tall lynx
CODE
/*
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 <= 105
-231 <= nums[i] <= 231 - 1
0 <= k <= 105
*/
#include <iostream>
#include <vector>
class Solution {
public:
void rotate(std::vector<int> &nums, int k) {
size_t length_of_vector = nums.size();
std::vector<int> temp_array = {0};
int j = 0;
// first copy last k elements since they would be overwritten by (i-k)th
// element of the vector.
for (size_t i = length_of_vector; i > length_of_vector - k; i++) {
temp_array[j++] = nums[i];
}
// overwrite elements from k to end of the vector.
for (size_t i = k; i < length_of_vector; i++) {
nums[i] = nums[i - k];
}
// finally copy last k eleemtns from vector which is stored in temp_array
// back to nums.
for (size_t i = 0; i < k; i++) {
nums[i] = temp_array[i];
}
}
};
int main(int argc, char *argv[]) {
std::vector<int> nums = {7, 1, 2, 3, 4, 5, 6};
Solution s;
s.rotate(nums, 3);
for (auto it : nums) {
std::cout << it;
}
return 0;
}
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