Solution as discussed in video at neetcode channel

cpp/153-Find-Minimum-In-Rotated-Sorted-Array.cpp

@@ -1,36 +1,41 @@
 /*
- Given sorted array after some rotation, find min value
- Ex. nums = [3,4,5,1,2] -> 1, nums = [4,5,6,7,0,1,2] -> 0
+ * @lc app=leetcode id=153 lang=cpp
+ *
+ * [153] Find Minimum in Rotated Sorted Array
+ */
 
- Binary search + ensuring we never disqualify possible min value
+// @lc code=start
+class Solution
+{
+public:
+ int findMin(vector<int> &nums)
+ {
+ // Neetcode solution Ologn time O1 space binary search
+ int res = nums[0];
+ int l = 0;
+ int r = nums.size() - 1;
 
- Time: O(log n)
- Space: O(1)
-*/
+ while (l <= r)
+ {
+ if (nums[l] < nums[r])
+ {
+ res = min(res, nums[l]);
+ break;
+ }
+ int mid = l + (r - l) / 2;
+ res = min(res, nums[mid]);
 
-class Solution {
-public:
- int findMin(vector<int>& nums) {
- int low = 0;
- int high = nums.size() - 1;
-
- // not low <= high since not searching for target
- while (low < high) {
- int mid = low + (high - low) / 2;
- // ex. [3,4,5,6,7,8,9,1,2], mid = 4, high = 8
- // nums[mid] > nums[high], min must be right
- if (nums[mid] > nums[high]) {
- // never consider mid bc know for sure not min
- low = mid + 1;
- // ex. [8,9,1,2,3,4,5,6,7], mid = 4, high = 8
- // nums[mid] <= nums[right], min must be left
- } else {
- // consider mid still bc could be min
- high = mid;
+ if (nums[mid] >= nums[l]) // mid present in left sorted array
+ {
+ l = mid + 1; // try to move closer to right sorted array
+ }
+ else
+ {
+ r = mid - 1;
  }
  }
-
- // low lands on correct value, never disqualified mins
- return nums[low];
+
+ return res;
  }
 };
+// @lc code=end