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:

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

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 <bits/stdc++.h> 
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)

This article is contributed by Sahil Rajput. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

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

C++

Java

C#

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