diff --git a/find_first_and_last_position_of_element_in_sorted_array.java b/find_first_and_last_position_of_element_in_sorted_array.java
new file mode 100644
index 00000000..0ad30900
--- /dev/null
+++ b/find_first_and_last_position_of_element_in_sorted_array.java
@@ -0,0 +1,69 @@
+// Time Complexity : log(n) + log(m) ( log(m) becuase on the right part will wont be doing the binary search from 0 to N, but we can ignore log m becasue only higher order term matter)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : yes i bit, how we are reaching in first index and the last index is a bit compusing to me still but i'll know how we do it in terms of theory code is were i am stipping a bit
+
+/*
+ * Approach
+ * in order to find the first and the last index we are basically doing the binary serach 2 times, one is to find the starts index of the number
+ * and then second time to find the index where the element occurs for the last time.
+ * 
+ * we reduce the search space for the last element by using the first occurance index as low  
+ * 
+ */
+
+class Solution {
+    public int rightbinarysearch(int[] nums, int target, int low, int high) {
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            if (nums[mid] == target) {
+                if (mid == nums.length - 1 || nums[mid] < nums[mid + 1]) {
+                    return mid;
+                } else {
+                    low = mid + 1;
+                }
+            } else if (nums[mid] > target) {
+                high = mid - 1;
+            } else {
+                low = mid + 1;
+            }
+        }
+        return -1;
+    }
+
+    public int leftbinarysearch(int[] nums, int target, int low, int high) {
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            if (nums[mid] == target) {
+                if (mid == 0 || nums[mid] > nums[mid - 1]) {
+                    return mid;
+                } else {
+                    high = mid - 1;
+                }
+            } else if (nums[mid] > target) {
+                high = mid - 1;
+            } else {
+                low = mid + 1;
+            }
+        }
+        return -1;
+    }
+
+    public int[] searchRange(int[] nums, int target) {
+        int low = 0;
+        int high = nums.length - 1;
+        if (nums == null || nums.length == 0) {
+            return new int[] { -1, -1 };
+        }
+        if (target > nums[high]) {
+            return new int[] { -1, -1 };
+        }
+        if (target < nums[0]) {
+            return new int[] { -1, -1 };
+        }
+        int inital = leftbinarysearch(nums, target, low, high);
+        int last = rightbinarysearch(nums, target, inital, high);
+        return new int[] { inital, last };
+
+    }
+}
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diff --git a/find_minimum_in_rotated_sorted_array.java b/find_minimum_in_rotated_sorted_array.java
new file mode 100644
index 00000000..847af674
--- /dev/null
+++ b/find_minimum_in_rotated_sorted_array.java
@@ -0,0 +1,33 @@
+// Time Complexity : log(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+/*
+ * Approach
+ * if the low value is smaller then high, that means our element is at low but if that id not the case
+ * we check is mids might me the high lowest by comparing mid -1 and mid + 1 to the mids if that also not the case then
+ * we need to eleminate one half of the side how we do that is before we 
+ * we check if low it smaller then the mid if that is true that side is sorted and we eleminate that
+ * 
+ */
+
+class Solution {
+    public int findMin(int[] nums) {
+        int low = 0;
+        int high = nums.length - 1;
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            if (nums[low] <= nums[high]) {
+                return nums[low];
+            } else if ((mid > 0 && nums[mid - 1] > nums[mid]) && (mid < nums.length - 1 && nums[mid] < nums[mid + 1])) {
+                return nums[mid];
+            } else if (nums[low] <= nums[mid]) {
+                low = mid + 1;
+            } else {
+                high = mid - 1;
+            }
+        }
+        return -1;
+    }
+}
\ No newline at end of file
diff --git a/find_peak_element.java b/find_peak_element.java
new file mode 100644
index 00000000..18c12720
--- /dev/null
+++ b/find_peak_element.java
@@ -0,0 +1,29 @@
+// Time Complexity : log(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+/*
+ * Approach
+ * idea here is to find the number that a peak meaning the number before and after that number are smaller then that number
+ * even though it is a sorted array we can apply binary search becasue of the conditiion that nums[i] != nums[i+1]
+ * we move the search space where we find the greater element when comparing elements
+ */
+
+class Solution {
+    public int findPeakElement(int[] nums) {
+        int low = 0;
+        int high = nums.length - 1;
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            if ((mid == 0 || nums[mid] > nums[mid - 1]) && (mid == nums.length - 1 || nums[mid] > nums[mid + 1])) {
+                return mid;
+            } else if (mid > 0 && nums[mid - 1] > nums[mid]) {
+                high = mid - 1;
+            } else {
+                low = mid + 1;
+            }
+        }
+        return 999;
+    }
+}
\ No newline at end of file