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| 1 | +package LeetCodeJava.Array; |
| 2 | + |
| 3 | +// https://leetcode.com/problems/rotate-array/description/ |
| 4 | + |
| 5 | +import java.util.ArrayList; |
| 6 | +import java.util.Deque; |
| 7 | +import java.util.LinkedList; |
| 8 | +import java.util.List; |
| 9 | + |
| 10 | +/** |
| 11 | + * 189. Rotate Array |
| 12 | + * Solved |
| 13 | + * Medium |
| 14 | + * Topics |
| 15 | + * Companies |
| 16 | + * Hint |
| 17 | + * Given an integer array nums, rotate the array to the right by k steps, where k is non-negative. |
| 18 | + * |
| 19 | + * |
| 20 | + * |
| 21 | + * Example 1: |
| 22 | + * |
| 23 | + * Input: nums = [1,2,3,4,5,6,7], k = 3 |
| 24 | + * Output: [5,6,7,1,2,3,4] |
| 25 | + * Explanation: |
| 26 | + * rotate 1 steps to the right: [7,1,2,3,4,5,6] |
| 27 | + * rotate 2 steps to the right: [6,7,1,2,3,4,5] |
| 28 | + * rotate 3 steps to the right: [5,6,7,1,2,3,4] |
| 29 | + * Example 2: |
| 30 | + * |
| 31 | + * Input: nums = [-1,-100,3,99], k = 2 |
| 32 | + * Output: [3,99,-1,-100] |
| 33 | + * Explanation: |
| 34 | + * rotate 1 steps to the right: [99,-1,-100,3] |
| 35 | + * rotate 2 steps to the right: [3,99,-1,-100] |
| 36 | + * |
| 37 | + * |
| 38 | + * Constraints: |
| 39 | + * |
| 40 | + * 1 <= nums.length <= 105 |
| 41 | + * -231 <= nums[i] <= 231 - 1 |
| 42 | + * 0 <= k <= 105 |
| 43 | + * |
| 44 | + * |
| 45 | + * Follow up: |
| 46 | + * |
| 47 | + * Try to come up with as many solutions as you can. There are at least three different ways to solve this problem. |
| 48 | + * Could you do it in-place with O(1) extra space? |
| 49 | + * |
| 50 | + * |
| 51 | + */ |
| 52 | +public class RotateArray { |
| 53 | + |
| 54 | + // V0 |
| 55 | + // IDEA: DEQEUE |
| 56 | + public void rotate(int[] nums, int k) { |
| 57 | + // edge |
| 58 | + if (nums == null || nums.length == 0) { |
| 59 | + return; |
| 60 | + } |
| 61 | + |
| 62 | + Deque<Integer> dq = new LinkedList<>(); |
| 63 | + for (int n : nums) { |
| 64 | + dq.add(n); |
| 65 | + } |
| 66 | + |
| 67 | + int i = 0; |
| 68 | + for (i = 0; i < k; i++) { |
| 69 | + /** |
| 70 | + * NOTE !!! below |
| 71 | + * |
| 72 | + * we get dequeue last element, |
| 73 | + * then append it to first idx |
| 74 | + */ |
| 75 | + dq.addFirst(dq.pollLast()); |
| 76 | + } |
| 77 | + |
| 78 | + // System.out.println(">>> dq = " + dq); |
| 79 | + |
| 80 | + int j = 0; |
| 81 | + while (!dq.isEmpty()) { |
| 82 | + nums[j] = dq.pollFirst(); |
| 83 | + j += 1; |
| 84 | + } |
| 85 | + } |
| 86 | + |
| 87 | + // V0-1 |
| 88 | + // IDEA: ARRAY OP (TLE) |
| 89 | + public void rotate_0_1(int[] nums, int k) { |
| 90 | + // edge |
| 91 | + if (nums == null || nums.length == 0) { |
| 92 | + return; |
| 93 | + } |
| 94 | + |
| 95 | + List<Integer> list = new ArrayList<>(); |
| 96 | + for (int n : nums) { |
| 97 | + list.add(n); |
| 98 | + } |
| 99 | + |
| 100 | + int i = 0; |
| 101 | + for (i = 0; i < k; i++) { |
| 102 | + // int last = nums[nums.length - 1]; |
| 103 | + int last = list.get(list.size() - 1); |
| 104 | + list.remove(list.size() - 1); |
| 105 | + list.add(0, last); |
| 106 | + } |
| 107 | + |
| 108 | + for (int j = 0; j < list.size(); j++) { |
| 109 | + nums[j] = list.get(j); |
| 110 | + } |
| 111 | + } |
| 112 | + |
| 113 | + // V1 |
| 114 | + // IDEA: rotate ( `%` op) (gpt) |
| 115 | + public void rotate_1(int[] nums, int k) { |
| 116 | + // Edge case |
| 117 | + if (nums == null || nums.length == 0) { |
| 118 | + return; |
| 119 | + } |
| 120 | + |
| 121 | + // Adjust k in case it's greater than the array length |
| 122 | + k = k % nums.length; |
| 123 | + if (k == 0) { |
| 124 | + return; |
| 125 | + } |
| 126 | + |
| 127 | + // Reverse the entire array |
| 128 | + reverse(nums, 0, nums.length - 1); |
| 129 | + |
| 130 | + // Reverse the first k elements |
| 131 | + reverse(nums, 0, k - 1); |
| 132 | + |
| 133 | + // Reverse the remaining elements |
| 134 | + reverse(nums, k, nums.length - 1); |
| 135 | + } |
| 136 | + |
| 137 | + private void reverse(int[] nums, int start, int end) { |
| 138 | + while (start < end) { |
| 139 | + int temp = nums[start]; |
| 140 | + nums[start] = nums[end]; |
| 141 | + nums[end] = temp; |
| 142 | + start++; |
| 143 | + end--; |
| 144 | + } |
| 145 | + } |
| 146 | + |
| 147 | + // V2 |
| 148 | + |
| 149 | +} |
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