# 246. Strobogrammatic Number

### Description

Given a string `num` which represents an integer, return `true` *if* `num` *is a **strobogrammatic number***.

A **strobogrammatic number** is a number that looks the same when rotated `180` degrees (looked at upside down).

### Constraints

* `1 <= num.length <= 50`
* `num` consists of only digits.
* `num` does not contain any leading zeros except for zero itself.

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/strobogrammatic-number/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** num = "69"

**Output:** true
{% endtab %}

{% tab title="Example 2" %}
**Input:** num = "88"

**Output:** true
{% endtab %}

{% tab title="Example 3" %}
**Input:** num = "962"

**Output:** false
{% endtab %}

{% tab title="Example 4" %}
**Input:** num = "1"

**Output:** true
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(N)
 * Space complexity : O(N)
 */

class Solution {

    public boolean isStrobogrammatic(String num) {
        
        // Note that using a String here and appending to it would be
        // poor programming practice.
        StringBuilder rotatedStringBuilder = new StringBuilder();
        
        // Remember that we want to loop backwards through the string
        for (int i = num.length() - 1; i >= 0; i--) {
            char c = num.charAt(i);
            if (c == '0' || c == '1' || c == '8') {
                rotatedStringBuilder.append(c);
            } else if (c == '6') {
                rotatedStringBuilder.append('9');
            } else if (c == '9') {
                rotatedStringBuilder.append('6');
            } else { // This must be an invalid digit.
                return false;
            }
        }
        
        String rotatedString = rotatedStringBuilder.toString();
        return num.equals(rotatedString);
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : O(N)
 * Space complexity : O(1)
 */
 
 class Solution {
    public boolean isStrobogrammatic(String num) {
        Map<Character, Character> dict = Map.of('0', '0',
                                                '1', '1',
                                                '6', '9',
                                                '8', '8',
                                                '9', '6');
        int left = 0, right = num.length()-1;
        
        while(left <= right) {
            if(!dict.containsKey(num.charAt(left)) || 
               !dict.get(num.charAt(left)).equals(num.charAt(right))) {
                return false;
            }
            left++;
            right--;
        }
        
        return true;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*