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2200. Find All K-Distant Indices in an Array

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

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 $O(1)$.

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 $nums$.

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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Rust

• 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;
      }
  }
  
  
  // Solution 2
  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;
      }
  }
  
  
  
  // Solution 3
  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;
      }
  }
  
  

• 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;
      }
  };
  
  
  // Solution 2
  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;
      }
  };
  
  
  
  // Solution 3
  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;
      }
  };
  
  

• 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
  
  
  # Solution 2
  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
  
  
  
  # Solution 3
  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
  
  

• 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
  }
  
  
  // Solution 2
  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
  }
  
  
  
  // Solution 3
  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
  }
  
  

• 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;
  }
  
  
  // Solution 2
  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;
  }
  
  
  
  // Solution 3
  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;
  }
  
  

• impl Solution {
      pub fn find_k_distant_indices(nums: Vec<i32>, key: i32, k: i32) -> Vec<i32> {
          let n = nums.len();
          let mut ans = Vec::new();
          for i in 0..n {
              for j in 0..n {
                  if (i as i32 - j as i32).abs() <= k && nums[j] == key {
                      ans.push(i as i32);
                      break;
                  }
              }
          }
          ans
      }
  }
  
  
  // Solution 2
  impl Solution {
      pub fn find_k_distant_indices(nums: Vec<i32>, key: i32, k: i32) -> Vec<i32> {
          let n = nums.len();
          let mut idx = Vec::new();
          for i in 0..n {
              if nums[i] == key {
                  idx.push(i as i32);
              }
          }
  
          let search = |x: i32| -> usize {
              let (mut l, mut r) = (0, idx.len());
              while l < r {
                  let mid = (l + r) >> 1;
                  if idx[mid] >= x {
                      r = mid;
                  } else {
                      l = mid + 1;
                  }
              }
              l
          };
  
          let mut ans = Vec::new();
          for i in 0..n {
              let l = search(i as i32 - k);
              let r = search(i as i32 + k + 1) as i32 - 1;
              if l as i32 <= r {
                  ans.push(i as i32);
              }
          }
  
          ans
      }
  }
  
  
  
  // Solution 3
  impl Solution {
      pub fn find_k_distant_indices(nums: Vec<i32>, key: i32, k: i32) -> Vec<i32> {
          let n = nums.len();
          let mut ans = Vec::new();
          let mut j = 0;
          for i in 0..n {
              while j < i.saturating_sub(k as usize) || (j < n && nums[j] != key) {
                  j += 1;
              }
              if j < n && j <= i + k as usize {
                  ans.push(i as i32);
              }
          }
          ans
      }
  }
  
  

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