658. Find K Closest Elements

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658. Find K Closest Elements

Description

Difficulty: Medium

Related Topics: Array, Two Pointers, Binary Search, Sliding Window, Sorting, Heap (Priority Queue)

Given a sorted integer array arr, two integers k and x, return the k closest integers to x in the array. The result should also be sorted in ascending order.

An integer a is closer to x than an integer b if:

  • |a - x| < |b - x|, or
  • |a - x| == |b - x| and a < b

Example 1:

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Input: arr = [1,2,3,4,5], k = 4, x = 3
Output: [1,2,3,4]

Example 2:

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Input: arr = [1,2,3,4,5], k = 4, x = -1
Output: [1,2,3,4]

Constraints:

  • 1 <= k <= arr.length
  • 1 <= arr.length <= 104
  • arr is sorted in ascending order.
  • -104 <= arr[i], x <= 104

Hints/Notes

  • 2023/11/16
  • since we need the smaller index, find the left boundary
  • Leetcode solution(checked)

Solution

Language: C++

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class Solution {
public:
    vector<int> findClosestElements(vector<int>& arr, int k, int x) {
        int index = findLeft(arr, x);
        int left = index - 1, right = index;
        list<int> res;
        while (k > 0) {
            if (left < 0) {
                res.push_back(arr[right++]);
            } else if (right == arr.size()) {
                res.push_front(arr[left--]);
            } else {
                int ldelta = abs(arr[left] - x);
                int rdelta = abs(arr[right] - x);
                if (ldelta <= rdelta) {
                    res.push_front(arr[left--]);
                } else {
                    res.push_back(arr[right++]);
                }
            }
            k--;
        }
        return vector<int>(res.begin(), res.end());
    }

    int findLeft(vector<int>& arr, int x) {
        int left = 0, right = arr.size();
        // left - 1 < x
        // right >= x
        while (left < right) {
            int mid = (right - left) / 2 + left;
            if (arr[mid] == x) {
                right = mid;
            } else if (arr[mid] > x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return right;
    }
};