diff --git a/explanations/189/en.md b/explanations/189/en.md new file mode 100644 index 0000000..fbbd799 --- /dev/null +++ b/explanations/189/en.md @@ -0,0 +1,130 @@ +# 189. Rotate Array + +**Difficulty:** Medium +**Link:** https://leetcode.com/problems/rotate-array/ + +## Problem Description + +Given an integer array `nums`, rotate the array to the right by `k` steps, where `k` is non-negative. + +**Example 1:** +``` +Input: nums = [1,2,3,4,5,6,7], k = 3 +Output: [5,6,7,1,2,3,4] +Explanation: +rotate 1 steps to the right: [7,1,2,3,4,5,6] +rotate 2 steps to the right: [6,7,1,2,3,4,5] +rotate 3 steps to the right: [5,6,7,1,2,3,4] +``` + +**Example 2:** +``` +Input: nums = [-1,-100,3,99], k = 2 +Output: [3,99,-1,-100] +Explanation: +rotate 1 steps to the right: [99,-1,-100,3] +rotate 2 steps to the right: [3,99,-1,-100] +``` + +**Constraints:** +- `1 <= nums.length <= 10^5` +- `-2^31 <= nums[i] <= 2^31 - 1` +- `0 <= k <= 10^5` + +**Follow up:** +- Try to come up with as many solutions as you can. There are at least **three** different ways to solve this problem. +- Could you do it in-place with `O(1)` extra space? + +## Explanation + +### Strategy + +This is an **array rotation problem** that requires shifting elements to the right by `k` positions. The key insight is that rotating by `k` positions is equivalent to moving the last `k` elements to the front. + +**Key observations:** +- Rotating by `k` positions moves the last `k` elements to the front +- If `k` is larger than the array length, we can use modulo to reduce it +- We can solve this using a three-step reversal approach +- The reversal approach works in-place with O(1) extra space + +**High-level approach:** +1. **Normalize k**: Use `k % len(nums)` to handle cases where `k` is larger than array length +2. **Use three reversals**: + - Reverse the entire array + - Reverse the first `k` elements + - Reverse the remaining elements +3. **Result**: This gives us the rotated array + +### Steps + +Let's break down the solution step by step: + +**Step 1: Normalize the rotation amount** +- Calculate `k = k % len(nums)` to handle cases where `k` exceeds array length +- If `k` is 0, no rotation is needed + +**Step 2: Reverse the entire array** +- This moves the last `k` elements to the front (but in reverse order) + +**Step 3: Reverse the first k elements** +- This corrects the order of the elements that were moved to the front + +**Step 4: Reverse the remaining elements** +- This corrects the order of the elements that were shifted back + +**Example walkthrough:** +Let's trace through the first example: + +``` +nums = [1,2,3,4,5,6,7], k = 3 + +Step 1: Normalize k +k = 3 % 7 = 3 + +Step 2: Reverse entire array +[1,2,3,4,5,6,7] → [7,6,5,4,3,2,1] + +Step 3: Reverse first k elements (first 3) +[7,6,5,4,3,2,1] → [5,6,7,4,3,2,1] + +Step 4: Reverse remaining elements (last 4) +[5,6,7,4,3,2,1] → [5,6,7,1,2,3,4] + +Result: [5,6,7,1,2,3,4] +``` + +> **Note:** The three-reversal approach is elegant because it works in-place and has O(n) time complexity. The key insight is that reversing parts of the array in the right order gives us the desired rotation. + +### Solution + +```python +class Solution: + def rotate(self, nums: List[int], k: int) -> None: + """ + Do not return anything, modify nums in-place instead. + """ + n = len(nums) + k = k % n # Normalize k + + if k == 0: + return + + # Reverse the entire array + self.reverse(nums, 0, n - 1) + + # Reverse the first k elements + self.reverse(nums, 0, k - 1) + + # Reverse the remaining elements + self.reverse(nums, k, n - 1) + + def reverse(self, nums: List[int], start: int, end: int) -> None: + """Helper function to reverse a portion of the array""" + while start < end: + nums[start], nums[end] = nums[end], nums[start] + start += 1 + end -= 1 +``` + +**Time Complexity:** O(n) - we visit each element a constant number of times +**Space Complexity:** O(1) - we modify the array in-place without extra space \ No newline at end of file diff --git a/solutions/189/01.py b/solutions/189/01.py new file mode 100644 index 0000000..0d24e29 --- /dev/null +++ b/solutions/189/01.py @@ -0,0 +1,29 @@ +from typing import List + + +class Solution: + def rotate(self, nums: List[int], k: int) -> None: + """ + Do not return anything, modify nums in-place instead. + """ + n = len(nums) + k = k % n # Normalize k + + if k == 0: + return + + # Reverse the entire array + self.reverse(nums, 0, n - 1) + + # Reverse the first k elements + self.reverse(nums, 0, k - 1) + + # Reverse the remaining elements + self.reverse(nums, k, n - 1) + + def reverse(self, nums: List[int], start: int, end: int) -> None: + """Helper function to reverse a portion of the array""" + while start < end: + nums[start], nums[end] = nums[end], nums[start] + start += 1 + end -= 1

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