# Maximize the decimal equivalent by flipping only a contiguous set of 0s

• Last Updated : 16 Nov, 2021

Given a binary number in the form of a string, the task is to print a binary equivalent obtained by flipping only one contiguous set of 0s such that the decimal equivalent of this binary number is maximum.
Note: Do not assume any trailing zeroes in the start of the binary number i.e. “0101” is given as “101”.
Examples:

Input: s = “10101”
Output: 11101
Explanation:
Here we can only flip the 2nd character of the string “10101” ( = 21) that will change it to “11101” (= 29). Since we are allowed to flip a continuous subarray, any more flipping will lead to decrease in decimal equivalent.
Input: s = “1000”
Output: 1111
Explanation:
If we flip the continuous characters starting from position 1 till 3 we will get 1111 which is the maximum number possible in 4 bits i.e. 15.

Approach: To solve the problem mentioned above we know that we have to increase the value of the binary equivalent. Therefore, we must increase the number of 1’s at higher position. Clearly, we can increase the value of the number by only flipping the initially occurring zeroes. Traverse the string and flip the first occurrence of zeroes until a 1 occurs, in this case, the loop must break. Print the resultant string.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to Maximize the value of``// the decimal equivalent given in the binary form``#include ``using` `namespace` `std;` `// Function to print the binary number``void` `flip(string& s)``{``    ``for` `(``int` `i = 0; i < s.length(); i++) {` `        ``// Check if the current number is 0``        ``if` `(s[i] == ``'0'``) {` `            ``// Find the continuous 0s``            ``while` `(s[i] == ``'0'``) {` `                ``// Replace initially``                ``// occurring 0 with 1``                ``s[i] = ``'1'``;``                ``i++;``            ``}` `            ``// Break out of loop if 1 occurs``            ``break``;``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``string s = ``"100010001"``;``    ``flip(s);` `    ``cout << s;``    ``return` `0;``}`

## Java

 `// Java implementation to maximize the value of``// the decimal equivalent given in the binary form``import` `java.util.*;` `class` `GFG{` `// Function to print the binary number``static` `void` `flip(String s)``{``    ``StringBuilder sb = ``new` `StringBuilder(s);``    ``for``(``int` `i = ``0``; i < sb.length(); i++)``    ``{``       ` `       ``// Check if the current number is 0``       ``if` `(sb.charAt(i) == ``'0'``)``       ``{``           ` `           ``// Find the continuous 0s``           ``while` `(sb.charAt(i) == ``'0'``)``           ``{``               ` `               ``// Replace initially``               ``// occurring 0 with 1``               ``sb.setCharAt(i, ``'1'``);``               ``i++;``           ``}``           ` `           ``// Break out of loop if 1 occurs``           ``break``;``       ``}``    ``}``    ``System.out.println(sb.toString());``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String s = ``"100010001"``;``    ``flip(s);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 implementation to``# Maximize the value of the``# decimal equivalent given``# in the binary form` `# Function to print the binary``# number``def` `flip(s):``    ``s ``=` `list``(s)``    ``for` `i ``in` `range``(``len``(s)):` `        ``# Check if the current number``        ``# is 0``        ``if``(s[i] ``=``=` `'0'``):` `            ``# Find the continuous 0s``            ``while``(s[i] ``=``=` `'0'``):` `                ``# Replace initially``                ``# occurring 0 with 1``                ``s[i] ``=` `'1'``                ``i ``+``=` `1``            ``s ``=` `''.join(``map``(``str``, s))` `            ``# return the string and``            ``# break the loop``            ``return` `s` `# Driver code``s ``=` `"100010001"``print``(flip(s))` `# This code is contributed by avanitrachhadiya2155`

## C#

 `// C# implementation to maximize the value of``// the decimal equivalent given in the binary form``using` `System;` `class` `GFG{` `// Function to print the binary number``static` `String flip(``char` `[]s)``{``    ``for``(``int` `i = 0; i < s.Length; i++)``    ``{``       ` `       ``// Check if the current number is 0``       ``if` `(s[i] == ``'0'``)``       ``{``           ` `           ``// Find the continuous 0s``           ``while` `(s[i] == ``'0'``)``           ``{``               ` `               ``// Replace initially``               ``// occurring 0 with 1``               ``s[i] = ``'1'``;``               ``i++;``           ``}``           ` `           ``// Break out of loop if 1 occurs``           ``break``;``       ``}``    ``}``    ``return` `new` `String(s);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String s = ``"100010001"``;``    ` `    ``Console.WriteLine(flip(s.ToCharArray()));``}``}` `// This code is contributed by Rohit_ranjan`

## Javascript

 ``
Output:
`111110001`

Time Complexity: O(|s|)

Auxiliary Space: O(1)

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