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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 $O(1)$.

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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;
}

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;
}

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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