Solution of Special Numbers is added

Author: Ishitamukherjee2004

dsa-solutions/gfg-solutions/Easy problems/Special-Number.md

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+---
+id: special-numbers
+title: Special Numbers
+sidebar_label: Special-Numbers
+tags:
+  - Mathematical
+  - Algorithms
+description: "This tutorial covers the solution to the Special Numbers problem from the GeeksforGeeks."
+---
+## Problem Description
+
+A number is a special number if it’s digits only consist `0`, `1`, `2`, `3`, `4` or `5`. Given a number N and we have to find N-th Special Number. 
+
+## Examples
+
+**Example 1:**
+
+```
+Input:
+N = 6
+Output: 5
+Explanation: First 6 numbers are
+( 0, 1, 2, 3, 4, 5 )
+```
+
+**Example 2:**
+
+```
+Input: 
+N = 7
+Output: 10
+Explanation: First 7 numbers are
+( 0, 1, 2, 3, 4, 5, 10 )
+```
+
+
+Expected Time Complexity: O(logN)
+
+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 findNthSpecial(int N) {
+    int count = 0;
+    int num = 0;
+    while (true) {
+        num++;
+        if (isSpecial(num)) {
+            count++;
+            if (count == N) {
+                return num;
+            }
+        }
+    }
+}
+
+bool isSpecial(int num) {
+    while (num > 0) {
+        int digit = num % 10;
+        if (digit > 5) {
+            return false;
+        }
+        num /= 10;
+    }
+    return true;
+}
+
+
+
+```
+
+```java
+public int findNthSpecial(int N) {
+    int count = 0;
+    int num = 0;
+    while (true) {
+        num++;
+        if (isSpecial(num)) {
+            count++;
+            if (count == N) {
+                return num;
+            }
+        }
+    }
+}
+
+public boolean isSpecial(int num) {
+    while (num > 0) {
+        int digit = num % 10;
+        if (digit > 5) {
+            return false;
+        }
+        num /= 10;
+    }
+    return true;
+}
+
+
+```
+
+```python
+
+def find_nth_special(N):
+    count = 0
+    num = 0
+    while True:
+        num += 1
+        if is_special(num):
+            count += 1
+            if count == N:
+                return num
+
+def is_special(num):
+    while num > 0:
+        digit = num % 10
+        if digit > 5:
+            return False
+        num //= 10
+    return True
+
+
+```
+
+```javascript
+function findNthSpecial(N) {
+    let count = 0;
+    let num = 0;
+    while (true) {
+        num++;
+        if (isSpecial(num)) {
+            count++;
+            if (count === N) {
+                return num;
+            }
+        }
+    }
+}
+
+function isSpecial(num) {
+    while (num > 0) {
+        const digit = num % 10;
+        if (digit > 5) {
+            return false;
+        }
+        num = Math.floor(num / 10);
+    }
+    return true;
+}
+
+
+
+```
+
+## Solution Logic:
+1. Initialize a counter count to 0 and a number num to 0.
+2. Iterate through numbers starting from 1.
+3. For each number, check if it's special (i.e., its digits only consist of 0, 1, 2, 3, 4, or 5) using the isSpecial function.
+4. If the number is special, increment the count.
+5. If the count reaches N, return the current number.
+6. Repeat steps 2-5 until the N-th special number is found.
+
+## Time Complexity
+
+- The `isSpecial` function has a time complexity of O(log num), where num is the input number, because it iterates through the digits of the number.
+
+- The main function findNthSpecial has a time complexity of O(N log M), where N is the input number and M is the N-th special number, because it iterates through numbers and calls the isSpecial function for each number.
+
+
+
+## Space Complexity
+
+- The solution uses a constant amount of space to store the count and num variables, so the space complexity is O(1).