Last Updated: 30 Jul, 2022
Ninja and Numbers
Moderate
Asked in companies
Ninja has a number ‘A’ containing ‘B’ digits. ‘A’ can be represented by a string ‘S’ where ‘S[i]’ denotes the ‘ith’ digit of ‘A’. You are also given an integer ‘K’.
Ninja thinks that a number is stable if the following condition is satisfied:
For every ‘ith’ digit where (0 <= ‘i’ <= ‘B-1’) ‘S[i] = S[i%K]’. Here, ‘X%Y’ represents the modulo operations. The remainder when ‘X’ is divided by ‘Y’.
Your task is to find the smallest number which is stable and whose value is greater than or equal to ‘A’. Zero-based indexing is used everywhere.
Example :
‘B’ = 4, ‘S’ = “4321”, ‘K’ = 3.
The given number is not stable as ‘S[3]’ is not the same as ‘S[0]’ but 3%3 = 0 same as 0%3. ‘S[3] = 1’ and ‘S[0] = 4’. But the number “4324” is stable. As, for all ‘i’, ‘S[i]’ = ‘S[i%K]’ and “4324” is also greater than the given number. It can be proved that this is the best possible answer.
Hence, the answer is “4324”.
Input Format :
The first line contains a single integer ‘T’ denoting the number of test cases, then the test case follows.
The first line of each test case contains two single space-separated integers, ‘B’ and ‘K'.
The second line contains a string with ‘B’ characters representing the string ‘S’.
Output Format :
For each test case, return the minimum stable number whose value is the same or exceeds the value of ‘A’.
Output for each test case will be printed on a separate line.
Note :
You are not required to print anything; it has already been taken care of. Just implement the function.
Constraints :
1 ≤ T ≤ 10
1 ≤ B ≤ 10^5
1 ≤ K ≤ 10^9
It’s guaranteed that sum of ‘B’ over all cases does not exceed 10^5.
The given number will not contain leading zeros.
Time limit: 1 sec
Approaches
Approach:
- The first ‘K’ digits of the given number are the most important ones from 0 to ‘K-1’. The other digits are just formed using them because if ‘i >= K’, then ‘S[i] = S[i%K]’ and ‘i%K < K’. Based on the definition of stable numbers.
- Let us check if the stable number form using given ‘K’ digits is greater or equal to the given number.
- For that create new number where for all (0 <= ‘i’ <= ‘B-1’) ‘CURR[i] = S[i%K]’. If ‘CURR’ is not smaller than ‘S’, the ‘CURR’ will be the answer.
- Otherwise, just add 1 to the first ‘K’ digit:
- To do that, find the last digit among the ‘K’ digits that is not ‘9’ and add 1 to it.
- After adding 1, set all the digits on the right side to ‘0’.
- Finally, form the complete answer by copying the other digits from the first ‘K’ digits. Just do ‘S[i] = S[i%K]’ for all index.
Algorithm:
- Declare a variable ‘n’ and set it to the size of the given string.
- Create a string ‘currStable’ of length ‘n’ where ‘currStable[i] = s[i%k]’.
- Now, if ‘currStable’ is greater or equal to ‘s’, then return ‘currStable’.
- Declare a variable ‘ind’ = ‘k-1’.
- While ‘s[ind]’ == ‘9’:- ‘ind- -’
- Finally, we found an index where ‘s[ind]’ != ‘9’, incrementing this digit.
- ‘s[ind]’++
- Form the rest of the answer by looping through ‘i’ from ‘k’ to ‘n-1’ and setting ‘s[i] = s[i%k]’.
- Return the final answer ‘s’.