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corejava-programming: Program to find consecutive 1 and 0 in binary equivalent

While studying decimal to binary conversion Levin felt funny seeing the occurrence of consecutive 1´s and 0´s. He asked you to help him sort out a program that will find out the number of occurrences of k consecutive 0´s or 1´s in the binary equivalent of n where k and n are user specified inputs.

Input Specifications :

First input n specifies the decimal number to be converted to binary where 0<=n<=10000 Second input k specifies the consecutive 0´s or 1´s which is supposed to occur in the binary(n)

Output Specifications :

Output contains 2 line First line contains number of occurrences of K consecutive 1´s Second line contains number of occurrences of K consecutive 0´s

Sample Input :

7 2

Sample output :



corejava x 353
programming x 168
Posted On : 2017-12-07 17:41:02.0
profile Divesh - Divesh


Here i´ll recommend to use Pattern, Matcher. It´s a bit tricky as in this problem we want to count overlapping sequences too as for input 7, binary equivalent 111 returns 2 for pattern 11.
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class PatternCounter {
public static void main(String[] args) throws Exception{
Scanner input = new Scanner(;
int n=input.nextInt();
int k=input.nextInt();
if(0<=n && n<=10000){
String binaryEquivalent = Integer.toBinaryString(n);

String zeroPattern=formPattern(k, "0");
String onePattern=formPattern(k, "1");

System.out.println(getPatternCount(onePattern, binaryEquivalent));
System.out.println(getPatternCount(zeroPattern, binaryEquivalent));
throw new IllegalAccessException("Invalid Inputs!");

public static String formPattern(int k, String sequence){
StringBuilder sb = new StringBuilder();
for(int i=0;i<k;i++)
return sb.toString();

public static int getPatternCount(String inputPattern, String inputSequence){
Pattern pattern = Pattern.compile(inputPattern);
Matcher matcher = pattern.matcher(inputSequence);
int count=0;
int pos = 0;
while (matcher.find(pos)){
pos = matcher.start() + 1;
return count;

Posted On : 2017-12-10 00:04:59
Satisfied : 2 Yes  0 No
profile Rishi Kumar - Rishi Kumar
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