# C++ 4ms solution, based on findMin and binarySearch with explaination

• Firstly, the leetcode problem "Find Minimum in Rotated Sorted Array" solution here

When we found the min index, we call binarySearch(0, index-1) and binarySearch(index, nums.size() -1), and return the bigger one. here is the code

``````class Solution {
int findMin(vector<int>& nums, int l, int r){
if(nums.size() == 1)
return 0;
if (l < 0 || r < 0 || l >= nums.size() || r >= nums.size())
return -1;
int mid = (l + r) / 2;
if (l <= r){
if (mid == 0)
return nums[mid] > nums[mid + 1] ? mid + 1 : -1;
if (mid == nums.size() - 1)
return nums[mid] < nums[mid - 1] ? mid : -1;

if (nums[mid] <= nums[mid - 1] && nums[mid] <= nums[mid + 1])
return mid;
else
return max(findMin(nums, l, mid - 1), findMin(nums, mid + 1, r));
}

return -1;
}

int binarySearch(vector<int>& nums, int l, int r, int& target){
int m = (l + r)/2;
while(l <= r){
if(nums[m] == target)
return m;
else if(nums[m] > target){
r = m - 1;
m = (l+r)/2;
}else{
l = m + 1;
m = (l+r)/2;
}
}
return -1;
}

public:
int search(vector<int>& nums, int target) {
int index = findMin(nums, 0, nums.size() - 1);
if(index == -1)
return binarySearch(nums, 0, nums.size() - 1, target);
return max(binarySearch(nums, 0, index - 1, target), binarySearch(nums, index, nums.size() - 1, target));
}
};
``````

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