Rotated Binary Search

PeakIndex.java

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+//https://leetcode.com/problems/peak-index-in-a-mountain-array/
+
+class Solution {
+    public int peakIndexInMountainArray(int[] arr) {
+        int start = 0;
+        int end = arr.length - 1;
+
+        while(start < end) {
+            int mid = start + (end - start) / 2;
+            if(arr[mid] > arr[mid+1]) {
+                // you are in dec part of the array
+                // this may be the ans, but look at left
+                end = mid;
+            } else {
+                // arr[mid + 1] > arr[mid]
+                // you are in inc part of the array
+                start = mid + 1;
+                   }
+        } 
+
+        // In the end, start == end and pointing to the largest number because of the 2 checks 
+        // start and end are always trying to find max element in the above 2 checks
+        // hence, when they are pointing to just one element, that is the max element
+        // At every point of time of start and end, there may be a best possible ans at that particular time
+        return end;
+    }
+}

RotatedBinarySearch.java

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+public class RBS {
+  public static void main(String[] args) {
+
+
+}
+
+static int search(int[] arr, int target, int s, int e) {
+       if(s > e) {
+          return -1;
+        }    
+
+       int m = s + (e-s) / 2;
+       if(arr[m] == target) {
+              return m;
+          }
+
+       if(arr[s] <= arr[m]) {
+            if(target >= arr[s] && target <= arr[m]) {
+                   return search(arr, target, s, m-1);
+               } else {
+                   return search(arr, target, m+1, e);
+                     }
+         }
+
+       if(target >= arr[m] && target <= arr[e]) {
+             return search(arr, target, m+1, e);
+          } 
+
+        return search(arr, target, s, m-1);
+     }
+}