diff --git a/P1_Better.java b/P1_Better.java
new file mode 100644
index 00000000..c51917a5
--- /dev/null
+++ b/P1_Better.java
@@ -0,0 +1,61 @@
+class Solution {
+
+    private int lowerBound(int[] nums, int target) {
+        int low = 0, high = nums.length - 1;
+        int ans = nums.length; 
+
+        while(low <= high) {
+        int mid = low + (high - low) / 2;
+
+        if(nums[mid] >= target) {
+                ans = mid;  
+                high = mid - 1;  
+            }
+        else {
+                low = mid + 1; 
+            }
+        }
+        return ans;
+    }
+
+    private int upperBound(int[] nums, int target) {
+        int low = 0, high = nums.length - 1;
+        int ans = nums.length;
+
+        while(low <= high) {
+        int mid = low + (high - low) / 2;
+
+        if(nums[mid] > target) {
+                ans = mid;  
+                high = mid - 1;
+        }
+        else {
+                low = mid + 1;
+            }
+        }
+        return ans;
+    }
+
+    public int[] searchRange(int[] nums, int target) {
+   int firstOcc = lowerBound(nums, target);  
+
+        // Check if the target is present in the array or not
+        if(firstOcc == nums.length || nums[firstOcc] != target) return new int[]{-1, -1}; 
+
+        // Function call to find the last occurrence (upper bound)
+        int lastOcc = upperBound(nums, target) - 1;  
+        
+        return new int[]{firstOcc, lastOcc};  
+    }
+}
+
+// Time Complexity :    2*O(logn)
+// Space Complexity :   O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this :    None
+
+/* Your code here along with comments explaining your approach in three sentences only - 
+
+We use lower bound to find the first occurence of the element and the upper bound - 1 to get the last occurence of the element 
+
+*/
diff --git a/P1_Brute.java b/P1_Brute.java
new file mode 100644
index 00000000..afd83f85
--- /dev/null
+++ b/P1_Brute.java
@@ -0,0 +1,27 @@
+class Solution {
+    public int[] searchRange(int[] nums, int target) {
+    int first = -1, last = -1;
+    int[] ans = new int[2];
+    int n = nums.length;
+
+    for (int i = 0; i < n; i ++) {
+        if (nums[i] == target) {
+            if (first == -1) first = i;
+            last = i;
+        }
+    }
+
+    ans[0] = first;
+    ans[1] = last;
+    return ans;
+    }
+}
+
+// Time Complexity :    O(n)
+// Space Complexity :   O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this :    None
+
+/* Your code here along with comments explaining your approach in three sentences only - Will initise first and last variable to -1 and traverse the array to check if we have our target. 
+    Once the target is met we store its index in both first and last variable and keep searching and updating the last variable with indexes.
+*/
diff --git a/P1_Optimal.java b/P1_Optimal.java
new file mode 100644
index 00000000..4909eb84
--- /dev/null
+++ b/P1_Optimal.java
@@ -0,0 +1,82 @@
+class Solution {
+    // Function to find the first occurrence of the target
+    private int firstOccurrence(int[] nums, int target) {
+        int low = 0, high = nums.length - 1;
+        int first = -1;
+
+        // Applying Binary Search Algorithm
+        while(low <= high) {
+            int mid = low + (high - low) / 2;
+
+            /*  If the target element is found, we 
+                update the first variable to mid and
+                eliminate the right half to look for 
+                more smaller index where target is present */
+            if(nums[mid] == target) {
+                first = mid;
+                high = mid - 1;  
+            } 
+
+            /*  If middle element is smaller,
+                we eliminate the left half  */
+            else if(nums[mid] < target) {
+                low = mid + 1;  
+            } 
+            
+            /*  If middle element is greater,
+                we eliminate the right half  */
+            else {
+                high = mid - 1;  
+            }
+        }
+        return first;
+    }
+
+    // Function to find the last occurrence of the target
+    private int lastOccurrence(int[] nums, int target) {
+        int low = 0, high = nums.length - 1;
+        int last = -1;
+
+        // Applying Binary Search Algorithm
+        while(low <= high) {
+            int mid = low + (high - low) / 2;
+
+            /*  If the target element is found, we 
+                update the first variable to mid and
+                eliminate the right half to look for 
+                more greater index where target is present */
+            if(nums[mid] == target) {
+                last = mid;
+                low = mid + 1;
+            } 
+            
+            /*  If middle element is smaller,
+                we eliminate the left half  */
+            else if(nums[mid] < target) {
+                low = mid + 1;  
+            } 
+            
+            /*  If middle element is greater,
+                we eliminate the right half  */
+            else {
+                high = mid - 1; 
+            }
+        }
+        return last;
+    }
+
+    // Function to find the first and last occurrences of the target
+    public int[] searchRange(int[] nums, int target) {
+
+        // Function call to find the first occurence of target
+        int first = firstOccurrence(nums, target); 
+        
+        // If the target is not present in the array
+        if(first == -1) return new int[]{-1, -1};  
+
+        // Function call to find the last occurence of target
+        int last = lastOccurrence(nums, target);  
+
+        return new int[]{first, last};  
+    }
+}
\ No newline at end of file
diff --git a/P2.java b/P2.java
new file mode 100644
index 00000000..3160b522
--- /dev/null
+++ b/P2.java
@@ -0,0 +1,21 @@
+class Solution {
+    public int findMin(int[] nums) {
+        int n = nums.length;
+        int low = 0;
+        int high = n - 1;
+
+        while (low < high) { 
+            int mid = low + (high - low) / 2;
+
+            // If the middle element is greater than the last element, 
+            // the minimum is in the right part
+            if (nums[mid] > nums[high]) {
+                low = mid + 1;
+            } else { 
+                // Otherwise, the minimum is in the left part (including mid)
+                high = mid;
+            }
+        }
+        return nums[low]; // At the end of the loop, `low == high`, pointing to the minimum
+    }
+}
diff --git a/P3.java b/P3.java
new file mode 100644
index 00000000..345c2c2e
--- /dev/null
+++ b/P3.java
@@ -0,0 +1,20 @@
+class Solution {
+    public int findPeakElement(int[] nums) {
+        int n = nums.length;
+        int low = 0;
+        int high = n - 1;
+
+    while (low <= high) {
+        int mid = low + (high - low)/2;
+        if ((mid == 0 || nums[mid] > nums[mid - 1]) 
+        && (mid == n - 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 1;
+    }
+}