Skip to main content

Encoded String codechef January long challenge solution || Encoded String codechef January long challenge solution Editorial 2021

 Encoded String codechef January long challenge solution || Encoded String codechef January long challenge solution Editorial 2021-

Problem statement-

An encoder encodes the first 16 lowercase English letters using 4 bits each. The first bit (from the left) of the code is 0 if the letter lies among the first 8 letters, else it is 1, signifying that it lies among the last 8 letters. The second bit of the code is 0 if the letter lies among the first 4 letters of those 8 letters found in the previous step, else it's 1, signifying that it lies among the last 4 letters of those 8 letters. Similarly, the third and the fourth bit each signify the half in which the letter lies.

For example, the letter j would be encoded as :

  • Among (a,b,c,d,e,f,g,h | i,j,k,l,m,n,o,p)j appears in the second half. So the first bit of its encoding is 1.
  • Now, among (i,j,k,l | m,n,o,p)j appears in the first half. So the second bit of its encoding is 0.
  • Now, among (i,j | k,l)j appears in the first half. So the third bit of its encoding is 0.
  • Now, among (i | j)j appears in the second half. So the fourth and last bit of its encoding is 1.

So j's encoding is 1001,

Given a binary encoded string S, of length at most 105, decode the string. That is, the first 4 bits are the encoding of the first letter of the secret message, the next 4 bits encode the second letter, and so on. It is guaranteed that the string's length is a multiple of 4.

Input:

  • The first line of the input contains an integer T, denoting the number of test cases.
  • The first line of each test case contains an integer N, the length of the encoded string.
  • The second line of each test case contains the encoded string S.

Output:

For each test case, print the decoded string, in a separate line.

Constraints

  • 1T10
  • 4N105
  • The length of the encoded string is a multiple of 4.
  • 0Si1

Subtasks

  • 100 points : Original constraints.

Sample Input:

3
4
0000
8
00001111
4
1001

Sample Output:

a
ap
j

Explanation:

  • Sample Case 1 :

The first bit is 0, so the letter lies among the first 8 letters, i.e., among a,b,c,d,e,f,g,h. The second bit is 0, so it lies among the first four of these, i.e., among a,b,c,d.

The third bit is 0, so it again lies in the first half, i.e., it's either a or b. Finally, the fourth bit is also 0, so we know that the letter is a.

  • Sample Case 2 :

Each four bits correspond to a character. Just like in sample case 10000 is equivalent to a. Similarly, 1111 is equivalent to p. So, the decoded string is ap.

  • Sample Case 3 :

The first bit is 1, so the letter lies among the last 8 letters, i.e., among i,j,k,l,m,n,o,p. The second bit is 0, so it lies among the first four of these, i.e., among i,j,k,l.

The third bit is 0, so it again lies in the first half, i.e., it's either i or j. Finally, the fourth bit is 1, so we know that the letter is j.

Encoded String codechef January long challenge solution || Encoded String codechef January long challenge solution Editorial 2021


Solution-

THis question is based on pattern finding question. You have to find a simple pattern  that 

if(s=="0000"){
return 'a';
}
if(s=="0001"){
return 'b';
}
if(s=="0010"){
return 'c';
}
if(s=="0011"){
return 'd';
}
if(s=="0100"){
return 'e';
}
if(s=="0101"){
return 'f';
}
if(s=="0110"){
return 'g';
}
if(s=="0111"){
return 'h';
}
if(s=="1000"){
return 'i';
}
if(s=="1001"){
return 'j';
}
if(s=="1010"){
return 'k';
}
if(s=="1011"){
return 'l';
}
if(s=="1100"){
return 'm';
}
if(s=="1101"){
return 'n';
}
if(s=="1110"){
return 'o';
}
if(s=="1111"){
return 'p';
} Now you have to iterate over string in set of 4 character. finally print ans. follow my code
Join telegram channel for code and editorial.-https://t.me/competitiveProgrammingDiscussion



Comments

Popular posts from this blog

codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster

 Codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster- Codeforces is one of the most popular platforms for competitive programmers and  codeforces rating matters a lot  .  Competitive Programming  teaches you to find the easiest solution in the quickest possible way. CP enhances your problem-solving and debugging skills giving you real-time fun. It's brain-sport. As you start solving harder and harder problems in live-contests your analytical and rational thinking intensifies. To have a good codeforces profile makes a good impression on the interviewer. If you have a good  codeforces profile so it is very easy to get a referral for product base company like amazon, google , facebook etc.So in this blog I have explained everything about codeforces rating system. What are different titles on codeforces- based on rating codeforces divide rating into 10 part. Newbie Pupil Specialist Expert Candidate Codemaster Master International Master Grandmaster Internat

Merge Two sorted array Without Extra Space

Problem statement-  Given two sorted arrays arr1[] and arr2[] of sizes N and M in non-decreasing order. Merge them in sorted order without using any extra space. Modify arr1 so that it contains the first N elements and modify arr2 so that it contains the last M elements.    Example 1: Input:  N = 4, arr1[] = [1 3 5 7]  M = 5, arr2[] = [0 2 6 8 9] Output:  arr1[] = [0 1 2 3] arr2[] = [5 6 7 8 9] Explanation: After merging the two  non-decreasing arrays, we get,  0 1 2 3 5 6 7 8 9.   Example 2: Input:  N = 2, arr1[] = [10, 12]  M = 3, arr2[] = [5 18 20] Output:  arr1[] = [5 10] arr2[] = [12 18 20] Explanation: After merging two sorted arrays  we get 5 10 12 18 20. Your Task: You don't need to read input or print anything. You only need to complete the function merge() that takes arr1, arr2, N and M as input parameters and modifies them in-place so that they look like the sorted merged array when concatenated. Expected Time Complexity:  O((n+m) log(n+m)) Expected Auxilliary Space: O(1

Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code

 Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code - Apple Division CSES Problem Solution Easy Explanation. Apple division is problem is taken form cses introductory problem set.Let's Read Problem statement first. Problem Statement- Time limit:  1.00 s   Memory limit:  512 MB There are  n n  apples with known weights. Your task is to divide the apples into two groups so that the difference between the weights of the groups is minimal. Input The first input line has an integer  n n : the number of apples. The next line has  n n  integers  p 1 , p 2 , … , p n p 1 , p 2 , … , p n : the weight of each apple. Output Print one integer: the minimum difference between the weights of the groups. Constraints 1 ≤ n ≤ 20 1 ≤ n ≤ 20 1 ≤ p i ≤ 10 9 1 ≤ p i ≤ 10 9 Example Input: 5 3 2 7 4 1 Output: 1 Explanation: Group 1 has weights 2, 3 and 4 (total weight 9), and group 2 has weights 1 and 7 (total weight 8). Join Telegram channel for code discussi