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 <bits/stdc++.h>using namespace std;// Function to print the binary numbervoid 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 codeint 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 formimport java.util.*;class GFG{// Function to print the binary numberstatic 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 codepublic 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 # numberdef 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 formusing System;class GFG{// Function to print the binary numberstatic 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 codepublic static void Main(String[] args){ String s = "100010001"; Console.WriteLine(flip(s.ToCharArray()));}}// This code is contributed by Rohit_ranjan |
Javascript
<script> // JavaScript implementation to maximize the value of // the decimal equivalent given in the binary form // Function to print the binary number function flip(s) { for(let 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 s.join(""); } let s = "100010001"; document.write(flip(s.split('')));</script> |
111110001
Time Complexity: O(|s|)
Auxiliary Space: O(1)
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