# C++ 4ms recursive solution using binary search

• the bsearch function is to find the index of minimum number;

if nums[mid] is smaller than both nums[mid - 1] and nums[mid + 1], then the index is mid

otherwise, recursively call bsearch(nums, l, mid -1) and bsearch(nums, mid +1 ,r)

if cannot find in one half, return -1, then the result is the larger one of two half

``````class Solution {
int bsearch(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(bsearch(nums, l, mid - 1), bsearch(nums, mid + 1, r));
}

return -1;
}
public:
int findMin(vector<int>& nums) {
int t = bsearch(nums, 0, nums.size() - 1);
return t == -1?nums[0] : nums[t];
}
};``````

• My nonrecursive version:

``````class Solution {
public:
int findMin(vector<int>& nums) {
if (nums[0] <= nums[nums.size() - 1]) {
return nums[0];
}
int i = 0, j = nums.size() - 1;
while (i + 1 < j) {
int k = (i + j) / 2;
if (nums[i] > nums[k]) {
j = k;
}
else {
i = k;
}
}
return nums[j];
}
};``````

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