# Length of Longest sub-string that can be removed

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 indexes 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.

**Implementation:**

## 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 character 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 character 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` |

## Python3

`# Python3 implementation of program to find the maximum length` `# that can be removed` `# 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 and second field stores` `# corresponding index of that character` `def` `longestNull(S):` ` ` `arr` `=` `[]` ` ` `# store {'@',-1} in arr , here this value will` ` ` `# work as base index` ` ` `arr.append([` `'@'` `, ` `-` `1` `])` ` ` `maxlen ` `=` `0` `# Initialize result` ` ` `# one by one iterate characters of String` ` ` `for` `i ` `in` `range` `(` `len` `(S)):` ` ` `# make pair of char and index , then store` ` ` `# them into arr` ` ` `arr.append([S[i], i])` ` ` `# now if last three elements of arr[] are making` ` ` `# sub-string"100" or not` ` ` `while` `(` `len` `(arr)>` `=` `3` `and` ` ` `arr[` `len` `(arr)` `-` `3` `][` `0` `]` `=` `=` `'1'` `and` ` ` `arr[` `len` `(arr)` `-` `2` `][` `0` `]` `=` `=` `'0'` `and` ` ` `arr[` `len` `(arr)` `-` `1` `][` `0` `]` `=` `=` `'0'` `):` ` ` `# if above condition is true then delete` ` ` `# sub-string"100" from arr[]` ` ` `arr.pop()` ` ` `arr.pop()` ` ` `arr.pop()` ` ` `# index of current last element in arr[]` ` ` `tmp ` `=` `arr[` `-` `1` `]` ` ` `# This is important, here 'i' is the index of` ` ` `# current character 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[` `1` `])` ` ` `return` `maxlen` `# Driver code` `print` `(longestNull(` `"1011100000100"` `))` `# This code is contributed by mohit kumar 29` |

## C#

`// C# implementation of program to find ` `// the maximum length that can be removed` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG ` `{ ` ` ` `// User defined class Pair` ` ` `class` `Pair` ` ` `{` ` ` `public` `char` `first; ` ` ` `public` `int` `second;` ` ` `public` `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)` ` ` `{` ` ` `List<Pair> arr = ` `new` `List<Pair>();` ` ` ` ` `// 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[i], i));` ` ` ` ` `// now if last three elements of []arr` ` ` `// are making sub-string "100" or not` ` ` `while` `(arr.Count >= 3 &&` ` ` `arr[arr.Count-3].first==` `'1'` `&&` ` ` `arr[arr.Count-2].first==` `'0'` `&&` ` ` `arr[arr.Count-1].first==` `'0'` `)` ` ` `{` ` ` `// if above condition is true then ` ` ` `// delete sub-string "100" from []arr` ` ` `arr.RemoveAt(arr.Count - 3);` ` ` `arr.RemoveAt(arr.Count - 2);` ` ` `arr.RemoveAt(arr.Count - 1);` ` ` `}` ` ` ` ` `// index of current last element in []arr` ` ` `int` `tmp = arr[arr.Count - 1].second;` ` ` ` ` `// This is important, here 'i' is the index` ` ` `// of current character 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 code` ` ` `public` `static` `void` `Main(String []args)` ` ` `{` ` ` `Console.WriteLine(longestNull(` `"1011100000100"` `));` ` ` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Javascript

`<script>` `// Javascript implementation of program to find ` `// the maximum length that can be removed` ` ` `/* 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*/` ` ` `function` `longestNull(str)` ` ` `{` ` ` `let arr = [];` ` ` ` ` `// store {'@',-1} in arr , here this value` ` ` `// will work as base index` ` ` `arr.push([` `'@'` `, -1]);` ` ` ` ` `let maxlen = 0; ` `// Initialize result` ` ` ` ` `// one by one iterate characters of string` ` ` `for` `(let i = 0; i < str.length; ++i)` ` ` `{` ` ` `// make pair of char and index , then ` ` ` `// store them into arr` ` ` `arr.push([str[i],i]);` ` ` ` ` `// now if last three elements of arr[]` ` ` `// are making sub-string "100" or not` ` ` `while` `(arr.length >= 3 &&` ` ` `arr[arr.length-3][0]==` `'1'` `&&` ` ` `arr[arr.length-2][0]==` `'0'` `&&` ` ` `arr[arr.length-1][0]==` `'0'` `)` ` ` `{` ` ` `// if above condition is true then ` ` ` `// delete sub-string "100" from arr[]` ` ` `arr.pop();` ` ` `arr.pop();` ` ` `arr.pop();` ` ` `}` ` ` ` ` `// index of current last element in arr[]` ` ` `let tmp = arr[arr.length-1][1];` ` ` ` ` `// This is important, here 'i' is the index` ` ` `// of current character 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` ` ` `document.write(longestNull(` `"1011100000100"` `));` ` ` ` ` `// This code is contributed by unknown2108` `</script>` |

**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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.