Longest subsequence possible that starts and ends with 1 and filled with 0 in the middle

Difficulty Level : Expert
Last Updated : 03 Jun, 2021

Given a binary string s, the task is to find the length of the longest subsequence that can be divided into three substrings such that the first and third substrings are either empty or filled with 1 and the substring at the middle is either empty or filled with 0.

Examples:

Input: s = “1001”
Output: 4
Explanation:
The entire string can be divided into the desired three parts: “1”, “00”, “1”.
Input: s = “010”
Output: 2
Explanation:
The subsequences “00”, “01” and “10” can be split into three desired parts {“”, “00”, “”}, {“”, “0”, “1”} and {“1”, “0”, “”}

Approach:
To solve the problem, we need to follow the steps given below:

Firstly, pre-compute and store in prefix arrays, the occurrences of ‘1’ and ‘0’ respectively.
Initialize two integers i and j, where i will be the point of partition between the first and second string and j will be the point of partition between the second and third strings.
Iterate over all possible values of i & j (0 <= i < j <=n) and find the maximum possible length of the subsequence possible which satisfies the given condition.

Below is the implementation of the above approach:

C++

// C++ Program to find the
// longest subsequence possible
// that starts and ends with 1
// and filled with 0 in the middle
 
#include <bits/stdc++.h>
using namespace std;
 
int longestSubseq(string s, int length)
{
    // Prefix array to store the
    // occurences of '1' and '0'
    int ones[length + 1], zeroes[length + 1];
 
    // Initialise prefix arrays with 0
    memset(ones, 0, sizeof(ones));
    memset(zeroes, 0, sizeof(zeroes));
 
    // Iterate over the length of the string
    for (int i = 0; i < length; i++) {
 
        // If current character is '1'
        if (s[i] == '1') {
            ones[i + 1] = ones[i] + 1;
            zeroes[i + 1] = zeroes[i];
        }
 
        // If current character is '0'
        else {
            zeroes[i + 1] = zeroes[i] + 1;
            ones[i + 1] = ones[i];
        }
    }
 
    int answer = INT_MIN;
    int x = 0;
 
    for (int i = 0; i <= length; i++) {
        for (int j = i; j <= length; j++) {
            // Add '1' available for
            // the first string
            x += ones[i];
 
            // Add '0' available for
            // the second string
            x += (zeroes[j] - zeroes[i]);
 
            // Add '1' available for
            // the third string
            x += (ones[length] - ones[j]);
 
            // Update answer
            answer = max(answer, x);
 
            x = 0;
        }
    }
 
    // Print the final result
    cout << answer << endl;
}
 
// Driver Code
int main()
{
 
    string s = "10010010111100101";
 
    int length = s.length();
 
    longestSubseq(s, length);
 
    return 0;
}

Java

// Java program to find the
// longest subsequence possible
// that starts and ends with 1
// and filled with 0 in the middle
import java.io.*;
 
class GFG{
 
static void longestSubseq(String s,
                          int length)
{
     
    // Prefix array to store the
    // occurences of '1' and '0'
    int[] ones = new int[length + 1];
    int[] zeroes = new int[length + 1];
 
    // Iterate over the length of
    // the string
    for(int i = 0; i < length; i++)
    {
         
        // If current character is '1'
        if (s.charAt(i) == '1')
        {
            ones[i + 1] = ones[i] + 1;
            zeroes[i + 1] = zeroes[i];
        }
 
        // If current character is '0'
        else
        {
            zeroes[i + 1] = zeroes[i] + 1;
            ones[i + 1] = ones[i];
        }
    }
 
    int answer = Integer.MIN_VALUE;
    int x = 0;
 
    for(int i = 0; i <= length; i++)
    {
        for(int j = i; j <= length; j++)
        {
             
            // Add '1' available for
            // the first string
            x += ones[i];
 
            // Add '0' available for
            // the second string
            x += (zeroes[j] - zeroes[i]);
 
            // Add '1' available for
            // the third string
            x += (ones[length] - ones[j]);
 
            // Update answer
            answer = Math.max(answer, x);
            x = 0;
        }
    }
 
    // Print the final result
    System.out.println(answer);
}
 
// Driver code
public static void main(String[] args)
{
    String s = "10010010111100101";
    int length = s.length();
 
    longestSubseq(s, length);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to find the
# longest subsequence possible
# that starts and ends with 1
# and filled with 0 in the middle
import sys
 
def longestSubseq(s, length):
     
    # Prefix array to store the
    # occurences of '1' and '0'
    # Initialise prefix arrays with 0
    ones = [0 for i in range(length + 1)]
    zeroes = [0 for i in range(length + 1)]
 
    # Iterate over the length of the string
    for i in range(length):
         
        # If current character is '1'
        if(s[i] == '1'):
            ones[i + 1] = ones[i] + 1
            zeroes[i + 1] = zeroes[i]
 
        # If current character is '0'
        else:
            zeroes[i + 1] = zeroes[i] + 1
            ones[i + 1] = ones[i]
 
    answer = -sys.maxsize - 1
    x = 0
 
    for i in range(length + 1):
        for j in range(i, length + 1):
             
            # Add '1' available for
            # the first string
            x += ones[i]
 
            # Add '0' available for
            # the second string
            x += (zeroes[j] - zeroes[i])
 
            # Add '1' available for
            # the third string
            x += (ones[length] - ones[j])
 
            # Update answer
            answer = max(answer, x)
            x = 0
 
    # Print the final result
    print(answer)
 
# Driver Code
S = "10010010111100101"
length = len(S)
 
longestSubseq(S, length)
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program to find the
// longest subsequence possible
// that starts and ends with 1
// and filled with 0 in the middle
using System;
 
class GFG{
 
static void longestSubseq(String s,
                          int length)
{
     
    // Prefix array to store the
    // occurences of '1' and '0'
    int[] ones = new int[length + 1];
    int[] zeroes = new int[length + 1];
 
    // Iterate over the length of
    // the string
    for(int i = 0; i < length; i++)
    {
         
        // If current character is '1'
        if (s[i] == '1')
        {
            ones[i + 1] = ones[i] + 1;
            zeroes[i + 1] = zeroes[i];
        }
 
        // If current character is '0'
        else
        {
            zeroes[i + 1] = zeroes[i] + 1;
            ones[i + 1] = ones[i];
        }
    }
 
    int answer = int.MinValue;
    int x = 0;
 
    for(int i = 0; i <= length; i++)
    {
        for(int j = i; j <= length; j++)
        {
             
            // Add '1' available for
            // the first string
            x += ones[i];
 
            // Add '0' available for
            // the second string
            x += (zeroes[j] - zeroes[i]);
 
            // Add '1' available for
            // the third string
            x += (ones[length] - ones[j]);
 
            // Update answer
            answer = Math.Max(answer, x);
            x = 0;
        }
    }
 
    // Print the readonly result
    Console.WriteLine(answer);
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "10010010111100101";
    int length = s.Length;
 
    longestSubseq(s, length);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

<script>
      // JavaScript program to find the
      // longest subsequence possible
      // that starts and ends with 1
      // and filled with 0 in the middle
      function longestSubseq(s, len) {
        // Prefix array to store the
        // occurences of '1' and '0'
        var ones = new Array(len + 1).fill(0);
        var zeroes = new Array(len + 1).fill(0);
 
        // Iterate over the length of
        // the string
        for (var i = 0; i < len; i++) {
          // If current character is '1'
          if (s[i] === "1") {
            ones[i + 1] = ones[i] + 1;
            zeroes[i + 1] = zeroes[i];
          }
 
          // If current character is '0'
          else {
            zeroes[i + 1] = zeroes[i] + 1;
            ones[i + 1] = ones[i];
          }
        }
 
        var answer = -2147483648;
        var x = 0;
 
        for (var i = 0; i <= len; i++) {
          for (var j = i; j <= len; j++) {
            // Add '1' available for
            // the first string
            x += ones[i];
 
            // Add '0' available for
            // the second string
            x += zeroes[j] - zeroes[i];
 
            // Add '1' available for
            // the third string
            x += ones[len] - ones[j];
 
            // Update answer
            answer = Math.max(answer, x);
            x = 0;
          }
        }
 
        // Print the readonly result
        document.write(answer);
      }
 
      // Driver code
      var s = "10010010111100101";
      var len = s.length;
 
      longestSubseq(s, len);
</script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

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