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Day20_LeetCode162.java

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+class Solution {
+    public int findPeakElement(int[] nums) {
+        if(nums.length==1){
+            return 0;
+        }else if ( nums[0]>nums[1]){
+            return 0;
+        }else if ( nums[nums.length-1]>nums[nums.length-2]){
+            return nums.length-1;
+        }else {
+            int start = 1 ;
+            int end = nums.length-2;
+            while(start<=end){
+                 int mid = ( start + end)/2;
+                 if(nums[mid]>nums[mid-1] && nums[mid]>nums[mid+1]){
+                     return mid ;
+                 }else if ( nums[mid]<nums[mid+1]){
+                     start = mid+1 ;
+                 }else{
+                     end = mid-1 ;
+                 }
+             }
+           return -1 ;
+        }
+    }
+}

Day20_LeetCode162.md

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+**Day 20: LeetCode 162 [Find Peak Element]**
+
+**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.
+
+Example 1:
+Input: nums = [1,2,3,1]
+Output: 2
+Explanation: 3 is a peak element and your function should return the index number 2.
+
+**Link to LeetCode Problem:**
+https://leetcode.com/problems/find-peak-element/description/
+
+**My thought process:**
+To implement binary search, we need to initialize the start and end indices to the beginning and end of the array, respectively. We then enter a while loop that continues until start is greater than end. Inside the loop, we calculate the middle index mid and check if nums[mid] is a peak. If it is, we return mid. If it's not, we need to determine on which side of mid the peak lies. If nums[mid] is less than nums[mid+1], it means that the peak lies to the right of mid, so we set start to mid+1. If nums[mid] is less than nums[mid-1], it means that the peak lies to the left of mid, so we set end to mid-1.