diff --git a/ FindPeakElement.java b/ FindPeakElement.java
new file mode 100644
index 00000000..4f7045c6
--- /dev/null
+++ b/ FindPeakElement.java	
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+/**
+ * LeetCode 162: Find Peak Element
+ * https://leetcode.com/problems/find-peak-element/
+ *
+ * This class provides a solution to the problem of finding a peak element in an array.
+ * A peak element is an element that is strictly greater than its neighbors.
+ * The algorithm uses binary search to achieve O(log n) time complexity.
+ */
+class Solution {
+    public int findPeakElement(int[] nums) {
+
+        int start = 0;
+        int end = nums.length - 1;
+        while (start <= end) {
+            int mid = start + (end - start) / 2;
+            if( (mid ==0 || nums[mid]> nums[mid -1]) && (mid == nums.length-1 || nums[mid]> nums[mid+1]))
+                return  mid;
+            else if(nums[mid+1] > nums[mid])
+                start = mid+1;
+            else
+                end = mid -1;
+
+        }
+
+        return -1;
+    }
+}
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diff --git a/FindMinimumInRotatedSortedArray.java b/FindMinimumInRotatedSortedArray.java
new file mode 100644
index 00000000..fd78427d
--- /dev/null
+++ b/FindMinimumInRotatedSortedArray.java
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+/**
+ * Leet code : https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/
+ * We wil use binary search to find the min in the rotated sorted array.
+ * Time Complexity = O(log n)
+ */
+
+class FindMinimumInRotatedSortedArray {
+
+    public int findMin(int[] nums) {
+        //check if array is not rotated
+        if (nums[0] <= nums[nums.length - 1]) return nums[0];
+
+        int start = 0;
+        int end = nums.length - 1;
+        int currentMin = nums[end];
+
+        while (start <= end) {
+            int mid = start + (end - start) / 2;
+            if (nums[mid] <= currentMin) {
+                currentMin = nums[mid];
+                end = mid - 1;
+            } else
+                start = mid + 1;
+        }
+        return currentMin;
+    }
+
+    public int findMin2(int[] nums) {
+        int n = nums.length;
+        int low = 0, high = n - 1;
+
+        while (low <= high) {
+            if (nums[low] <= nums[high]) {
+                return nums[low];
+            }
+
+            int mid = low + (high - low) / 2;
+            if ((mid == 0 || nums[mid] < nums[mid - 1]) && (mid == n - 1 || nums[mid] < nums[mid + 1])) {
+                return nums[mid];
+            } else if (nums[low] <= nums[mid]) { // left sorted range
+                low = mid + 1;
+            } else {
+                high = mid - 1;
+            }
+        }
+
+        return -1;
+    }
+}
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diff --git a/FirstLastPositionOfElementInSortedArray.java b/FirstLastPositionOfElementInSortedArray.java
new file mode 100644
index 00000000..abc13403
--- /dev/null
+++ b/FirstLastPositionOfElementInSortedArray.java
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+/**
+ * Leet Code: https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/description/
+ * 1. We use binary search to find the fist and last index.
+ * 2. Use one binary search to find first occurrence and one more binary search to find last occurrence.
+ * 3. Time Complexity : 0(log n)
+ */
+class FirstLastPositionOfElementInSortedArray {
+    public int[] searchRange(int[] nums, int target) {
+        int[] result = {-1, -1};
+        result[0] = binarySearch(nums, target, true);
+        result[1] = binarySearch(nums, target, false);
+
+        return result;
+    }
+
+    private int binarySearch(int[] nums, int target, boolean isSearchingLeftBound) {
+        int start = 0;
+        int end = nums.length - 1;
+        int idx = -1;
+        while (start <= end) {
+            int mid = start + (end - start) / 2;
+            if (nums[mid] == target) {
+                idx = mid;
+                if (isSearchingLeftBound) {
+                    if (mid == 0 || nums[mid - 1] != target) {
+                        return mid;
+                    } else {
+                        end = mid - 1;
+                    }
+                } else {
+                    if (mid == nums.length - 1 || nums[mid + 1] != target) {
+                        return mid;
+                    } else {
+                        start = mid + 1;
+                    }
+                }
+            } else if (nums[mid] > target)
+                end = mid - 1;
+            else
+                start = mid + 1;
+        }
+        return idx;
+    }
+}
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