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Array

2113. Elements in Array After Removing and Replacing Elements 🔒

中文文档

Description

You are given a 0-indexed integer array nums. Initially on minute 0, the array is unchanged. Every minute, the leftmost element in nums is removed until no elements remain. Then, every minute, one element is appended to the end of nums, in the order they were removed in, until the original array is restored. This process repeats indefinitely.

  • For example, the array [0,1,2] would change as follows: [0,1,2] → [1,2] → [2] → [] → [0] → [0,1] → [0,1,2] → [1,2] → [2] → [] → [0] → [0,1] → [0,1,2] → ...

You are also given a 2D integer array queries of size n where queries[j] = [timej, indexj]. The answer to the jth query is:

  • nums[indexj] if indexj < nums.length at minute timej
  • -1 if indexj >= nums.length at minute timej

Return an integer array ans of size n where ans[j] is the answer to the jth query.

 

Example 1:

Input: nums = [0,1,2], queries = [[0,2],[2,0],[3,2],[5,0]]
Output: [2,2,-1,0]
Explanation:
Minute 0: [0,1,2] - All elements are in the nums.
Minute 1: [1,2]   - The leftmost element, 0, is removed.
Minute 2: [2]     - The leftmost element, 1, is removed.
Minute 3: []      - The leftmost element, 2, is removed.
Minute 4: [0]     - 0 is added to the end of nums.
Minute 5: [0,1]   - 1 is added to the end of nums.

At minute 0, nums[2] is 2.
At minute 2, nums[0] is 2.
At minute 3, nums[2] does not exist.
At minute 5, nums[0] is 0.

Example 2:

Input: nums = [2], queries = [[0,0],[1,0],[2,0],[3,0]]
Output: [2,-1,2,-1]
Minute 0: [2] - All elements are in the nums.
Minute 1: []  - The leftmost element, 2, is removed.
Minute 2: [2] - 2 is added to the end of nums.
Minute 3: []  - The leftmost element, 2, is removed.

At minute 0, nums[0] is 2.
At minute 1, nums[0] does not exist.
At minute 2, nums[0] is 2.
At minute 3, nums[0] does not exist.

 

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 100
  • n == queries.length
  • 1 <= n <= 105
  • queries[j].length == 2
  • 0 <= timej <= 105
  • 0 <= indexj < nums.length

Solutions

Solution 1

Python3

class Solution:
    def elementInNums(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        n, m = len(nums), len(queries)
        ans = [-1] * m
        for j, (t, i) in enumerate(queries):
            t %= 2 * n
            if t < n and i < n - t:
                ans[j] = nums[i + t]
            elif t > n and i < t - n:
                ans[j] = nums[i]
        return ans

Java

class Solution {
    public int[] elementInNums(int[] nums, int[][] queries) {
        int n = nums.length, m = queries.length;
        int[] ans = new int[m];
        for (int j = 0; j < m; ++j) {
            ans[j] = -1;
            int t = queries[j][0], i = queries[j][1];
            t %= (2 * n);
            if (t < n && i < n - t) {
                ans[j] = nums[i + t];
            } else if (t > n && i < t - n) {
                ans[j] = nums[i];
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> elementInNums(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size(), m = queries.size();
        vector<int> ans(m, -1);
        for (int j = 0; j < m; ++j) {
            int t = queries[j][0], i = queries[j][1];
            t %= (n * 2);
            if (t < n && i < n - t) {
                ans[j] = nums[i + t];
            } else if (t > n && i < t - n) {
                ans[j] = nums[i];
            }
        }
        return ans;
    }
};

Go

func elementInNums(nums []int, queries [][]int) []int {
	n, m := len(nums), len(queries)
	ans := make([]int, m)
	for j, q := range queries {
		t, i := q[0], q[1]
		t %= (n * 2)
		ans[j] = -1
		if t < n && i < n-t {
			ans[j] = nums[i+t]
		} else if t > n && i < t-n {
			ans[j] = nums[i]
		}
	}
	return ans
}