') {
            if (i == n - 1)
                profit = 0;
            else
                profit = number_of_zero_suffix[i + 1];
            if (i == 0)
                loss = 0;
            else
                loss = number_of_one_prefix[i - 1];
 
            // Calculating net profit
            let net = profit - loss;
 
            // Storing maximum value of net
            // profit after performing
            // the operation.
            maxi = Math.max(maxi, net);
        }
    }
 
    // Calculating final answer.
    let answer = initial_answer + maxi;
 
    return answer;
}
 
// Driver code
// First test case
let S1 = "1110";
console.log(maximum_subseq(S1));
 
// Second test case
let S2 = "10110";
console.log(maximum_subseq(S2));
 
// Third test case
let S3 = "101100";
console.log(maximum_subseq(S3));
 
// This code is contributed by akashish__

Output

For string 1110 number of good pair are 3
For string 10110 number of good pair are 4

Time complexity: O(N),  
Auxiliary Space: O(N)

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