Check if a binary string has a 0 between 1s or not | Set 1 (General approach)
Last Updated :
18 Apr, 2023
Given a string of 0 and 1, we need to check that the given string is valid or not. The given string is valid when there is no zero present in between 1’s. For example, 1111, 0000111110, 1111000 are valid strings but 01010011, 01010, 101 are not.
One approach to solving the problem is discussed here, other using Regular expressions is given Set 2
Examples:
Input : 100
Output : VALID
Input : 1110001
Output : NOT VALID
There is 0 between 1s
Input : 00011
Output : VALID
Approach 1 : (Simple Traversal)
Suppose we have a string: 01111011110000 Now take two variables say A and B. run a loop for 0 to length of the string and point A to the first occurrence of 1, after that again run a loop from length of the string to 0and point second variable to the last occurrence of 1.So A = 1 (Position of first ‘1’ is the string) and B = 9 (last occurrence of ‘1’).Now run a loop from A to B and check that ‘0’ is present between 1 or not, if “YES” than set flag to 1 and break the loop, else set flag to 0.In this case flag will set to 1 as the given string is not valid and print “NOT VALID”.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
bool checkString(string s)
{
int len = s.length();
int first = s.size() + 1;
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
first = i;
break;
}
}
int last = 0;
for (int i = len - 1; i >= 0; i--) {
if (s[i] == '1') {
last = i;
break;
}
}
for (int i = first; i <= last; i++)
if (s[i] == '0')
return false;
return true;
}
int main()
{
string s = "00011111111100000";
checkString(s) ? cout << "VALID\n" : cout << "NOT VALID\n";
return 0;
}
|
Java
class Test {
static boolean checkString(String s)
{
int len = s.length();
int first = 0;
for (int i = 0; i < len; i++) {
if (s.charAt(i) == '1') {
first = i;
break;
}
}
int last = 0;
for (int i = len - 1; i >= 0; i--) {
if (s.charAt(i) == '1') {
last = i;
break;
}
}
for (int i = first; i <= last; i++)
if (s.charAt(i) == '0')
return false;
return true;
}
public static void main(String args[])
{
String s = "00011111111100000";
System.out.println(checkString(s) ? "VALID" : "NOT VALID");
}
}
|
Python
def checkString(s):
Len = len(s)
first = len(s) + 1
for i in range(Len):
if(s[i] == '1'):
first = i
break
last = 0
for i in range(Len - 1,
-1, -1):
if(s[i] == '1'):
last = i
break
for i in range(first,
last + 1):
if(s[i] == '0'):
return False
return True
s = "00011111111100000"
if(checkString(s)):
print("VALID")
else:
print("NOT VALID")
|
C#
using System;
class GFG {
static bool checkString(String s)
{
int len = s.Length;
int first = 0;
for (int i = 0; i < len; i++) {
if (s[i] == '1') {
first = i;
break;
}
}
int last = 0;
for (int i = len - 1; i >= 0; i--) {
if (s[i] == '1') {
last = i;
break;
}
}
for (int i = first; i <= last; i++)
if (s[i] == '0')
return false;
return true;
}
public static void Main()
{
string s = "00011111111100000";
Console.WriteLine(checkString(s) ?
"VALID" : "NOT VALID");
}
}
|
Javascript
<script>
function checkString(s)
{
let len = s.length;
let first = 0;
for (let i = 0; i < len; i++) {
if (s[i] == '1') {
first = i;
break;
}
}
let last = 0;
for (let i = len - 1; i >= 0; i--) {
if (s[i] == '1') {
last = i;
break;
}
}
for (let i = first; i <= last; i++)
if (s[i] == '0')
return false;
return true;
}
let s = "00011111111100000";
document.write(checkString(s) ?
"VALID" : "NOT VALID");
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1) since it is using constant space for variables
Approach 2 : (Using Two Pointers)
Will keep two pointers on the both end of the string and will check whether in the both side ‘1’ is found (track the index with another variable) or not. If in the both sides we find ‘1’ then will run another loop for the inner part using the other two variables, which will again be a part of two pointers algorithm.
Look at the code for better understanding,
C++
#include <bits/stdc++.h>
using namespace std;
bool checkString(string s)
{
int i = 0, j = s.size() - 1;
int si = -1, ei = -1;
while (i <= j) {
if (si == -1 and s[i] == '1')
si = i;
if (ei == -1 and s[j] == '1')
ei = j;
if (si != -1 and ei != -1) {
while (si <= ei) {
if (s[si] == '0' or s[ei] == '0')
return false;
si++;
ei--;
}
return true;
}
i++;
j--;
}
return true;
}
int main()
{
string s = "00011111111100000";
checkString(s) ? cout << "VALID\n"
: cout << "NOT VALID\n";
return 0;
}
|
Python3
def check_string(s):
i, j = 0, len(s) - 1
si, ei = -1, -1
while i <= j:
if si == -1 and s[i] == '1':
si = i
if ei == -1 and s[j] == '1':
ei = j
if si != -1 and ei != -1:
while si <= ei:
if s[si] == '0' or s[ei] == '0':
return False
si += 1
ei -= 1
return True
i += 1
j -= 1
return True
s = "00011111111100000"
print("VALID" if check_string(s) else "NOT VALID")
|
Java
import java.util.Scanner;
public class Main {
public static boolean checkString(String s)
{
int i = 0, j = s.length() - 1;
int si = -1, ei = -1;
while (i <= j) {
if (si == -1 && s.charAt(i) == '1')
si = i;
if (ei == -1 && s.charAt(j) == '1')
ei = j;
if (si != -1 && ei != -1) {
while (si <= ei) {
if (s.charAt(si) == '0'
|| s.charAt(ei) == '0')
return false;
si++;
ei--;
}
return true;
}
i++;
j--;
}
return true;
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
String s = "00011111111100000";
if (checkString(s)) {
System.out.println("VALID");
}
else {
System.out.println("NOT VALID");
}
}
}
|
C#
using System;
class Program {
static bool CheckString(string s)
{
int i = 0, j = s.Length - 1;
int si = -1, ei = -1;
while (i <= j) {
if (si == -1 && s[i] == '1')
si = i;
if (ei == -1 && s[j] == '1')
ei = j;
if (si != -1 && ei != -1) {
while (si <= ei) {
if (s[si] == '0' || s[ei] == '0')
return false;
si++;
ei--;
}
return true;
}
i++;
j--;
}
return true;
}
static void Main(string[] args)
{
string s = "00011111111100000";
if (CheckString(s))
Console.WriteLine("VALID");
else
Console.WriteLine("NOT VALID");
}
}
|
Javascript
function check_string(s) {
let i = 0, j = s.length - 1;
let si = -1, ei = -1;
while (i <= j) {
if (si == -1 && s[i] == '1') {
si = i;
}
if (ei == -1 && s[j] == '1') {
ei = j;
}
if (si != -1 && ei != -1) {
while (si <= ei) {
if (s[si] == '0' || s[ei] == '0') {
return false;
}
si += 1;
ei -= 1;
}
return true;
}
i += 1;
j -= 1;
}
return true;
}
let s = "00011111111100000";
console.log(check_string(s) ? "VALID" : "NOT VALID");
|
Time Complexity : O(N)
Auxiliary Space : O(1), since it is using constant space for variables.
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