@@ -40,6 +40,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Search Insert Position](problems/src/binary_search/SearchInsertPosition.java) (Easy)
 - [Median of Two Sorted Arrays](problems/src/binary_search/MedianOfTwoSortedArrays.java) (Hard)
 - [Pow(x, n)](problems/src/binary_search/PowXN.java) (Medium)
+- [Find Peak Element](problems/src/binary_search/FindPeakElement.java) (Medium)
 
 #### [Bit Manipulation](problems/src/bit_manipulation)
 
 
@@ -0,0 +1,55 @@
+package binary_search;
+
+/**
+ * Created by gouthamvidyapradhan on 10/07/2017.
+ * A peak element is an element that is greater than its neighbors.
+
+ Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.
+
+ The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
+
+ You may imagine that num[-1] = num[n] = -∞.
+
+ For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.
+
+ Note:
+ Your solution should be in logarithmic complexity.
+
+ Solution: O(log N) check if the first or the last element is the peak element, if yes then return this index.
+ Else binary search for the answer - check mid element if this is a peak element return this index, else if the
+ left element is greater than current element search left else search right.
+ */
+public class FindPeakElement {
+    public static void main(String[] args) throws Exception{
+        int[] nums = {3, 4, 3, 2, 1};
+        System.out.println(new FindPeakElement().findPeakElement(nums));
+    }
+
+    public int findPeakElement(int[] nums) {
+        if(nums.length == 1)return 0;
+        if(nums[0] > nums[1])
+            return 0;
+        else if(nums[nums.length - 1] > nums[nums.length - 2])
+            return nums.length - 1;
+
+        int l = 0, h = nums.length - 1;
+        int ans = 0;
+        while(l <= h){
+            int m = l + (h - l) / 2;
+            if(m - 1 >= 0 && m + 1 < nums.length){
+                if(nums[m] > nums[m - 1] && nums[m] > nums[m + 1]){
+                    return m;
+                }
+            }
+            if(m - 1 >= 0 && nums[m - 1] > nums[m]){ //search left
+                h = m - 1;
+            }
+            else {
+                ans = l; //mark this as the answer and search right
+                l = m + 1;
+            }
+        }
+        return ans;
+    }
+
+}