Completed Binary Search-2 Assignment

154_rotated_sorted_array_min.java

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+class Solution {
+    public int findMin(int[] nums) {
+        int low =0;
+        int n = nums.length;
+        int 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[mid]>=nums[low]){
+                low = mid + 1;
+            }else{
+                high = mid -1 ;
+            }
+        }
+
+        return -1;
+    }
+}
+
+/*There are few things to consider to solve this question - 
+(1) The minimum is always on the other side of the sorted half - thats an observation 
+(2) Whenever you see that its a sorted array - think of Binary search (also it is  O(logn))
+(3) One mistake that i made was while submitting it was to write the following statement - realised when dry run the attempt - if (nums[low]<=nums[high]) return nums[low];
+*/ 

162_find_peak_element.java

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+class Solution {
+    public int findPeakElement(int[] nums) {
+        // concept of slope was introduced here 
+        // where every there is slope - we use binary search 
+        // till now our thought process was binary search == sorted array but it changes a bit here 
+        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 (nums[mid]<nums[mid+1]) low = mid+1;
+            else high = mid - 1;
+        }
+        return -1;
+    }
+}

34_find_first_and_last_element.java

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+class Solution {
+    public int[] searchRange(int[] nums, int target) {
+        int first = binarySearchFirst(nums,0,nums.length-1,target);
+        if (first==-1) return new int[]{-1,-1};
+        int last = binarySearchLast(nums, first, nums.length - 1, target);
+        return new int[]{first, last};
+    }
+    private int binarySearchFirst(int[] nums,int low,int high, int target){
+        while(low<=high){
+            int mid = low + (high - low)/2;
+            if(nums[mid] == target){
+                if(mid == 0 || nums[mid-1] != target){
+                    return mid;
+                }else{
+                    high = mid - 1;
+                }}
+            else if (nums[mid]>target){
+                high = mid - 1;
+            }else low = mid+1; 
+        }
+        return -1;
+    }
+    private int binarySearchLast(int[] nums,int low,int high, int target){
+        while(low<=high){
+            int mid = low + (high - low)/2;
+            if(nums[mid] == target){
+                if(mid == nums.length-1 || nums[mid+1] != target) return mid;
+                else low = mid +1;
+            }
+            else if (nums[mid]>target){
+                high = mid - 1;
+            }else low = mid+1; 
+        }
+        return -1;
+    }
+}
+
+// We used two binary search indivually to find the values for the first and last 
+// also for values with just single value it should return the same postion twice - like if 5 appears at 0 -{0,0}