Longest subsequence of the form 0*1*0* in a binary string

Given a binary string, find the longest subsequence of the form (0)*(1)*(0)* in it. Basically we need to divide the string into 3 non overlapping strings (these strings might be empty) without changing the order of letters. First and third strings are made up of only 0 and the second string is made up of only 1. These strings could be made by deleting some characters in original string. What is the maximum size of string, we can get?

Examples –

Input : 000011100000 
Output : 12
Explanation : 
First part from 1 to 4.
Second part 5 to 7. 
Third part from 8 to 12 

Input : 100001100  
Output : 8
Explanation : 
Delete the first letter. 
First part from 2 to 4. 
Second part from 5 to 6. 
Last part from 7.

Input : 00000 
Output : 5
Explanation : 
Special Case of Only 0

Input : 111111
Output : 6
Explanation : 
Special Case of Only 1

Input : 0000001111011011110000 
Output : 20
Explanation : 
Second part is from 7 to 18. 
Remove all the 0 between indices 7 to 18.



A simple solution is to generate all subsequences of given sequence. For every subsequence, check if it is in given form. If yes, compare it with result so far and update the result if needed.

This problem can be efficiently solved by pre-computing below arrays in O(n^2) time.

Let pre_count_0[i] be the count of letter 0 in the prefix of string till index i.
Let pre_count_1[i] be the count of letter 1 in the prefix of string till length i.
Let post_count_0[i] be the count of letter 0 in the suffix string from index i till index n (here n is the size of string).

Now we fix two two positions i and j, 1 <=i <= j <=n. We will remove all 0 from substring which starts at index i and ends in index j. Thus this makes the second substring of only 1. In the prefix before the index i and in the postfix after the index j we will delete all 1 and thus it will make first and third part of the string.

then the maximum length of string attainable is max of
pre_count_0[i-1] + (pre_count_1[j]-pre_count_1[i-1]) + pre_count_1[j+1]

Special cases : When String is made of only 0s or 1s, ans is n where n is length of string.

C++

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// CPP program to find longest subsequence
// of the form 0*1*0* in a binary string
#include <bits/stdc++.h>
using namespace std;
  
// Returns length of the longest subsequence
// of the form 0*1*0*
int longestSubseq(string s)
{
    int n = s.length();
  
    // Precomputing values in three arrays
    // pre_count_0[i] is going to store count
    //             of 0s in prefix str[0..i-1]
    // pre_count_1[i] is going to store count
    //             of 1s in prefix str[0..i-1]
    // post_count_0[i] is going to store count
    //             of 0s in suffix str[i-1..n-1]
    int pre_count_0[n + 2];
    int pre_count_1[n + 1];
    int post_count_0[n + 1];
    pre_count_0[0] = 0;
    post_count_0[n + 1] = 0;
    pre_count_1[0] = 0;
    for (int j = 1; j <= n; j++)
    {
        pre_count_0[j] = pre_count_0[j - 1];
        pre_count_1[j] = pre_count_1[j - 1];
        post_count_0[n - j + 1] = post_count_0[n - j + 2];
  
        if (s[j - 1] == '0')
            pre_count_0[j]++;
        else
            pre_count_1[j]++;
        if (s[n - j] == '0')
           post_count_0[n - j + 1]++;
    }
  
    // string is made up of all 0s or all 1s
    if (pre_count_0[n] == n ||
        pre_count_0[n] == 0)
        return n;
  
  
    // Compute result using precomputed values
    int ans = 0;
    for (int i = 1; i <= n; i++)
        for (int j = i; j <= n; j++)
            ans = max(pre_count_0[i - 1]
                    + pre_count_1[j]
                    - pre_count_1[i - 1]
                    + post_count_0[j + 1],
                    ans);
    return ans;
}
  
// driver program
int main()
{
    string s = "000011100000";
    cout << longestSubseq(s);
    return 0;
}

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Python3

# Python 3 program to find longest subsequence
# of the form 0*1*0* in a binary string

# Returns length of the longest subsequence
# of the form 0*1*0*
def longestSubseq(s):
n = len(s)

# Precomputing values in three arrays
# pre_count_0[i] is going to store count
# of 0s in prefix str[0..i-1]
# pre_count_1[i] is going to store count
# of 1s in prefix str[0..i-1]
# post_count_0[i] is going to store count
# of 0s in suffix str[i-1..n-1]
pre_count_0 = [0 for i in range(n + 2)]
pre_count_1 = [0 for i in range(n + 1)]
post_count_0 = [0 for i in range(n + 2)]
pre_count_0[0] = 0
post_count_0[n + 1] = 0
pre_count_1[0] = 0
for j in range(1, n + 1):
pre_count_0[j] = pre_count_0[j – 1]
pre_count_1[j] = pre_count_1[j – 1]
post_count_0[n – j + 1] = post_count_0[n – j + 2]

if (s[j – 1] == ‘0’):
pre_count_0[j] += 1
else:
pre_count_1[j] += 1
if (s[n – j] == ‘0’):
post_count_0[n – j + 1] += 1

# string is made up of all 0s or all 1s
if (pre_count_0[n] == n or
pre_count_0[n] == 0):
return n

# Compute result using precomputed values
ans = 0
for i in range(1, n + 1):
for j in range(i, n + 1, 1):
ans = max(pre_count_0[i – 1] +
pre_count_1[j] –
pre_count_1[i – 1] +
post_count_0[j + 1], ans)
return ans

# Driver Code
if __name__ == ‘__main__’:
s = “000011100000”
print(longestSubseq(s))

# This code is contributed by
# Surendra_Gangwar


Output :

12

This article is contributed by nikhil ranjan 7. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : SURENDRA_GANGWAR