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+---
+id: powerof-four
+title: Power of Four
+sidebar_label: 342-Power of Four
+tags:
+  - Math
+  - Bit Manipulation
+  - LeetCode
+  - Java
+  - Python
+  - C++
+description: "This is a solution to the Power of Four problem on LeetCode."
+sidebar_position: 6
+---
+
+## Problem Description
+
+Given an integer `n`, return `true` if it is a power of four. Otherwise, return `false`.
+
+An integer `n` is a power of four if there exists an integer `x` such that `n == 4^x`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: n = 16
+Output: true
+```
+
+**Example 2:**
+
+```
+Input: n = 5
+Output: false
+```
+
+**Example 3:**
+
+```
+Input: n = 1
+Output: true
+```
+
+### Constraints
+
+- `-2^31 <= n <= 2^31 - 1`
+
+
+
+---
+
+## Solution for Power of Four Problem
+
+### Approach 1: Brute Force (Loop/Recursion)
+
+The brute force approach involves repeatedly dividing the number by `4` and checking if it becomes `1`.
+
+#### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        if (n < 1) return false;
+        while (n % 4 == 0) {
+            n /= 4;
+        }
+        return n == 1;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public boolean isPowerOfFour(int n) {
+        if (n < 1) return false;
+        while (n % 4 == 0) {
+            n /= 4;
+        }
+        return n == 1;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def isPowerOfFour(self, n: int) -> bool:
+        if n < 1:
+            return False
+        while n % 4 == 0:
+            n //= 4
+        return n == 1
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- **Time Complexity**: $O(\log n)$, as we keep dividing `n` by `4`.
+- **Space Complexity**: $O(1)$, constant space usage.
+
+### Approach 2: Optimized (Bit Manipulation)
+
+The optimized approach checks if `n` is a power of four without loops or recursion. We can use bit manipulation:
+
+1. `n > 0`: Ensure `n` is positive.
+2. `n & (n - 1) == 0`: Ensure `n` is a power of two (only one bit set).
+3. `(n - 1) % 3 == 0`: Ensure `n` is a power of four (using properties of powers of four).
+
+#### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    bool isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && (n - 1) % 3 == 0;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public boolean isPowerOfFour(int n) {
+        return n > 0 && (n & (n - 1)) == 0 && (n - 1) % 3 == 0;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def isPowerOfFour(self, n: int) -> bool:
+        return n > 0 and (n & (n - 1)) == 0 and (n - 1) % 3 == 0
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- **Time Complexity**: $O(1)$, constant time operations.
+- **Space Complexity**: $O(1)$, constant space usage.
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['ImmidiSivani'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>
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