Program to check the validity of a Password
Ninjas are trying to hack a system of a terrorist organization so that they can know where they will be going to attack next. But to hack the system and to get access to data they need a password that must satisfy certain conditions as described below:

1) Length of the password must be between 8 to 15 characters. 2) At least one digit (0-9), one lowercase letter (a-z), one uppercase letter (A-Z) and one special character (%, ^, &, #, *, %, etc) must be present. 3) Password must not contain any space.

You are given a string ‘STR’, help ninjas to find whether it's a valid password or not.

For example :

Given ‘STR’ = “Codingninja#1” As it satisfies all the above conditions so it is a valid password and therefore you have to return true.

Input Format

The first line of input contains an integer 'T' representing the number of test cases. The description of the next ‘T’ lines is as follows-. The first and the only line of each test case contains a single string 'STR' representing the password.

Output Format:

For each test case, return “Valid” if the given string ‘STR’ satisfies all the above conditions and “Not Valid” if it's not. The output of each test case will be printed in a separate line.

Constraints:

1

Question

Program to check the validity of a Password

Ninjas are trying to hack a system of a terrorist organization so that they can know where they will be going to attack next. But to hack the system and to get access to data they need a password that must satisfy certain conditions as described below:

1) Length of the password must be between 8 to 15 characters.
2) At least one digit (0-9), one lowercase letter (a-z), one uppercase letter (A-Z) and one special character (%, ^, &, #, *, %, etc) must be present.
3) Password must not contain any space.

You are given a string ‘STR’, help ninjas to find whether it's a valid password or not.

For example :

Given ‘STR’ =  “Codingninja#1” As it satisfies all the above conditions so it is a valid password and therefore you have to return true.

Input Format

The first line of input contains an integer 'T' representing the number of test cases.

The description of the next ‘T’ lines is as follows-.

The first and the only line of each test case contains a single string 'STR' representing the password.

Output Format:

For each test case, return “Valid” if the given string ‘STR’ satisfies all the above conditions and “Not Valid” if it's not. 

The output of each test case will be printed in a separate line.

Constraints:

1 <= T <= 5
1 <= |STR| <= 5*10^3

Time Limit: 1 second

Accepted Answer

1.) Maintained 4 flags as "false" for checking the capital letter, small letter, special character and number.
2.) Iterated the string and checking every character in it for the above 4 flags.
3.) If either of the flags were false, returned 'false'
4.) If the length of string was less than 'K', returned 'false'
5.) Returned 'true'

Accepted Answer

Conditional Implementation

The idea here is we will check all conditions one by one, if any condition fails at any instance we will directly return false and output Not valid password else we will output valid.

Every character has a particular number known as ASCII code so we will use the ASCII code to check whether all conditions are satisfied or not.

In the below table there are ASCII codes given of the required characters.

Ascii Code	Characters
48 - 57	Digits (0-9)
65 - 90	Uppercase (A-Z)
97 - 122	Lowercase (a-z)
33 - 47, 58 - 64, 91 - 96, 123 - 126	Special Characters

For a valid password, we need to check at least one character from each row must be present.

Algorithm:

Declare 6 variables ‘N’= STR.length, ‘CONTAIN_DIGIT’ = 0, ‘CONTAIN_UPPER’ = 0, ‘CONTAIN_LOWER’ = 0, ‘CONTAIN_SPECIAL’ = 0, ‘CONTAIN_SPACE’ = 0.
Check if ‘N’ is between 8 and 15. If it's not then return false.
Run a loop from i = 0 to ‘N’. Here ‘N’ is the length of the string.

If str[i] is equal to ‘ ‘(space)

'CONTAIN_SPACE’ = 1;

Check for all above ASCII code and change the corresponding variable.

Return true if(‘CONTAIN_SPACE’ = 0 and ‘CONTAIN_DIGIT’ = 1 and ‘CONTAIN_UPPER’ = 1 and ‘CONTAIN_LOWER’ = 1 and ‘CONTAIN_SPECIAL’ = 1) else return false.

Space Complexity: O(1)Explanation:

O(1).

Since we are not using any extra space to store anything.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the length of the string.

In the worst case, we need to check all characters of the string and so we need to traverse the whole string therefore the net time complexity will be O(N) .

C++ (g++ 5.4)

/*

	Time Complexity : O(N)

	Space Complexity: O(1)



	Where N is the length of the string.

*/



bool isValid(string &str)

{



	int n = str.length();



	// Case 1: Checking for length of password.

	if (n < 8 || n > 15)

	{

		return false;

	}



	int containDigit = 0, containSpace = 0, containUpper = 0, containLower = 0, containSpecial = 0;

	for (int i = 0; i < n; i++)

	{

		// Case 2: Checking for space.

		if (str[i] == ' ')

		{

			containSpace = 1;

			break;

		}

		

		// Case 3: Checking for digits.

		else if (48 <= str[i] && str[i] <= 57) 

		{

			containDigit = 1;

		}



		// Case 4: Checking for UpperCase.

		else if (65 <= str[i] && str[i] <= 90) 

		{

			containUpper = 1;

		}

		

		// Case 5: Checking for LowerCase.

		else if (97 <= str[i] && str[i] <= 122)

		{

			containLower = 1;

		}

		

		// Case 6: Checking for Special characters.

		else if ((33 <= str[i] && str[i] <= 47) ||

				 (58 <= str[i] && str[i] <= 64) ||

				 (91 <= str[i] && str[i] <= 96) ||

				 (123 <= str[i] && str[i] <= 126)) 

		{

			containSpecial = 1;

		}



	}



	if (containSpace == 0 && containDigit == 1 && containLower == 1 && containSpecial == 1 && containUpper == 1)

	{

		return true;

	}

	

	return false;

}

Python (3.5)

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

	Where N is the length of the string
'''

def isValid(string):
    
    n = len(string)
    
    # Case 1: Checking for length of password.
    if n < 8 or n > 15:
        return False
        
    containDigit = 0
    containSpace = 0
    containUpper = 0
    containLower = 0
    containSpecial = 0
    
    for i in range(n):
        # Case 2: Checking for space.
        if string[i] == ' ':
            containSpace = 1
            break
        
        # Case 3: Checking for digits.
        elif 48 <= ord(string[i]) and ord(string[i]) <= 57:
            containDigit = 1
            
        # Case 4: Checking for UpperCase.
        elif 65 <= ord(string[i]) and ord(string[i]) <= 90:
            containUpper = 1
            
        # Case 5: Checking for LowerCase.
        elif 97 <= ord(string[i]) and ord(string[i]) <= 122:
            containLower = 1
            
        elif (33 <= ord(string[i]) and ord(string[i]) <= 47) or (58 <= ord(string[i]) and ord(string[i]) <= 64) or (91 <= ord(string[i]) and ord(string[i]) <= 96) or (123 <= ord(string[i]) and ord(string[i]) <= 126):
            containSpecial = 1
            
    if containSpace == 0 and containDigit == 1 and containLower == 1 and containSpecial == 1 and containUpper == 1:
        return True
            
    return False

Java (SE 1.8)

/*
	Time Complexity : O(N)
	Space Complexity: O(1)

	Where N is the length of the string.
*/

public class Solution {
	public static boolean isValid(String str) {

		int n = str.length();

		// Case 1: Checking for length of password.
		if (n < 8 || n > 15) {
			return false;
		}

		int containDigit = 0, containSpace = 0, containUpper = 0, containLower = 0, containSpecial = 0;
		for (int i = 0; i < n; i++) {
			// Case 2: Checking for space.
			if (str.charAt(i) == ' ') {
				containSpace = 1;
				break;
			}

			// Case 3: Checking for digits.
			else if (48 <= str.charAt(i) && str.charAt(i) <= 57) {
				containDigit = 1;
			}

			// Case 4: Checking for UpperCase.
			else if (65 <= str.charAt(i) && str.charAt(i) <= 90) {
				containUpper = 1;
			}

			// Case 5: Checking for LowerCase.
			else if (97 <= str.charAt(i) && str.charAt(i) <= 122) {
				containLower = 1;
			}

			// Case 6: Checking for Special characters.
			else if ((33 <= str.charAt(i) && str.charAt(i) <= 47) || (58 <= str.charAt(i) && str.charAt(i) <= 64)
					|| (91 <= str.charAt(i) && str.charAt(i) <= 96) || (123 <= str.charAt(i) && str.charAt(i) <= 126)) {
				containSpecial = 1;
			}

		}

		if (containSpace == 0 && containDigit == 1 && containLower == 1 && containSpecial == 1 && containUpper == 1) {
			return true;
		}

		return false;
	}

}

Accepted Answer

String Based Search Approach

In this approach, we will use the find function to find a particular character if the character is present then it will return its position else it will return the end of the position of string(npos).

First, we will check the length of the password and then one by one all characters.

If all the characters present at least once then we will return true else false.