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Hackerrank recursive digit sum

Webdef superDigit(n:str, k:int): result = sum(map(int, n)) * k # base case if result <= 9: return result # recursive case return superDigit(str(result), 1) 0 Permalink

Recursive Digit Sum HackerRank

WebJan 30, 2024 · int doSuperSum(int sum) { if (sum 9) { res += (sum%10); sum=sum/10; } res+=sum; // last remainder digit from left (hightest weight) return res; } int superDigit(string n, int k) { unsigned long long sum=0; /* Unoptimized way to get sum of all concatinated string's digits for (int i=0; i9) { sd = doSuperSum(sd); } return sd; } … WebOct 30, 2024 · Recursive digit sum hackerrank javascript runtime error on 3 tests Ask Question Asked 4 months ago Modified 4 months ago Viewed 354 times -1 I am trying to solve Recurive Digit Sum, and I actually solved it, but I got 3 runtime errors on large inputs when submitting. I have optimized my code a lot, but still I am getting runtime errors. test 2022.uz natijalari https://senetentertainment.com

HackerRank Recursive Digit Sum problem solution

WebAug 25, 2024 · As discussed in this post, recursive sum of digits is 9 if number is multiple of 9, else n % 9. Since divisibility and modular arithmetic are compatible with multiplication, we simply find result for single occurrence, multiply result with x and again find the result. How does this work? Lets N = 24 and X = 3. So, sumUntilSingle (N) = 2 + 4 = 6. WebRecursion: Fibonacci Numbers HackerRank hackerrank.com Like Comment Comment WebRecursive Digit Sum Problem Submissions Leaderboard Discussions Editorial You are viewing a single comment's thread. Return to all comments → yzn20080 6 years ago n, k = map(int, input().split()) x = n * k % 9 print(x if x else 9) More Python solutions here. 78 … test 6 coding ninjas

HackerRank Recursive Digit Sum problem solution

Category:Recursive Digit Sum Discussions Algorithms HackerRank

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Hackerrank recursive digit sum

Recursive sum of digits of a number formed by repeated appends

Web2 years ago If it works in Python, it's slow on big numbers and a half of test cases are really big numbers. You shoud make the first pass and sum all the digits and then do the trick. You have a limit of 100.000.000.000 there which suits long type. 0 Parent Permalink Gambit420 2 years ago Because n here in String. 0 Parent Permalink WebJan 11, 2024 · HackerRank: Recursive Digit Sum (in Algorithms) Problem Statement. Given an integer, we need to find the super digit of the integer. We define super digit of an integer using the following rules: If \(x\) has only 1 digit, then its super digit is \(x\). Otherwise, the super digit of \(x\) is equal to the super digit of the digit-sum of \(x ...

Hackerrank recursive digit sum

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WebMar 28, 2024 · My Python code that passed all the test cases: def superDigit(n, k): p = int(str(n))%9 p = (p*k)%9 if p == 0: return 9 return p 0 Permalink annnguyen32 1 week … WebHackerRank solutions in C and C++ by Isaac Asante. They include data structures and algorithms to practice for coding interview questions.

WebAnother approach would be the use of tail recursion, which does not extend the stack for each calling function, because the function ends with the call without keeping a temporary value liek in above function, the actual numerical value +num[0] and the waiting for execution of the adition.. In this case, you could store the intermediate result along with … WebWe define super digit of an integer using the following rules: Given an integer, we need to find the super digit of the integer. If has only digit, then its super digit is . Otherwise, the … John is new to Mathematics and does not know how to calculate GCD of numbers. … public static int superDigit (String n, int k) {// Write your code here if (n. length == 1 …

WebSep 23, 2024 · This is the recursive function. It stops when we end up with a single digit which is indicated by having p <= 9. We do some processing which implies generating the sum of digits in ‘p’. We then issue a recursive call to process the new ‘p’. When done, we return the super digit. These last two functions passed all the tests at HackerRank! WebRecursive Digit Sum Problem Submissions Leaderboard Discussions Editorial Reveal solutions Hacker Rank Country Score redprogrammer1 01 100.00 divyanshsengarj1 01 100.00 laijason2 01 100.00 PurtiAgarwal 01 100.00 phtsanov 01 100.00 wishrao24 01 100.00 sattujaiswal 01 100.00 olivier_piron 01 100.00 kore3d 01 100.00 …

WebMar 17, 2024 · def sup_digits (x,k): a = digsum (x) return sup_digit (str (int (a)*k)) def sup_digit (x): if len (x) <= 1: return x else: return sup_digit ( digsum (x) ) def digsum (x): return str (sum ( (int (i) for i in list (x)))) n, k = …

Webx. x x. For example, the super digit of 9875 9875 will be calculated as: super_digit(9875) 9+8+7+5 = 29 super_digit(29) 2 + 9 = 11 super_digit(11) 1 + 1 = 2 super_digit(2) = 2. … test 7 coding ninjaWebThis hackerrank problem is a part of Problem Solving Practice Algorithms Recursion Recursive Digit Sum and solved in python. 🔔 Subscribe: http://bit.ly/hackersrealm 🗓️ 1:1... batman harperWebRecursive Digit Sum Problem Submissions Leaderboard Discussions Editorial Reveal solutions Hacker Rank Country Score PurtiAgarwal 01 100.00 snekparti 01 100.00 … batman harper rowWebMay 12, 2024 · All of the digits of p sum to 116. The digits of 116 sum to 8. and 8 is only one digit, so it is the super digit. Example 2 : n = 148. k = 3. The number p is created by concatenating the string n, k times so the … batman hardwareWebWe define super digit of an integer using the following rules: . If has only digit, then its super digit is .; Otherwise, the super digit of is equal to the super digit of the digit-sum of .Here, digit-sum of a number is defined as the sum of its digits. For example, super digit of will be calculated as:. super_digit(9875) = super_digit(9+8+7+5) = super_digit(29) = … batman handyWebNov 24, 2024 · I was trying to solve this problem on hackerrank. But I got some problem. Specific problem is: For example: The sum of digits 9875 will be calculate as: sum … test 7 coding ninja githubWebI just wanted to know why you used n.chars ().mapToLong (Character::getNumericValue).sum () * k + ""; I understand you are trying to sum all the characters and multiply by k and converting back to string but this notation i have never used. is this faster than say doing a for loop to find this out? doesnt this take O (x) time … batman hasbro dc