Length of Longest sub-string that can be removed

3.4

Given a binary string (consists of only 0 and 1). If there is “100” as a sub-string in the string, then we can delete this sub-string. The task is to find the length of longest sub-string which can be make removed?

Examples:

Input  : str = "1011100000100"
Output : 6
// Sub-strings present in str that can be make removed 
// 101{110000}0{100}. First sub-string 110000-->100-->null,
// length is = 6. Second sub-string 100-->null, length is = 3

Input  : str = "111011"
Output : 0
// There is no sub-string which can be make null

We can solve this problem using a container like vector in c++ or ArrayList in Java. Below is the algorithm to solve this problem :

  • Take a vector arr of pair type. Each element in arr stores two values character and it’s respective index in string.
  • Store pair(‘@’,-1) as a base in arr. Take variable maxlen = 0 which stores the final result.
  • Now one by one iterate for all characters in string, make pair of characters and it’s respective index and store it in arr. In parallel also check the condition if after inserting i’th character last three elements of ‘arr’ are making sub-string “100”.
  • If sub-string exist then delete it from ‘arr’. Repeat this loop by number of times till you are getting sub-string “100” in arr and make it null by deleting continuously.
  • The difference of indexs of i’th character and index of last element currently present in arr after deletion gives the length of sub-string that can be make null by continuous deletion of sub-string “100”, update maxlen.

C++

// C++ implementation of program to find the maximum length
// that can be removed
#include<bits/stdc++.h>
using namespace std;

// Function to find the length of longest sub-string that
// can me make removed
// arr  --> pair type of array whose first field store
//          character in string and second field stores
//          corresponding index of that character
int longestNull(string str)
{
    vector<pair<char,int> > arr;

    // store {'@',-1} in arr , here this value will
    // work as base index
    arr.push_back({'@', -1});

    int maxlen = 0;   // Initialize result

    // one by one iterate characters of string
    for (int i = 0; i < str.length(); ++i)
    {
        // make pair of char and index , then store
        // them into arr
        arr.push_back({str[i], i});

        // now if last three elements of arr[]  are making
        // sub-string "100" or not
        while (arr.size()>=3 &&
               arr[arr.size()-3].first=='1' &&
               arr[arr.size()-2].first=='0' &&
               arr[arr.size()-1].first=='0')
        {
            // if above condition is true then delete
            // sub-string "100" from arr[]
            arr.pop_back();
            arr.pop_back();
            arr.pop_back();
        }

        // index of current last element in arr[]
        int tmp = arr.back().second;

        // This is important, here 'i' is the index of
        // current charcater inserted into arr[]
        // and 'tmp' is the index of last element in arr[]
        // after continuous deletion of sub-string
        // "100" from arr[] till we make it null, difference
        // of these to 'i-tmp' gives the length of current
        // sub-string that can be make null by continuous
        // deletion of sub-string "100"
        maxlen = max(maxlen, i - tmp);
    }

    return maxlen;
}

// Driver program to run the case
int main()
{
    cout << longestNull("1011100000100");
    return 0;
}

Java

// Java implementation of program to find 
// the maximum length that can be removed
import java.util.ArrayList;

public class GFG 
{    
    // User defined class Pair
    static class Pair{
        char first; 
        int second;
        Pair(char first, int second){
            this.first = first;
            this.second = second;
        }
    }
    
    /* Function to find the length of longest 
    sub-string that can me make removed
     arr  --> pair type of array whose first 
              field store character in string
              and second field stores 
              corresponding index of that character*/
    static int longestNull(String str)
    {
        ArrayList<Pair> arr = new ArrayList<>();
     
        // store {'@',-1} in arr , here this value
        // will work as base index
        arr.add(new Pair('@', -1));
     
        int maxlen = 0;   // Initialize result
     
        // one by one iterate characters of string
        for (int i = 0; i < str.length(); ++i)
        {
            // make pair of char and index , then 
            // store them into arr
            arr.add(new Pair(str.charAt(i), i));
     
            // now if last three elements of arr[]
            // are making sub-string "100" or not
            while (arr.size() >= 3 &&
                   arr.get(arr.size()-3).first=='1' &&
                   arr.get(arr.size()-2).first=='0' &&
                   arr.get(arr.size()-1).first=='0')
            {
                // if above condition is true then 
                // delete sub-string "100" from arr[]
                arr.remove(arr.size() - 3);
                arr.remove(arr.size() - 2);
                arr.remove(arr.size() - 1);
            }
     
            // index of current last element in arr[]
            int tmp = arr.get(arr.size() - 1).second;
     
            // This is important, here 'i' is the index
            // of current charcater inserted into arr[]
            // and 'tmp' is the index of last element
            // in arr[] after continuous deletion of 
            // sub-string "100" from arr[] till we make 
            // it null, difference of these to 'i-tmp' 
            // gives the length of current sub-string 
            // that can be make null by continuous
            // deletion of sub-string "100"
            maxlen = Math.max(maxlen, i - tmp);
        }
     
        return maxlen;
    }
     
    // Driver program to run the case
    public static void main(String args[])
    {
        System.out.println(longestNull("1011100000100"));
    }
}
// This code is contributed by Sumit Ghosh


Output:

6

Time complexity : O(n)
Auxiliary space : O(n)

This article is contributed by Shashank Mishra ( Gullu ). 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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