README_EN.md

2200. Find All K-Distant Indices in an Array

中文文档

Description

You are given a 0-indexed integer array nums and two integers key and k. A k-distant index is an index i of nums for which there exists at least one index j such that |i - j| <= k and nums[j] == key.

Return a list of all k-distant indices sorted in increasing order.

 

Example 1:

Input: nums = [3,4,9,1,3,9,5], key = 9, k = 1
Output: [1,2,3,4,5,6]
Explanation: Here, nums[2] == key and nums[5] == key.
- For index 0, |0 - 2| > k and |0 - 5| > k, so there is no j where |0 - j| <= k and nums[j] == key. Thus, 0 is not a k-distant index.
- For index 1, |1 - 2| <= k and nums[2] == key, so 1 is a k-distant index.
- For index 2, |2 - 2| <= k and nums[2] == key, so 2 is a k-distant index.
- For index 3, |3 - 2| <= k and nums[2] == key, so 3 is a k-distant index.
- For index 4, |4 - 5| <= k and nums[5] == key, so 4 is a k-distant index.
- For index 5, |5 - 5| <= k and nums[5] == key, so 5 is a k-distant index.
- For index 6, |6 - 5| <= k and nums[5] == key, so 6 is a k-distant index.
Thus, we return [1,2,3,4,5,6] which is sorted in increasing order. 

Example 2:

Input: nums = [2,2,2,2,2], key = 2, k = 2
Output: [0,1,2,3,4]
Explanation: For all indices i in nums, there exists some index j such that |i - j| <= k and nums[j] == key, so every index is a k-distant index. 
Hence, we return [0,1,2,3,4].

 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000
• key is an integer from the array nums.
• 1 <= k <= nums.length

Solutions

Python3

class Solution:
    def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
        ans = []
        n = len(nums)
        for i in range(n):
            for j in range(n):
                if abs(i - j) <= k and nums[j] == key:
                    ans.append(i)
                    break
        return ans

Java

class Solution {
    public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
        int n = nums.length;
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (Math.abs(i - j) <= k && nums[j] == key) {
                    ans.add(i);
                    break;
                }
            }
        }
        return ans;
    }
}

TypeScript

function findKDistantIndices(nums: number[], key: number, k: number): number[] {
    const n = nums.length;
    let ans = [];
    for (let j = 0; j < n; j++) {
        if (nums[j] == key) {
            for (let i = j - k; i <= j + k; i++) {
                if (i >= 0 && i < n && !ans.includes(i)) {
                    ans.push(i);
                }
            }
        }
    }
    return ans;
}

C++

class Solution {
public:
    vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
        int n = nums.size();
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (abs(i - j) <= k && nums[j] == key) {
                    ans.push_back(i);
                    break;
                }
            }
        }
        return ans;
    }
};

Go

func findKDistantIndices(nums []int, key int, k int) []int {
	n := len(nums)
	var ans []int
	for i := 0; i < n; i++ {
		for j, v := range nums {
			if abs(i-j) <= k && v == key {
				ans = append(ans, i)
				break
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

...