Solution of Count Digits from gfg is added

Author: Ishitamukherjee2004

dsa-solutions/gfg-solutions/Easy problems/count-digits.md

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+---
+id: count-digits
+title: Count Digits
+sidebar_label: Count-Digits
+tags:
+  - Modular Arithmetic
+  - Algorithms
+description: "This tutorial covers the solution to the Count Digits problem from the GeeksforGeeks."
+---
+## Problem Description
+
+Given a number `n`. Count the number of digits in `n` which evenly divide n. Return an integer, total number of digits of n which divides n evenly.
+
+Note :- Evenly divides means whether `n` is divisible by a digit i.e. leaves a remainder `0` when divided.
+
+## Examples
+
+**Example 1:**
+
+```
+Input: n = 12
+Output: 2
+Explanation: 1, 2 when both divide 12 leaves remainder 0.
+```
+
+**Example 2:**
+
+```
+Input: n = 2446
+Output: 1
+Explanation: Here among 2, 4, 6 only 2 divides 2446 evenly while 4 and 6 do not.
+```
+
+
+Expected Time Complexity: O(N)
+
+Expected Auxiliary Space: O(1)
+
+## Constraints
+
+* `1 ≤ N ≤ 10^5`
+
+## Problem Explanation
+
+The task is to traverse the number and count the digits.
+
+## Code Implementation
+
+### C++ Solution
+
+```cpp
+int countDigits(int n) {
+  int count = 0;
+  int temp = n;
+  while (temp != 0) {
+    int digit = temp % 10;
+    if (n % digit == 0) {
+      count++;
+    }
+    temp /= 10;
+  }
+  return count;
+}
+
+
+```
+
+```java
+
+public int countDigits(int n) {
+  int count = 0;
+  int temp = n;
+  while (temp != 0) {
+    int digit = temp % 10;
+    if (n % digit == 0) {
+      count++;
+    }
+    temp /= 10;
+  }
+  return count;
+}
+
+
+```
+
+```python
+def count_digits(n):
+  count = 0
+  temp = n
+  while temp != 0:
+    digit = temp % 10
+    if n % digit == 0:
+      count += 1
+    temp //= 10
+  return count
+
+```
+
+```javascript
+function countDigits(n) {
+  let count = 0;
+  let temp = n;
+  while (temp !== 0) {
+    const digit = temp % 10;
+    if (n % digit === 0) {
+      count++;
+    }
+    temp = Math.floor(temp / 10);
+  }
+  return count;
+}
+
+
+```
+
+## Solution Logic:
+1. Initialize a variable count to 0, which will store the number of digits that evenly divide n.
+2. Initialize a variable temp to n, which will be used to iterate through each digit of n.
+3. Use a while loop to iterate through each digit of n. In each iteration, do the following:
+    - Calculate the current digit by taking the remainder of temp divided by 10 (temp % 10).
+    - Check if n is divisible by the current digit by checking if n % digit == 0. If it is, increment count.
+    - Update temp by dividing it by 10 (temp /= 10).
+4. Return count at the end of the function.
+
+
+
+## Time Complexity
+
+* The time complexity is $O(log(n))$ where n is the input number. This is because we are iterating through each digit of n using a while loop, and the number of digits in n is proportional to the logarithm of n.
+
+
+## Space Complexity
+
+* The auxiliary space complexity is $O(1)$ due to the only extra memory used is for temporary variables while swapping two values in Array.