Program to count the Occurrence of a Digit in a given Number in Python
Occurrence of a Digit in a given Number in Python
Here, in this page we will discuss the program to find the Occurrence of a Digit in a given Number in Python programming language. We will discuss different methods to solve this problem.
Example :
Input : Enter a number : 897982Enter the digit : 9
Output : 2
Explanation : The digit 9 occurs twice
Method Discussed :
- Method 1 : Using loops
- Method 2 : Using String
Method 1 :
- Declare variable count that will count the required number of occurrences
- Take a while loop.
- Check if(n%10 ==d)
- If yes, then increment the value of count by 1.
- Set, n=n//10
- Print the value of count.
Method 1 : Code in Python
Run
#Write a program to print the Occurrence of a Digit in a given Number in Python
def countOccurrances(n, d):
count = 0
# Loop to find the digits of N
while (n > 0):
# check if the digit is D
if(n % 10 == d):
count = count + 1
n = n // 10
return count
# Driver code
d = 2
n = 828282
print(countOccurrances(n, d))
Output:
3
Method 2 :
In this method we will convert the number and digit into string, and then count them.
Method 2 : Code in Python
Run
#Write a program to print the Occurrence of a Digit in a given Number in Python #Write a program to print the Occurrence of a Digit in a given Number in Python Number = 828282 Digit = 2 #initialize Strings String1 = str() String2 = str() #typecast int to str String1 = str(Number) String2 = str(Digit) #count and print the occurrence #Count function will return int value #so change it's type to string and concatenate it print(String1.count(String2))
Output:
3
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6*6*6 2*2*2
def count_digit(number,digit):
count=0
for i in str(number):
if i==str(digit):
count+=1
print(f”Count of {digit} is {count}”)
count_digit(98219,9)
n,x=[x for x in input(“Enter the number and digit: “).split()]
print(n.count(x))
a=input(“enter the value :”)
b=input(“enter the value of digit:”)
print(a.count(b))
———————————————
n = str(input(‘Enter Number: ‘))
k = str(input(‘Enter the digit: ‘))
print(“Digit count is: “,n.count(k))
n=input(“Enter a number:”)
m=input(“Enter Digit to check:”)
list1=list(n)
list2=[]
for i in range(len(list1)):
if(list1[i]==m):
list2.append(list1[i])
print(len(list2))
Code in Java :
import java.util.Scanner;
public class Frequency_of_Digit
{ public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
System.out.print(“Enter number: “);
String number = sc.nextLine();
System.out.print(“Enter digit: “);
char digit = sc.nextLine().charAt(0);
sc.close();
int count=0;
for(int i=0; i<number.length(); i++)
if(number.charAt(i)==digit)
count++;
System.out.println(count);
}
}
a=input(“enter the value :”)
b=input(“enter the value of digit:”)
print(a.count(b))
program in c given below
#include
void main()
{
int n,d,count=0,r;
printf(“enter the number”);
scanf(“%d”,&n);
printf(“enter digit”);
scanf(“%d”,&d);
while(n>0)
{
r=n%10;
if(r==d)
{
count++;
}
n=n/10;
}
printf(“%d occurs %d times”,d,count);
}
#include
int main()
{
int n,count=0;
printf(“enter last number:”);
scanf(“%d”,&n);
for(int i=1;i0){
if(m%10==3){
count++;
}
m=m/10;
}
}
printf(“%d”,count);
return 0;
}