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Find All K-Distant Indices in an Array

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

Solution Explanations for Finding K-Distant Indices

This problem asks to find all indices in an array that are within a distance k from any index containing the value key. We'll explore three approaches with varying time complexities:

Approach 1: Brute Force Enumeration (O(n^2) Time Complexity)

This approach directly implements the problem's definition. For each index i, we check every other index j to see if |i - j| <= k and nums[j] == key. If such a j exists, i is added to the result.

Time Complexity: O(n^2) because of the nested loops iterating through all pairs of indices. Space Complexity: O(n) in the worst case to store the result list (all indices could be k-distant).

Code Examples:

  • Python:
def findKDistantIndices(nums, key, k):
    result = []
    n = len(nums)
    for i in range(n):
        found = False
        for j in range(n):
            if abs(i - j) <= k and nums[j] == key:
                found = True
                break
        if found:
            result.append(i)
    return result
  • Java:
import java.util.ArrayList;
import java.util.List;
 
class Solution {
    public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
        List<Integer> result = new ArrayList<>();
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            boolean found = false;
            for (int j = 0; j < n; j++) {
                if (Math.abs(i - j) <= k && nums[j] == key) {
                    found = true;
                    break;
                }
            }
            if (found) {
                result.add(i);
            }
        }
        return result;
    }
}
  • C++:
#include <vector>
#include <cmath>
 
std::vector<int> findKDistantIndices(const std::vector<int>& nums, int key, int k) {
    std::vector<int> result;
    int n = nums.size();
    for (int i = 0; i < n; ++i) {
        bool found = false;
        for (int j = 0; j < n; ++j) {
            if (std::abs(i - j) <= k && nums[j] == key) {
                found = true;
                break;
            }
        }
        if (found) {
            result.push_back(i);
        }
    }
    return result;
}

Other languages (JavaScript, Go, etc.) would follow a similar structure.

Approach 2: Preprocessing and Binary Search (O(n log n) Time Complexity)

This approach first finds all indices where nums[i] == key. Then, for each index i, it performs a binary search on this preprocessed list to check if there's an index within the k distance.

Time Complexity: O(n log n) due to the binary search operation repeated for each index. The preprocessing step is O(n). Space Complexity: O(n) to store the preprocessed list of indices.

Code Examples (Illustrative Python):

import bisect
 
def findKDistantIndices(nums, key, k):
    key_indices = [i for i, num in enumerate(nums) if num == key]
    result = []
    for i in range(len(nums)):
        left = bisect.bisect_left(key_indices, i - k)
        right = bisect.bisect_right(key_indices, i + k)
        if left < right:  # Found a key index within k distance
            result.append(i)
    return result

Approach 3: Two Pointers (O(n) Time Complexity)

This is the most efficient approach. We use two pointers: i iterates through all indices, and j tracks the nearest index containing key. We only need to move j forward when it's too far from i or doesn't point to key.

Time Complexity: O(n) because both pointers traverse the array at most once. Space Complexity: O(n) to store the result list.

Code Examples (Illustrative Python):

def findKDistantIndices(nums, key, k):
    result = []
    j = 0
    for i in range(len(nums)):
        while j < len(nums) and (j < i - k or nums[j] != key):
            j += 1
        if j < len(nums) and abs(i - j) <= k:
            result.append(i)
    return result

The Java, C++, and other language implementations would be very similar in structure to the Python example provided here. The core idea of using a single pass with two pointers remains the same. Remember to handle edge cases appropriately (e.g., when j goes beyond the array bounds).