33. Search in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

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Explanation

Let’s say nums looks like this: [12, 13, 14, 15, 16, 17, 18, 19, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

Because it’s not fully sorted, we can’t do normal binary search. But here comes the trick:

If target is let’s say 14, then we adjust nums to this, where “inf” means infinity:

[12, 13, 14, 15, 16, 17, 18, 19, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf]

If target is let’s say 7, then we adjust nums to this:

[-inf, -inf, -inf, -inf, -inf, -inf, -inf, -inf, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

And then we can simply do ordinary binary search.

Of course we don’t actually adjust the whole array but >instead adjust only on the fly only the elements we look >at. And the adjustment is done by comparing both the >target and the actual element against nums[0].

int search(vector<int>& nums, int target) {
int lo = 0, hi = nums.size();
while (lo < hi) {
int mid = (lo + hi) / 2;

double num = (nums[mid] < nums[0]) == (target < nums[0])
? nums[mid]
: target < nums[0] ? -INFINITY : INFINITY;

if (num < target)
lo = mid + 1;
else if (num > target)
hi = mid;
else
return mid;
}
return -1;
}