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# LUCKFOUR - editorial

Author: Sergey Kulik
Tester: Roman Furko
Editorialist: Balajiganapathi Senthilnathan

Cakewalk

Basic string

# PROBLEM

For each number, you have to count the number of times the digit 4 occurs in it.

# QUICK EXPLANATION

Input the number as a string and count the number of times the character '4' occurs in it.

# EXPLANATION

We may be tempted to input the number as an int, but this is really a string related task. We have to count the number of times a particular character appears in the string. So, taking the input as a string helps us avoid all the complicated stuff we will have to do if we input as a number.

We just have to loop through all characters of the string and if it is '4', increment the answer.

It is also instructive to see how we can solve this if we treat the input as a number.

Say we have a number x = 12345. How can we extract its digits? Observe that we can easily get the rightmost digit(unit's place) - if we divide x by 10 and take the reminder, it will be the rightmost digit. Further, the quotient we get is the rest of the number. For example, 12345 = 1234 * 10 + 5. So we get both the unit's place digit and the rest of the number. Now to get the next digit, we continue in the same fashion. When should we stop? Suppose x is a single digit, say x = 1. Now the quotient after dividing by 10 is 0, and we can stop since there are no more digits to process.

Pseudocode:

ans = 0
while x > 0:
remainder = x % 10
quotient = x / 10
if remainder == 4 ans++
x = quotient


# Complexity

The number of times the while loop will run is equal to the number of digits. Say d is the number of digits, the complexity will therefore be $O(d)$. Also, if we treat the input as integer, we can see that $d = \lceil log_{10}$ $n \rceil$. So, the complexity will be $\mathcal{O}(log$ $n)$.

# AUTHOR'S, TESTER'S and Editorialist's SOLUTIONS:

75542742
accept rate: 77%

2561314

 0 Even if the input is given as an integer (for example in TopCoder SRMs, you have to implement a function instead of reading inputs from STDIO), you could still use sprintf(C++) or str()(Python) or whatever methods your familiar language has and turn the number to string. This will reduce the possibility of making errors. answered 22 Feb '15, 14:11 196●1●2●5 accept rate: 0%
 0 yes . Actaully just check s[i]=='4' . answered 22 Feb '15, 14:13 131●5 accept rate: 14%
 0 I used numeric input rather than string and got AC answered 22 Feb '15, 14:14 893●2●11●35 accept rate: 10%
 0 for _ in xrange(input()): print raw_input().count('4')  Gotta love python. :) answered 22 Feb '15, 14:14 757●2●11 accept rate: 16%

# include<conio.h>

struct count
{
char a[50];
}e[5];
void main() {
char i,j=0,c=0;

printf("enter the strings");
for(i=0;i<5;i++)
{
gets(e[i].a);
}

for(i=0;i<5;i++)
{
for(j=0;j<50;j++)
{
if(e[i].a[j]=='4')
c=c+1;
}
}
printf("number of 4s is %d",c);

getch();
}

112
accept rate: 0%

 0 Another way the difference between the length of original string and string.replace("4","") will give you the answer public class NumberOfFours { private static StringBuilder sb= new StringBuilder(); public static void main(String[] args) { Scanner input= new Scanner(System.in); int numberOfTestCases = input.nextInt(); while(numberOfTestCases-- >0){ String inputString = input.next(); Solve(inputString); } System.out.println(sb); } private static void Solve(String inputString) { String s = inputString.replaceAll("4", ""); int diff = inputString.length()-s.length(); sb.append(Integer.toString(diff)).append("\n"); }  } answered 24 Feb '15, 21:53 21●1 accept rate: 0%
 0 For reference .... I have tried this and it worked for me import java.lang.; import java.util.; class four { public static void main(String[] args) { Scanner obj= new Scanner(System.in); int x,z; x= obj.nextInt(); if ((x>0)&&(x<=100000)) { int[] y; y = new int[x]; for(int i=0; i
 0 Is the second method is more efficient than the first method? answered 01 Feb '16, 02:21 1 accept rate: 0%
 0 Hi, I am not able to get following line : (Subtask 1): 0 ≤ Numbers from the list ≤ 9 - 33 points. (Subtask 2): 0 ≤ Numbers from the list ≤ 109 - 67 points. can someone elaborate it? Thanks answered 05 Jun '16, 23:59 0★vijnux 1 accept rate: 0%
 0 i actually managed to receive 33 points so i guess i got Subtask 1, but i am not able to fulfill Subtask 2 although its the same as Sub1 but with different values. Can someone help me ? I am quite new to programming at all so my solution is not that good i guess :D Thats what i got so far : (comments are in german, sorry) https://www.codechef.com/viewsolution/12076756 answered 11 Nov '16, 19:07 0★coppice 1 accept rate: 0%
 0 Hello Guys..!! I have this C Code which I wrote for this problem. It runs fine and gives me the right answers in my Dev-C++ Environment. But the CodeChef Online Judge says Wrong answer and doesn't award me points. Can anyone please tell me what is the problem here...?? PS: Code Below #include int main() { int T , rem , c , e ; scanf("%d" , &T) ; for ( int i = 0 ; i < T ; ++i ) { scanf("%d" , &e) ; while (e != 0) { rem = e % 10 ; if (rem == 4) { ++c ; } e = e / 10 ; } printf("%d\n" , c) ; c = 0 ; } }  answered 27 Dec '16, 14:04 58●4 accept rate: 7%
 0 @sagar19122 I think you should define your c=0, because it will give wrong output when number is 0. answered 23 Mar '18, 19:40 4★elli0t 57●4 accept rate: 4%
 0 How about this: #include using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t = 0; string number = ""; cin >> t; for (int i = 0; i < t; i++) { cin >> number; cout << count(number.begin(), number.end(), '4') << "\n"; } }  answered 24 Aug '18, 23:00 2★salil03 30●6 accept rate: 0%
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question asked: 18 Feb '15, 23:05

question was seen: 5,740 times

last updated: 24 Aug '18, 23:00