##### 1. Search in Rotated Sorted Array

Let's formalize the problem: we need to find a number in array A[n]:

• A[0 .. i] is strictly increasing
• A[i+1 .. n-1] is strictly increasing
• A > A[n-1]

Utilising the above special properties of A[] we can use binary search as follow:

• Case 1: A[mid] == target => result is mid obviously, return.
• Case 2: mid ∈ [left,i] <=> A[mid] >= A
``````- if (A[mid] > target >= A[left]) right = mid - 1;
- else left = mid + 1;``````
• Case 3: mid ∈ (i,right] <=> A[mid] < A
``````- if (A[mid] < target <= A[right]) left = mid + 1;
- else right = mid - 1;``````

=> Runtime O(logn)

``````    int search(vector<int>& nums, int target) {
int n = nums.size();
if (n == 0) return -1;
int left = 0, right = n - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] >= nums) {
if (nums[mid] > target && target >= nums[left])
right = mid - 1;
else
left = mid + 1;
} else {
if (nums[mid] < target && target <= nums[right])
left = mid + 1;
else
right = mid - 1;
}
}
return -1;
}
``````
##### 2. Find Minimum in Rotated Sorted Array

Similar to above problem, instead of finding a target number, we find the rotated point.

``````int findMin(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
int rotatedPoint = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (mid + 1 < nums.size() && nums[mid] > nums[mid+1]) {
rotatedPoint = mid;
break;
}
if (nums[mid] >= nums[left])
left = mid + 1;
else
right = mid - 1;
}
return nums[rotatedPoint + 1];
}
``````
##### 3. Search in Rotated Sorted Array II

Let's formalize the problem: we need to find a number in array A[n]:

• A[0 .. i] is increasing
• A[i+1 .. n-1] is increasing
• A >= A[n-1]
• Case 1: a[mid] == target => result is mid obviously, return.
• Case 2: mid ∈ (j, i] <=> a[mid] > a
``````- if A[left] <= target < A[mid] right = mid - 1
- else left = mid + 1
``````
• Case 3: mid ∈ (i, k) <=> A[mid] < A[n-1]
``````- if (A[mid] < target <= A[right] left = mid + 1
- else right = mid - 1;
``````
• Case 4: A[left] == A[mid] or A[right] == A[mid] => we don't know which part mid belong to => we cannot reduce the search space by half as previous cases. We have to reduce the search space by 1: either from left or right end.

Suppose that I choose left side: first we need to check if A[left] == target, if yes => return left, otherwise left++.

``````    bool search(vector<int>& nums, int target) {
int left = 0, right = nums.size() - 1;
int n = right;
if (right < 0) return false;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return true;
if (nums[mid] > nums) {
if (target < nums[mid] && target >= nums[left])
right = mid - 1;
else
left = mid + 1;
} else if (nums[mid] < nums[n]) {
if (target > nums[mid] && target <= nums[right])
left = mid + 1;
else
right = mid - 1;
} else {
if (nums[left] == target) return true;
left++;
}
}
return false;
}
``````

Runtime: O(n) in worst case (all elements are equal, we reduce the search space by 1 each time)

##### 4. Find Minimum in Rotated Sorted Array II

Similar to problem 3, this time we find the rotated point

``````    int findMin(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
int n = nums.size();
int res = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (mid + 1 < nums.size() && nums[mid] > nums[mid+1]) {
res = mid;
break;
}

if (nums[mid] > nums) {
left = mid + 1;
} else if (nums[mid] < nums[n-1]) {
right = mid - 1;
} else {
if (left + 1 < nums.size() && nums[left] > nums[left+1]) {
res = left;
break;
}
left++;
}

}
return nums[res+1];
}
``````

Runtime: O(n) in worst case (all elements are equal, we reduce the search space by 1 each time)