add confusing-number article

articles/confusing-number.md

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+## 1. Invert and Reverse
+
+::tabs-start
+
+```python
+class Solution:
+    def confusingNumber(self, n: int) -> bool:
+        # Use 'invertMap' to invert each valid digit.
+        invert_map = {"0":"0", "1":"1", "8":"8", "6":"9", "9":"6"}
+        rotated_number = []
+
+        # Iterate over each digit of 'n'.
+        for ch in str(n):
+            if ch not in invert_map:
+                return False
+
+            # Append the inverted digit of 'ch' to the end of 'rotated_number'. 
+            rotated_number.append(invert_map[ch])
+
+        rotated_number = "".join(rotated_number)
+
+        # Check if the reversed 'rotated_number' equals 'n'.
+        return int(rotated_number[::-1]) != n
+```
+
+```java
+class Solution {
+    public boolean confusingNumber(int n) {
+        // Use 'invertMap' to invert each valid digit.
+        Map<Character, Character> invertMap = new HashMap<>() {{
+            put('0', '0');
+            put('1', '1');
+            put('6', '9');
+            put('8', '8');
+            put('9', '6');
+        }};
+        StringBuilder sb = new StringBuilder();
+
+        // Iterate over each digit of 'n'.
+        for (char ch : String.valueOf(n).toCharArray()) {
+            if (!invertMap.containsKey(ch)) {
+                return false;
+            }
+
+            // Append the inverted digit of 'ch' to the end of 'rotatedNumber'. 
+            sb.append(invertMap.get(ch));
+        }
+
+        // Check if the reversed 'rotatedNumber' equals 'n'.
+        sb.reverse();
+        return Integer.parseInt(sb.toString()) != n;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    bool confusingNumber(int n) {
+        // Use 'invertMap' to invert each valid digit.
+        unordered_map<char, char> invertMap = {{'0','0'}, {'1','1'}, {'6','9'}, {'8','8'}, {'9','6'}};
+        string rotatedNumber;
+
+        // Iterate over each digit of 'n'.
+        for (auto ch : to_string(n)) {
+            if (invertMap.find(ch) == invertMap.end()) {
+                return false;
+            }
+
+            // Append the inverted digit of 'ch' to the end of 'rotatedNumber'. 
+            rotatedNumber += invertMap[ch];
+        }
+
+        // Check if the reversed 'rotatedNumber' equals 'n'.
+        reverse(begin(rotatedNumber), end(rotatedNumber));
+        return stoi(rotatedNumber) != n;
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number} n
+     * @return {boolean}
+     */
+    confusingNumber(n) {
+        // Use 'invertMap' to invert each valid digit.
+        const invertMap = {
+            '0': '0',
+            '1': '1',
+            '6': '9',
+            '8': '8',
+            '9': '6'
+        };
+
+        let rotatedNumber = '';
+        // Iterate over each digit of 'n'.
+        for (const ch of String(n)) {
+            if (!(ch in invertMap)) {
+                return false;
+            }
+            // Append the inverted digit of 'ch' to the end of 'rotatedNumber'. 
+            rotatedNumber += invertMap[ch];
+        }
+        // Check if the reversed 'rotatedNumber' equals 'n'.
+        rotatedNumber = rotatedNumber.split('').reverse().join('');
+        return parseInt(rotatedNumber) !== n;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+- Time complexity: $O(L)$
+- Space complexity: $O(L)$ extra space used
+
+>  Where $L$ is the maximum number of digits $n$ can have ($L = \log_{10} n$).
+
+---
+
+## 2. Use the remainder
+
+::tabs-start
+
+```python
+class Solution:
+    def confusingNumber(self, n: int) -> bool:
+        # Use 'invert_map' to invert each valid digit. Since we don't want to modify
+        # 'n', we create a copy of it as 'nCopy'.
+        invert_map = {0:0, 1:1, 8:8, 6:9, 9:6}
+        invert_number = 0
+        n_copy = n
+
+        # Get every digit of 'n_copy' by taking the remainder of it to 10.
+        while n_copy:
+            res = n_copy % 10
+            if res not in invert_map:
+                return False
+
+            # Append the inverted digit of 'res' to the end of 'rotated_number'. 
+            invert_number = invert_number * 10 + invert_map[res]
+            n_copy //= 10
+
+        # Check if 'rotated_number' equals 'n'.
+        return  invert_number != n
+```
+
+```java
+class Solution {
+    public boolean confusingNumber(int n) {
+        // Use 'invertMap' to invert each valid digit. Since we don't want to modify
+        // 'n', we create a copy of it as 'nCopy'.
+        Map<Integer, Integer> invertMap = new HashMap<>() {{
+            put(0, 0);
+            put(1, 1);
+            put(6, 9);
+            put(8, 8);
+            put(9, 6);
+        }};
+        int nCopy = n, rotatedNumber = 0;
+
+        // Get every digit of 'nCopy' by taking the remainder of it to 10.
+        while (nCopy > 0) {
+            int res = nCopy % 10;
+            if (!invertMap.containsKey(res)) {
+                return false;
+            }
+
+            // Append the inverted digit of 'res' to the end of 'rotatedNumber'. 
+            rotatedNumber = rotatedNumber * 10 + invertMap.get(res);
+            nCopy /= 10;
+        }
+
+        // Check if 'rotatedNumber' equals 'n'.
+        return rotatedNumber != n;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    bool confusingNumber(int n) {
+        // Use 'invertMap' to invert each valid digit. Since we don't want to modify
+        // 'n', we create a copy of it as 'nCopy'. 
+        map<int, int> invertMap = {{0, 0}, {1, 1}, {6, 9}, {8, 8}, {9, 6}};
+        int rotatedNumber = 0, nCopy = n;
+
+        // Get every digit of 'nCopy' by taking the remainder of it to 10.
+        while (nCopy > 0) {
+            int res = nCopy % 10;
+            if (invertMap.find(res) == invertMap.end()) {
+                return false;
+            }
+
+            // Append the inverted digit of 'res' to the end of 'rotatedNumber'. 
+            rotatedNumber = rotatedNumber * 10 + invertMap[res];
+            nCopy /= 10;
+        }
+
+        // Check if 'rotatedNumber' equals 'n'.
+        return rotatedNumber != n; 
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number} n
+     * @return {boolean}
+     */
+    confusingNumber(n) {
+        // Use 'invertMap' to invert each valid digit. Since we don't want to modify
+        // 'n', we create a copy of it as 'nCopy'.
+        const invertMap = {
+            0: 0,
+            1: 1,
+            6: 9,
+            8: 8,
+            9: 6
+        };
+        let nCopy = n;
+        let rotatedNumber = 0;
+
+        // Get every digit of 'nCopy' by taking the remainder of it to 10.
+        while (nCopy > 0) {
+            let res = nCopy % 10;
+            if (!(res in invertMap)) {
+                return false;
+            }
+            // Append the inverted digit of 'res' to the end of 'rotatedNumber'. 
+            rotatedNumber = rotatedNumber * 10 + invertMap[res];
+            nCopy = Math.floor(nCopy / 10);
+        }
+        // Check if 'rotatedNumber' equals 'n'.
+        return rotatedNumber !== n;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+- Time complexity: $O(L)$
+- Space complexity: $O(L)$ extra space used
+
+>  Where $L$ is the maximum number of digits $n$ can have.