# Check if a binary string has a 0 between 1s or not | Set 1 (General approach)

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 is present in between 1’s. For example, 1111, 0000111110, 1111000 are valid strings but 01010011, 01010, 101 are not.
One approach to solve 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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Algorithm:

1. Find first occurrence of 1 in string. Let it be first.
2. Find last occurrence of 1 in string. Let it be last.
3. Run a loop from first to last and return false if there is a 0. If no 0 found, return true.
4. Explanation:
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 0
and 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”.

## C++

 `// C++ program to check if a string is valid or not. ` `#include ` `using` `namespace` `std; ` ` `  `// Function returns 1 when string is valid ` `// else returns 0 ` `bool` `checkString(string s) ` `{ ` `    ``int` `len = s.length(); ` ` `  `    ``// Find first occurrence of 1 in s[] ` `    ``int` `first = s.size() + 1; ` `    ``for` `(``int` `i = 0; i < len; i++) { ` `        ``if` `(s[i] == ``'1'``) { ` `            ``first = i; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Find last occurrence of 1 in s[] ` `    ``int` `last = 0; ` `    ``for` `(``int` `i = len - 1; i >= 0; i--) { ` `        ``if` `(s[i] == ``'1'``) { ` `            ``last = i; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Check if there is any 0 in range ` `    ``for` `(``int` `i = first; i <= last; i++) ` `        ``if` `(s[i] == ``'0'``) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"00011111111100000"``; ` `    ``checkString(s) ? cout << ``"VALID\n"` `: cout << ``"NOT VALID\n"``; ` `    ``return` `0; ` `} `

Output:

```VALID
```

## Java

 `// Java program to check if a string is valid or not. ` ` `  `class` `Test { ` `    ``// Method returns 1 when string is valid ` `    ``// else returns 0 ` `    ``static` `boolean` `checkString(String s) ` `    ``{ ` `        ``int` `len = s.length(); ` ` `  `        ``// Find first occurrence of 1 in s[] ` `        ``int` `first = ``0``; ` `        ``for` `(``int` `i = ``0``; i < len; i++) { ` `            ``if` `(s.charAt(i) == ``'1'``) { ` `                ``first = i; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// Find last occurrence of 1 in s[] ` `        ``int` `last = ``0``; ` `        ``for` `(``int` `i = len - ``1``; i >= ``0``; i--) { ` `            ``if` `(s.charAt(i) == ``'1'``) { ` `                ``last = i; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// Check if there is any 0 in range ` `        ``for` `(``int` `i = first; i <= last; i++) ` `            ``if` `(s.charAt(i) == ``'0'``) ` `                ``return` `false``; ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver method ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``String s = ``"00011111111100000"``; ` `        ``System.out.println(checkString(s) ? ``"VALID"` `: ``"NOT VALID"``); ` `    ``} ` `} `

## C#

 `// C# program to check if a  ` `// string is valid or not. ` `using` `System; ` ` `  `class` `GFG {  ` `     `  `    ``// Method returns 1 when string is valid ` `    ``// else returns 0 ` `    ``static` `bool` `checkString(String s) ` `    ``{ ` `        ``int` `len = s.Length; ` ` `  `        ``// Find first occurrence of 1 in s[] ` `        ``int` `first = 0; ` `        ``for` `(``int` `i = 0; i < len; i++) { ` `            ``if` `(s[i] == ``'1'``) { ` `                ``first = i; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// Find last occurrence of 1 in s[] ` `        ``int` `last = 0; ` `        ``for` `(``int` `i = len - 1; i >= 0; i--) { ` `            ``if` `(s[i] == ``'1'``) { ` `                ``last = i; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// Check if there is any 0 in range ` `        ``for` `(``int` `i = first; i <= last; i++) ` `            ``if` `(s[i] == ``'0'``) ` `                ``return` `false``; ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver method ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``string` `s = ``"00011111111100000"``; ` `        ``Console.WriteLine(checkString(s) ? ` `                   ``"VALID"` `: ``"NOT VALID"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

Time Complexity : O(n)

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Improved By : Sam007

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