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In regular binary search we always go to the left sub array if pivot is greater than search element otherwise to the right sub array.
For rotated binary search, the idea is to check in which sub-array we need to continue the binary search based on certain boundary value checks. The below binary search works for both regular
and rotated arrays.
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 4);
- sunnyday July 12, 2015java.lang.Integer res48 = -2147483648
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 5);
java.lang.Integer res49 = 6
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 3);
java.lang.Integer res50 = 5
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 7);
java.lang.Integer res51 = 0
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 9);
java.lang.Integer res52 = 1
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 77);
java.lang.Integer res53 = -2147483648
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 4);
java.lang.Integer res54 = -2147483648
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 2);
java.lang.Integer res55 = -2147483648
java> rotatedBinarySearch(new int [] {7, 9, 10, 11, 1, 3, 5}, 0, 6, 2);
java.lang.Integer res56 = -2147483648