Find Minimum in Rotated Sorted Array | LeetCode 153 | C++, Java, Python3

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153. Find Minimum in Rotated Sorted Array

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[vaibhavmalik]

As you are using,
while (l<r) {
}
But the last case when nums[mid] == nums [r] will never occur bcoz l==r won't exist in loop.

[guptashashwat]

Nice and logical and simple as always, best channel for leetcode solutions. pen down

[yulsic]

Why do we have to compare to array[right] but not array[left]?

[medhak]

How to come up with such solutions taking into consideration the different scenarios??

[chandreshsikarwar]

How can we do all this by comparing mid by left, instead of comparing it with the right. As this is not obvious to compare it with right alone, please suggest how to do it by comparing mid with left.
I gave a try, but it fails in some cases:

public int findMin(int[] nums, int left, int right) {
if(left == right) return nums[right];
int mid = left + (right - left) / 2;

if(nums[mid] > nums[left]){
if(nums[mid] < nums[right])
return findMin(nums, left, mid-1);
if(nums[mid] > nums[right])
return findMin(nums, mid+1, right);
}

if(nums[mid] < nums[left]) return findMin(nums, left, mid);
return findMin(nums, left+1, right);
}

Please help me do it using left, as I started off with left, and this can happen at the time of the interview. Thanks a lot for your videos.

[kanikagupta]

easy solution..
class Solution {
public:
int findMin(vector<int>& nums ) {

int l=0;
int h=nums.size() - 1;
while(l<h){
int mid= l + (h-l)/2;
if (nums[mid]<nums[h]){
h=mid;
}
else{
l=mid+1;
}
}
return nums[l];
}
};