diff --git a/dsa-problems/leetcode-problems/0600-0699.md b/dsa-problems/leetcode-problems/0600-0699.md
index fd7ca7675..6bf7bf55a 100644
--- a/dsa-problems/leetcode-problems/0600-0699.md
+++ b/dsa-problems/leetcode-problems/0600-0699.md
@@ -573,7 +573,7 @@ export const problems = [
         "problemName": "693. Binary Number with Alternating Bits",
         "difficulty": "Easy",
         "leetCodeLink": "https://leetcode.com/problems/binary-number-with-alternating-bits",
-        "solutionLink": "#"
+        "solutionLink": "/dsa-solutions/lc-solutions/0600-0699/binary-number-with-alternating-bits"
     },
     {
         "problemName": "694. Number of Distinct Islands",
diff --git a/dsa-solutions/lc-solutions/0600-0699/0693-binary-number-with-alternating-bits.md b/dsa-solutions/lc-solutions/0600-0699/0693-binary-number-with-alternating-bits.md
new file mode 100644
index 000000000..4da1f13ca
--- /dev/null
+++ b/dsa-solutions/lc-solutions/0600-0699/0693-binary-number-with-alternating-bits.md
@@ -0,0 +1,147 @@
+---
+id: binary-number-with-alternating-bits
+title: Binary Number with Alternating Bits
+sidebar_label: 0693 - Binary Number with Alternating Bits
+tags:
+  - Bit Manipulation
+  - String
+  - Bitmask
+description: "This is a solution to the Binary Number with Alternating Bits problem on LeetCode."
+---
+
+## Problem Description
+
+Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: n = 5
+Output: true
+Explanation: The binary representation of 5 is: 101
+```
+
+**Example 2:**
+
+```
+Input: n = 7
+Output: false
+Explanation: The binary representation of 7 is: 111.
+```
+
+### Constraints
+
+- $$1 \leq n \leq 2^{31} - 1$$
+
+## Solution for Binary Number with Alternating Bits
+
+### Approach 1: Convert to String
+#### Intuition and Algorithm
+
+Let's convert the given number into a string of binary digits. Then, we should simply check that no two adjacent digits are the same.
+
+## Code in Different Languages
+
+<Tabs>
+<TabItem value="java" label="Java">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```java
+class Solution {
+    public boolean hasAlternatingBits(int n) {
+        String bits = Integer.toBinaryString(n);
+        for (int i = 0; i < bits.length() - 1; i++) {
+            if (bits.charAt(i) == bits.charAt(i+1)) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="python" label="Python">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```python
+class Solution(object):
+    def hasAlternatingBits(self, n):
+        bits = bin(n)
+        return all(bits[i] != bits[i+1]
+                   for i in xrange(len(bits) - 1))
+```
+</TabItem>
+</Tabs>
+
+## Complexity Analysis
+
+### Time Complexity: $O(1)$
+
+> **Reason**:  For arbitrary inputs, we do $O(w)$ work, where w is the number of bits in `n`. However, $w \leq 32$.
+
+### Space Complexity: $O(1)$
+
+> **Reason**: or alternatively $O(w)$.
+
+### Approach 2: Divide By Two
+#### Algorithm
+
+We can get the last bit and the rest of the bits via `n % 2` and `n // 2` operations. Let's remember `cur`, the last bit of `n`. If the last bit ever equals the last bit of the remaining, then two adjacent bits have the same value, and the answer is `False`. Otherwise, the answer is `True`.
+
+Also note that instead of `n % 2` and `n // 2`, we could have used operators `n & 1` and `n >>= 1` instead.
+
+## Code in Different Languages
+
+<Tabs>
+<TabItem value="java" label="Java">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```java
+class Solution {
+    public boolean hasAlternatingBits(int n) {
+        int cur = n % 2;
+        n /= 2;
+        while (n > 0) {
+            if (cur == n % 2) return false;
+            cur = n % 2;
+            n /= 2;
+        }
+        return true;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="python" label="Python">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```python
+class Solution(object):
+    def hasAlternatingBits(self, n):
+        n, cur = divmod(n, 2)
+        while n:
+            if cur == n % 2: return False
+            n, cur = divmod(n, 2)
+        return True
+```
+</TabItem>
+</Tabs>
+
+## Complexity Analysis
+
+### Time Complexity: $O(1)$
+
+> **Reason**:  For arbitrary inputs, we do $O(w)$ work, where w is the number of bits in n. However, $w \leq 32$.
+
+### Space Complexity: $O(1)$
+
+> **Reason**: constant space is used.
+
+## References
+
+- **LeetCode Problem**: [Binary Number with Alternating Bits](https://leetcode.com/problems/binary-number-with-alternating-bits/description/)
+
+- **Solution Link**: [Binary Number with Alternating Bits](https://leetcode.com/problems/binary-number-with-alternating-bits/solutions/)