Hello Programmers, In this post, you will learn how to solve HackerRank Beautiful Binary String Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t straight away look for the solutions, first try to solve the problems by yourself. If you find any difficulty after trying several times, then look for the solutions. We are going to solve the HackerRank Algorithms problems using C, CPP, JAVA, PYTHON, JavaScript & SCALA Programming Languages.
HackerRank Beautiful Binary String
Task
Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn’t contain the substring ”010”.
In one step, Alice can change a 0 to a 1 or vice versa. Count and print the minimum number of steps needed to make Alice see the string as beautiful.
Example
b = 010
She can change any one element and have a beautiful string.
Function Description
Complete the beautifulBinaryString function in the editor below.
beautifulBinaryString has the following parameter(s):
- string b: a string of binary digits
Returns
- int: the minimum moves required
Input Format
The first line contains an integer n, the length of binary string.
The second line contains a single binary string b.
Constraints
- 1 <= n <= 100
- b[i] ∈ {0, 1}
Output Format
Print the minimum number of steps needed to make the string beautiful.
Sample Input 0
STDIN Function
—– ——–
7 length of string n = 7
0101010 b = ‘0101010‘
Sample Output 0
2
Explanation 0:
In this sample, b = “0101010”
The figure below shows a way to get rid of each instance of “010”:
Make the string beautiful by changing 2 characters (b[2] and b[5]).
Sample Input 1
5
01100
Sample Output 1
0
Explanation 1
The substring “010” does not occur in b, so the string is already beautiful in 0 moves.
Sample Input 2
10
0100101010
Sample Output 2
3
Explanation 2
In this sample b = “0100101010”
One solution is to change the values of b[2], b[5], and b[9] to form a beautiful string.
HackerRank Beautiful Binary String Solution
HackerRank Beautiful Binary String Solution in C
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int n;
scanf("%d",&n);
char* B = (char *)malloc(10240 * sizeof(char));
scanf("%s",B);
int i=0,count=0;
while(B[i]){
if(B[i]=='0'&&B[i+1]=='1'&&B[i+2]=='0'){
B[i+2]='1';
count++;
}
i++;
}
printf("%d",count);
return 0;
}HackerRank Beautiful Binary String Solution in Cpp
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
char B[105];
int main(){
int n;
int ans=0;
scanf("%d", &n);
scanf("%s", B);
for(int i=2; B[i]; i++){
if(B[i-2] == '0' && B[i-1] == '1' && B[i] == '0') B[i] = '1', ans++;
}
printf("%d", ans);
return 0;
}HackerRank Beautiful Binary String Solution in Java
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
String B = in.next();
int i = 0;
int total = 0;
while (i < B.length()-2) {
if (B.substring(i,i+3).equals("010")) {
total++;
i+=3;
} else {
i++;
}
}
System.out.println(total);
}
}HackerRank Beautiful Binary String Solution in Python
#!/bin/python
import sys
n = int(raw_input().strip())
B = raw_input().strip()
if '010' not in B:
print 0
else:
print B.count('010')HackerRank Beautiful Binary String Solution using JavaScript
process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});
function readLine() {
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
function main() {
var n = parseInt(readLine());
var B = readLine();
console.log((B.match(/010/g)||[]).length);
}HackerRank Beautiful Binary String Solution in Scala
object Solution {
def main(args: Array[String]): Unit = {
val initTime = System.currentTimeMillis()
val n = lines.next().toInt
val binString = lines.next()
var p1 = 0
val l = binString.length
if(binString.length < 2) {
out.println(0)
} else {
var count = 0
while(p1 < l-2) {
if(binString(p1) == '0' && binString(p1+1) == '1' && binString(p1+2) == '0') {
count += 1
p1 += 3
} else {
p1 += 1
}
}
out.println(count)
}
printTime(initTime, System.currentTimeMillis())
out.flush()
}
private val file = new java.io.File("C:\\Users\\User\\Desktop\\hackerrank\\input_competition.txt")
private val lines = {
if(file.exists()) {
io.Source.fromFile(file).getLines
} else {
io.Source.stdin.getLines()
}
}
private val out = new java.io.PrintWriter(System.out)
private def printTime(init: Long, end:Long):Unit =
if(file.exists()) out.println(s"${end-init} ms")
}HackerRank Beautiful Binary String Solution in Pascal
uses math; var s:ansistring; i,n,ans:longint; begin readln(n); readln(s); n:=length(s); for i:=2 to n-1 do if (s[i]='1') and (s[i-1]='0' ) and (s[i+1]='0' ) then begin inc(ans); s[i+1]:='1'; end; writeln(ans); end.
Disclaimer: This problem (Beautiful Binary String) is generated by HackerRank but the solution is provided by Chase2learn. This tutorial is only for Educational and Learning purposes.
