Binary-search-2 problems solution

problem1Solution.java

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+//Find First and Last position of Element in Sorted Array
+//Time Complexity : O(log n)
+//Space Complexity : O(1)
+
+public class problem1Solution {
+    private int firstSearch(int low, int high, int[] nums, int target) {
+        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;
+    }
+
+    private int lastSearch(int low, int high, int[] nums, int target) {
+        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[] searchRange(int[] nums, int target) {
+        int n = nums.length;
+        int first = firstSearch(0, n-1, nums,target);
+        if(first == -1) return new int[]{-1,-1};
+        int last = lastSearch(first, n-1, nums, target);
+        return new int[]{first,last};
+    }
+}

problem2Solution.java

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+//Find Minimum in Rotated Sorted Array
+//Time Complexity : O(log n)
+//Space Complexity : O(1)
+
+public class problem2Solution {
+    public int findMin(int[] nums) {
+        int low = 0;
+        int high = nums.length-1;
+        int result = Integer.MAX_VALUE;
+        while(low<=high){
+            if(nums[low] <= nums[high]) return nums[low];
+            int mid =  low + (high - low)/2;
+            System.out.print("mid" + mid);
+            if(mid > 0 && nums[mid] < nums[mid-1] && nums[mid]<nums[mid+1]) return nums[mid];
+            if(nums[low] <= nums[mid]) {
+                low = mid +1;
+            }
+            else {
+                high = mid - 1;
+            }
+        }
+        return result;
+    }
+}

problem3Solution.java

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+//Find Peak Element
+//Time Complexity : O(log n)
+//Space Complexity : O(1)
+
+public class problem3Solution {
+    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;
+            if(nums[mid+1] > nums[mid]) low = mid + 1;
+            else high = mid - 1;
+        }
+        return -1;
+    }
+}