# Count number of 0’s with given conditions

Given a binary string (containing 1’s and 0’s), write a program to count the number of 0’s in a given string such that the following conditions hold:

• 1’s and 0’s are in any order in a string
• Use of conditional statements like if, if..else, and switch are not allowed
• Use of  addition/subtraction is not allowed

Examples:

Input : S = “101101”
Output : 2

Input : S = “00101111000”
Output : 6

Recommended: Please try your approach on {IDE}  first, before moving on to the solution.

Approach: The above problem can be solved with the below idea:

If counter and conditional statements are not allowed. We are having option of Error handling.

Steps involved in the implementation of the approach:

• Take a string and a static variable counter which maintain the count of zero.
• Traverse through each character and convert it into an integer.
• Take this number and use it as a denominator of the number “0”.
• Use the exception handling method try..catch in a loop such that whenever we got the Arithmetic exception, the counter will increment.
• Print the counter value as a result.

Below is the implementation of the above approach:

## Java

 `import` `java.io.*;` ` `  `class` `GFG {` ` `  `    ` `    ``public` `static` `int` `cnt = ``0``;` ` `  `    ``public` `static` `void` `main(String[] args)` `    ``{` ` `  `        ``String s = ``"01001001110"``;` ` `  `        ` `        ``countZero(s);` `    ``}` ` `  `    ` `    ` `    ``static` `void` `countZero(String s)` `    ``{` ` `  `        ``for` `(``int` `i = ``0``; i < s.length(); i++) {` `            ``try` `{` `                ``int` `div = Character.getNumericValue(` `                    ``s.charAt(i));` `                ` `                ` `                ` `                ` `                ``int` `val = ``0` `/ div;` `            ``}` ` `  `            ` `            ``catch` `(Exception exception) {` `                ``cnt++;` `            ``}` `        ``}` ` `  `        ` `        ``System.out.println(cnt);` `    ``}` `}`

Time Complexity: O(N)
Auxiliary Space: O(1)

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