Create 0153-find-minimum-in-rotated-sorted-array.md

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+---
+id: find-minimum-in-rotated-sorted-array
+title: Find Minimum in Rotated Sorted Array(LeetCode)
+sidebar_label: 0153-Find Minimum in Rotated Sorted Array
+tags:
+  - Array
+  - Binary Search
+description: Given the sorted rotated array nums of unique elements, return the minimum element of this array.
+---
+
+## Problem Statement
+
+Suppose an array of length `n` sorted in ascending order is rotated between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:
+
+`[4,5,6,7,0,1,2]` if it was rotated 4 times.
+`[0,1,2,4,5,6,7]` if it was rotated 7 times.
+Notice that rotating an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.
+
+Given the sorted rotated array `nums` of unique elements, return the minimum element of this array.
+
+You must write an algorithm that runs in `O(log n) time`.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input: nums = [3,4,5,1,2]
+Output: 1
+Explanation: The original array was [1,2,3,4,5] rotated 3 times.
+```
+
+**Example 2:**
+
+```plaintext
+Input: nums = [4,5,6,7,0,1,2]
+Output: 0
+Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
+```
+
+**Example 3:**
+
+```plaintext
+Input: nums = [11,13,15,17]
+Output: 11
+Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
+```
+
+### Constraints
+
+- `n == nums.length`
+- `1 <= n <= 5000`
+- `-5000 <= nums[i] <= 5000`
+- All the integers of `nums` are unique.
+- `nums` is sorted and rotated between `1` and `n` times.
+
+## Solution
+
+This problem aims to find the minimum element in a rotated sorted array. The provided solution uses a binary search approach to efficiently find the minimum element.
+
+### Approach
+
+#### Algorithm
+
+1. Initialize two pointers, `low` at the beginning of the array and `high` at the end of the array.
+2. Use a `while` loop with the condition `low < high`:
+* Calculate the middle index `mid` as `low + (high - low) / 2`.
+* Compare the element at index `mid` with the element at index `high`.
+* If `num[mid] < num[high]`, it means the minimum element is in the left part, so update `high = mid`.
+* If `num[mid] > num[high]`, it means the minimum element is in the right part, so update `low = mid + 1`.
+3. When the `while` loop ends, low will be pointing to the minimum element, so return `num[low]`.
+
+#### Implementation
+
+```C++
+class Solution {
+public:
+    int findMin(vector<int> &num) {
+        int low = 0, high = num.size() - 1;
+        // loop invariant: 1. low < high
+        //                 2. mid != high and thus A[mid] != A[high] (no duplicate exists)
+        //                 3. minimum is between [low, high]
+        // The proof that the loop will exit: after each iteration either the 'high' decreases
+        // or the 'low' increases, so the interval [low, high] will always shrink.
+        while (low < high) {
+            auto mid = low + (high - low) / 2;
+            if (num[mid] < num[high])
+                // the mininum is in the left part
+                high = mid;
+            else if (num[mid] > num[high])
+                // the mininum is in the right part
+                low = mid + 1;
+        }
+
+        return num[low];
+    }
+};
+```
+
+### Complexity Analysis
+
+- **Time complexity**: O(log n)
+- **Space complexity**: O(1)
+
+### Conclusion
+
+This algorithm effectively finds the minimum element in a rotated sorted array by using binary search. It maintains the loop invariant that the minimum is between the `low` and `high` indices, and ensures the interval `[low, high]` shrinks in each iteration. This solution is efficient and provides a clear way to handle such problems.