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 <= 10001 <= nums[i] <= 1000key is an integer from the array nums.1 <= k <= nums.length
Solutions#
Solution 1: Enumeration#
We enumerate the index $i$ in the range $[0, n)$, and for each index $i$, we enumerate the index $j$ in the range $[0, n)$. If $|i - j| \leq k$ and $nums[j] = key$, then $i$ is a K-nearest neighbor index. We add $i$ to the answer array, then break the inner loop and enumerate the next index $i$.
The time complexity is $O(n^2)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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| class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
ans = []
n = len(nums)
for i in range(n):
if any(abs(i - j) <= k and nums[j] == key for j in range(n)):
ans.append(i)
return ans
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| 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;
}
}
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| 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;
}
};
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| func findKDistantIndices(nums []int, key int, k int) (ans []int) {
for i := range nums {
for j, x := range nums {
if abs(i-j) <= k && x == key {
ans = append(ans, i)
break
}
}
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
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| function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (Math.abs(i - j) <= k && nums[j] === key) {
ans.push(i);
break;
}
}
}
return ans;
}
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Solution 2: Preprocessing + Binary Search#
We can preprocess to get the indices of all elements equal to $key$, recorded in the array $idx$. All index elements in the array $idx$ are sorted in ascending order.
Next, we enumerate the index $i$. For each index $i$, we can use binary search to find elements in the range $[i - k, i + k]$ in the array $idx$. If there are elements, then $i$ is a K-nearest neighbor index. We add $i$ to the answer array.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
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| class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
idx = [i for i, x in enumerate(nums) if x == key]
ans = []
for i in range(len(nums)):
l = bisect_left(idx, i - k)
r = bisect_right(idx, i + k) - 1
if l <= r:
ans.append(i)
return ans
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| class Solution {
public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
List<Integer> idx = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i] == key) {
idx.add(i);
}
}
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < nums.length; ++i) {
int l = Collections.binarySearch(idx, i - k);
int r = Collections.binarySearch(idx, i + k + 1);
l = l < 0 ? -l - 1 : l;
r = r < 0 ? -r - 2 : r - 1;
if (l <= r) {
ans.add(i);
}
}
return ans;
}
}
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| class Solution {
public:
vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
vector<int> idx;
int n = nums.size();
for (int i = 0; i < n; ++i) {
if (nums[i] == key) {
idx.push_back(i);
}
}
vector<int> ans;
for (int i = 0; i < n; ++i) {
auto it1 = lower_bound(idx.begin(), idx.end(), i - k);
auto it2 = upper_bound(idx.begin(), idx.end(), i + k) - 1;
if (it1 <= it2) {
ans.push_back(i);
}
}
return ans;
}
};
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| func findKDistantIndices(nums []int, key int, k int) (ans []int) {
idx := []int{}
for i, x := range nums {
if x == key {
idx = append(idx, i)
}
}
for i := range nums {
l := sort.SearchInts(idx, i-k)
r := sort.SearchInts(idx, i+k+1) - 1
if l <= r {
ans = append(ans, i)
}
}
return
}
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| function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
const idx: number[] = [];
for (let i = 0; i < n; i++) {
if (nums[i] === key) {
idx.push(i);
}
}
const search = (x: number): number => {
let [l, r] = [0, idx.length];
while (l < r) {
const mid = (l + r) >> 1;
if (idx[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
const l = search(i - k);
const r = search(i + k + 1) - 1;
if (l <= r) {
ans.push(i);
}
}
return ans;
}
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Solution 3: Two Pointers#
We enumerate the index $i$, and use a pointer $j$ to point to the smallest index that satisfies $j \geq i - k$ and $nums[j] = key$. If $j$ exists and $j \leq i + k$, then $i$ is a K-nearest neighbor index. We add $i$ to the answer array.
The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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| class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
ans = []
j, n = 0, len(nums)
for i in range(n):
while j < i - k or (j < n and nums[j] != key):
j += 1
if j < n and j <= (i + k):
ans.append(i)
return ans
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| 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, j = 0; i < n; ++i) {
while (j < i - k || (j < n && nums[j] != key)) {
++j;
}
if (j < n && j <= i + k) {
ans.add(i);
}
}
return ans;
}
}
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| class Solution {
public:
vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
int n = nums.size();
vector<int> ans;
for (int i = 0, j = 0; i < n; ++i) {
while (j < i - k || (j < n && nums[j] != key)) {
++j;
}
if (j < n && j <= i + k) {
ans.push_back(i);
}
}
return ans;
}
};
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| func findKDistantIndices(nums []int, key int, k int) (ans []int) {
n := len(nums)
for i, j := 0, 0; i < n; i++ {
for j < i-k || (j < n && nums[j] != key) {
j++
}
if j < n && j <= i+k {
ans = append(ans, i)
}
}
return
}
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| function findKDistantIndices(nums: number[], key: number, k: number): number[] {
const n = nums.length;
const ans: number[] = [];
for (let i = 0, j = 0; i < n; ++i) {
while (j < i - k || (j < n && nums[j] !== key)) {
++j;
}
if (j < n && j <= i + k) {
ans.push(i);
}
}
return ans;
}
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