Sherlock and the Valid String in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct]

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Algorithm Hackerrank Solution | Hackerrank Java Solution

Hello Programmers/Coders, Today we are going to share solutions of Programming problems of HackerRank, Algorithm Solutions of Problem Solving Section in Java. At Each Problem with Successful submission with all Test Cases Passed, you will get an score or marks. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for Sherlock and the Valid String in Java-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.

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Introduction To Algorithm

The word Algorithm means “a process or set of rules to be followed in calculations or other problem-solving operations”. Therefore Algorithm refers to a set of rules/instructions that step-by-step define how a work is to be executed upon in order to get the expected results. 

Advantages of Algorithms:

  • It is easy to understand.
  • Algorithm is a step-wise representation of a solution to a given problem.
  • In Algorithm the problem is broken down into smaller pieces or steps hence, it is easier for the programmer to convert it into an actual program.

Link for the Problem – Sherlock and the Valid String – Hacker Rank Solution

Sherlock and the Valid String – Hacker Rank Solution

Problem:

Sherlock considers a string to be valid if all characters of the string appear the same number of times. It is also valid if he can remove just  character at  index in the string, and the remaining characters will occur the same number of times. Given a string , determine if it is valid. If so, return YES, otherwise return NO.

Example

This is a valid string because frequencies are .

This is a valid string because we can remove one  and have  of each character in the remaining string.

This string is not valid as we can only remove  occurrence of . That leaves character frequencies of .

Function Description

Complete the isValid function in the editor below.

isValid has the following parameter(s):

  • string s: a string

Returns

  • string: either YES or NO

Input Format

A single string .

Constraints

  • Each character 

Sample Input 0

aabbcd

Sample Output 0

NO

Explanation 0

Given , we would need to remove two characters, both c and d  aabb or a and b  abcd, to make it valid. We are limited to removing only one character, so  is invalid.

Sample Input 1

aabbccddeefghi

Sample Output 1

NO

Explanation 1

Frequency counts for the letters are as follows:

{'a': 2, 'b': 2, 'c': 2, 'd': 2, 'e': 2, 'f': 1, 'g': 1, 'h': 1, 'i': 1}

There are two ways to make the valid string:

  • Remove  characters with a frequency of : .
  • Remove  characters of frequency : .

Neither of these is an option.

Sample Input 2

abcdefghhgfedecba

Sample Output 2

YES

Explanation 2

All characters occur twice except for  which occurs  times. We can delete one instance of  to have a valid string.

Sherlock and the Valid String – Hacker Rank Solution
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
import java.util.Set;

/**
 * @author Kanahaiya Gupta
 *
 */
public class SherlockAndTheValidString {
	static String isValid(String s) {
		int a[] = new int[26];
		for (int i = 0; i < s.length(); i++) {
			int index = s.charAt(i) - 'a';
			a[index]++;
		}
		Map<Integer, Integer> map = new HashMap<Integer, Integer>();
		for (int i = 0; i < 26; i++) {
			if (a[i] != 0) {
				if (map.containsKey(a[i])) {
					map.put(a[i], map.get(a[i]) + 1);
				} else {
					map.put(a[i], 1);
				}
			}
		}
		if (map.size() == 1)
			return "YES";
		if (map.size() == 2) {
			Set<Entry<Integer, Integer>> entrySet = map.entrySet();
			Iterator it = entrySet.iterator();
			Entry<Integer, Integer> e1 = (Entry<Integer, Integer>) it.next();
			int key1 = e1.getKey();
			int value1 = e1.getValue();
			Entry<Integer, Integer> e2 = (Entry<Integer, Integer>) it.next();
			int key2 = e2.getKey();
			int value2 = e2.getValue();
			if (value1 == 1 || value2 == 1 && Math.abs(key1 - key2) == 1)
				return "YES";
		}
		return "NO";
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String s = in.next();
		String result = isValid(s);
		System.out.println(result);
	}
}

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