Merge pull request #2010 from UtkarshBirla28/main

Added leetcode solution for problem number 162

Author: ajay-dhangar

dsa-problems/leetcode-problems/0100-0199.md

@@ -386,7 +386,7 @@ export const problems = [
         "problemName": "162. Find Peak Element",
         "difficulty": "Medium",
         "leetCodeLink": "https://leetcode.com/problems/find-peak-element/",
-        "solutionLink": "#"
+        "solutionLink": "/dsa-solutions/lc-solutions/0100-0199/Find-Peak-Element"
     },
     {
         "problemName": "163. Missing Ranges Premium",

dsa-solutions/lc-solutions/0100-0199/0162-Find-Peak-Element.md

@@ -0,0 +1,133 @@
+---
+id: Find-Peak-Element
+title: Find Peak Element
+sidebar_label: 0162-Find-Peak-Element
+tags:
+  - Array
+  - Binary Search
+description: "A peak element is an element that is strictly greater than its neighbors.
+
+Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.
+
+You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.
+
+You must write an algorithm that runs in O(log n) time."
+---
+
+
+### Problem Description
+
+A peak element is an element that is strictly greater than its neighbors.
+
+Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.
+
+You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.
+
+You must write an algorithm that runs in O(log n) time.
+
+### Examples
+
+#### Example 1
+
+- **Input:** `nums = [1,2,3,1]`
+- **Output:** `2`
+
+#### Example 2
+
+- **Input:** `nums = [1,2,1,3,5,6,4]`
+- **Output:** `5`
+
+### Constraints
+
+- `1 <= nums.length <= 1000`
+- `-231 <= nums[i] <= 231 - 1`
+- `nums[i] != nums[i + 1]` for all valid `i`.
+
+### Solution Code
+
+#### Python
+
+```python
+class Solution:
+    def findPeakElement(self, nums: list[int]) -> int:
+        left, right = 0, len(nums) - 1
+
+        while left < right:
+            mid = (left + right) >> 1
+
+            if nums[mid - 1] <= nums[mid] >= nums[mid + 1]:
+                return mid
+            elif nums[mid] < nums[mid + 1]:
+                left = mid + 1
+            else:
+                right = mid
+
+        return left
+```
+
+#### Java
+
+```java
+class Solution {
+    public int findPeakElement(int[] nums) {
+    int n=nums.length;
+    if(n==1)return 0;
+    if(nums[0]>nums[1])return 0;
+    if(nums[n-1]>nums[n-2])return n-1;
+    int low=1,high=n-2;
+    while(low<=high){
+        int mid=(low+high)/2;
+        if(nums[mid]>nums[mid-1] && nums[mid]>nums[mid+1])return mid;
+        else if(nums[mid]>nums[mid-1])low=mid+1;
+        else high=mid-1;
+    }
+    return -1;    
+    }
+}
+```
+
+#### C++
+
+```cpp
+
+class Solution {
+public:
+    int findPeakElement(vector<int>& nums) {
+        int low = 0, high = nums.size() - 1;
+        while (low < high) {
+            int mid = low + (high - low) / 2;
+            if (nums[mid] < nums[mid + 1]) {
+                low = mid + 1;
+            } else
+                high = mid;
+        }
+        return low;
+    }
+};
+```
+#### Javascript
+
+```javascript
+class Solution {
+    public int findPeakElement(int[] nums) {
+        int n = nums.length;
+        if(n == 1)
+            return 0;
+        else if(nums[0] > nums[1])
+            return 0;
+        else if(nums[n-1] > nums[n-2])
+            return n-1;
+        int low = 1, high = n-2;
+        while(low <= high) {
+            int mid = low + (high - low)/2;
+            if(nums[mid] > nums[mid-1] && nums[mid] > nums[mid+1])
+                return mid;
+            else if(nums[mid] < nums[mid-1])
+                high = mid-1;
+            else
+                low = mid+1;
+        }
+        return -1;
+    }
+}
+```