Count all digits of a number

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You are given an integer n. You need to return the number of digits in the number.

The number will have no leading zeroes, except when the number is 0 itself.

Examples:

Input: n = 4

Output: 1

Explanation: There is only 1 digit in 4.

Input: n = 14

Output: 2

Explanation: There are 2 digits in 14.

Input: n = 234

Constraints

0 <= n <= 5000
n will contain no leading zeroes except when it is 0 itself.

Company Tags

TCS Cognizant Accenture Infosys Capgemini Wipro IBM HCL Tech Mahindra MindTree

Editorial

Intuition:

Given a number, all the digits in the number can be counted by counting one by one every digit which can be done by extracting the last digit successively from the right until there are no more digits left to extract.

Approach:

Initialise a counter to keep the count of digits in the number. Keep dividing the number by 10 until no more digits are left to extract.
For every digit extracted from the number, increment the counter by 1.
Once the iterations are over, the number of digits is stored in the counter.

Edge Case:

What if the given number is zero?
Return 1, because the number of digits in zero is 1.

Dry Run:

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    // Function to count all digits in n
    int countDigit(int n) {
        // Edge case
        if(n == 0) return 1;
        
        // Set counter to 0
        int cnt = 0;
        
        // Iterate until n is greater than zero
        while(n > 0) {
            // Increment count of digits
            cnt = cnt + 1; 
            n = n / 10;
        }
        
        return cnt;
    }
};

int main()
{
    int n = 6678;
    
    /* Creating and instance of 
    Solution class */
    Solution sol; 
    
    // Function call to get count of digits in n
    int ans = sol.countDigit(n);
    cout << "The count of digits in the given number is: " << ans;
    
    return 0;
}
class Solution {
    // Function to count all digits in n
    public int countDigit(int n) {
        // Edge case
        if (n == 0) return 1;

        // Set counter to 0
        int cnt = 0;

        // Iterate until n is greater than zero
        while (n > 0) {
            // Increment count of digits
            cnt = cnt + 1;
            n = n / 10;
        }

        return cnt;
    }
}

class Main {
    public static void main(String[] args) {
        int n = 6678;

        /* Creating an instance of
        Solution class */
        Solution sol = new Solution();

        // Function call to get count of digits in n
        int ans = sol.countDigit(n);
        System.out.println("The count of digits in the given number is: " + ans);
    }
}
class Solution:
    # Function to count all digits in n
    def countDigit(self, n):
        # Edge case
        if n == 0:
            return 1

        # Set counter to 0
        cnt = 0

        # Iterate until n is greater than zero
        while n > 0:
            # Increment count of digits
            cnt = cnt + 1
            n = n // 10

        return cnt

# Input number
n = 6678

# Creating an instance of Solution class
sol = Solution()

# Function call to get count of digits in n
ans = sol.countDigit(n)
print(f"The count of digits in the given number is: {ans}")
class Solution {
    // Function to count all digits in n
    countDigit(n) {
        // Edge case
        if (n === 0) return 1;

        // Set counter to 0
        let cnt = 0;

        // Iterate until n is greater than zero
        while (n > 0) {
            // Increment count of digits
            cnt = cnt + 1;
            n = Math.floor(n / 10);
        }

        return cnt;
    }
}

// Input number
let n = 6678;

/* Creating an instance of
Solution class */
let sol = new Solution();

// Function call to get count of digits in n
let ans = sol.countDigit(n);
console.log("The count of digits in the given number is: " + ans);

Complexity Analysis:

Time Complexity: O(log₁₀(N)) – In every iteration we are dividing N by 10.

Space Complexity: O(1) – Using a couple of variables i.e., constant space.

Intuition:

Given a number, the number of digits can be found directly from the logarithm (base 10) of the number.

Approach:

Instead of iterating on the number for every digit, the count of digits in the number can be found using the following mathematical formula: count = log₁₀(N) + 1.

Edge Case:

What if the given number is zero?
Return 1, because the number of digits in zero is 1.

Solution:

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    /* Function to count the 
    number of odd digits in N */
    int countDigit(int n) {
        // Edge case
        if(n == 0) return 1;
        
        int count = log10(n) + 1;
        return count;
    }
};

int main()
{
    int n = 6678;
    
    /* Creating an instance of 
    Solution class */
    Solution sol; 
    
    // Function call to get count of digits in n
    int ans = sol.countDigit(n);
    cout << "The count of digits in the given number is: " << ans;
    
    return 0;
}
class Solution {
    /* Function to count the 
    number of odd digits in N */
    public int countDigit(int n) {
        // Edge case
        if (n == 0) return 1;
        
        int count = (int)(Math.log10(n) + 1);
        return count;
    }

    public static void main(String[] args) {
        int n = 6678;

        /* Creating an instance of 
        Solution class */
        Solution sol = new Solution(); 

        // Function call to get count of digits in n
        int ans = sol.countDigit(n);
        System.out.println("The count of digits in the given number is: " + ans);
    }
}
import math

class Solution:
    # Function to count the
    # number of odd digits in N
    def countDigit(self, n):
        # Edge case
        if n == 0:
            return 1
            
        count = int(math.log10(n) + 1)
        return count

# Input number
n = 6678

# Creating an instance of Solution class
sol = Solution()

# Function call to get count of digits in n
ans = sol.countDigit(n)
print("The count of digits in the given number is:", ans)
class Solution {
    /* Function to count the 
    number of odd digits in N */
    countDigit(n) {
        // Edge case
        if (n === 0) return 1;
        
        let count = Math.floor(Math.log10(n) + 1);
        return count;
    }
}

// Input number
let n = 6678;

// Creating an instance of Solution class
let sol = new Solution();

// Function call to get count of digits in n
let ans = sol.countDigit(n);
console.log("The count of digits in the given number is: " + ans);

Complexity Analysis:

Time Complexity: O(1) – Constant time taking statements were used.

Space Complexity: O(1) – Because only a couple of variables were used.

Notes

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Problem Categories

Count all digits of a number

Examples:

Constraints

Company Tags

Editorial

Intuition:

Approach:

Edge Case:

Dry Run:

Complexity Analysis:

Intuition:

Approach:

Edge Case:

Solution:

Complexity Analysis:

Notes

Code

Facts