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In this post, you will find the solution for Weighted Uniform Strings 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 – Weighted Uniform Strings– Hacker Rank Solution
Weighted Uniform Strings – Hacker Rank Solution
Problem:
A weighted string is a string of lowercase English letters where each letter has a weight. Character weights are to from to as shown below:
![Weighted Uniform Strings in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct] 2 image](https://s3.amazonaws.com/hr-challenge-images/0/1484319110-9529e3b407-uniform.png)
- The weight of a string is the sum of the weights of its characters. For example:
- A uniform string consists of a single character repeated zero or more times. For example,
ccc
anda
are uniform strings, butbcb
andcd
are not.
Given a string, , let be the set of weights for all possible uniform contiguous substrings of string . There will be queries to answer where each query consists of a single integer. Create a return array where for each query, the value is Yes
if . Otherwise, append No
.
Note: The symbol denotes that is an element of set .
Example
.
Working from left to right, weights that exist are:
string weight a 1 b 2 bb 4 c 3 cc 6 ccc 9 d 4 dd 8 ddd 12 dddd 16
Now for each value in , see if it exists in the possible string weights. The return array is ['Yes', 'No', 'No', 'Yes', 'No']
.
Function Description
Complete the weightedUniformStrings function in the editor below.
weightedUniformStrings has the following parameter(s):
– string s: a string
– int queries[n]: an array of integers
Returns
– string[n]: an array of strings that answer the queries
Input Format
The first line contains a string , the original string.
The second line contains an integer , the number of queries.
Each of the next lines contains an integer , the weight of a uniform subtring of that may or may not exist.
Constraints
- will only contain lowercase English letters, ascii[a-z].
Sample Input 0
abccddde 6 1 3 12 5 9 10
Sample Output 0
Yes Yes Yes Yes No No
Explanation 0
The weights of every possible uniform substring in the string abccddde
are shown below:
![Weighted Uniform Strings in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct] 4 image](https://s3.amazonaws.com/hr-challenge-images/0/1485248159-0ecbb3c67b-uniform4.png)
We print Yes
on the first four lines because the first four queries match weights of uniform substrings of . We print No
for the last two queries because there are no uniform substrings in that have those weights.
Note that while de
is a substring of that would have a weight of , it is not a uniform substring.
Note that we are only dealing with contiguous substrings. So ccc
is not a substring of the string ccxxc
.
Sample Input 1
aaabbbbcccddd 5 9 7 8 12 5
Sample Output 1
Yes No Yes Yes No
Weighted Uniform Strings – Hacker Rank Solution
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); int len = s.length(); int n = in.nextInt(); Set<Integer> set = new HashSet<Integer>(); int i=0; while(i<len){ int j=i; int sum =0; while( j<len && s.charAt(i)==s.charAt(j) ){ sum += (s.charAt(i)-'a') +1; set.add(sum); j++; } i = j; } for(int a0 = 0; a0 < n; a0++){ int x = in.nextInt(); if (set.contains(x)){ System.out.println("Yes"); } else{ System.out.println("No"); } } } }