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LC - M - 658. Find K Closest Elements (multimap, vector, sort)

Solutions in C++

Solution 1

Complexity Analysis:

• $T=O(n\log n)$$T = O(n\log n)$, where $n$$n$ is the number of elements in arr

• $S=O(n)$$S = O(n)$, additional memory space used for multimap

Solution:

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// #include <map>
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class Solution {
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public:
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    vector<int> findClosestElements(vector<int>& arr, int k, int x) {
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        multimap<int, int> m;
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        vector<int> ret;
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        // Loop through the 'arr', insert each pair {|arr[i] - x|, arr[i]} into multimap
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        // T = O(nlogn), where n is the size of 'arr'
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        for (int i = 0; i < arr.size(); i++)
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        {
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            int diff = abs(arr[i] - x);
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            m.insert({diff, arr[i]});   // T = O(logn)
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        }
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        // T = O(k), where k <= n
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        for (auto it = m.begin(); k-- > 0; it++)
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            ret.push_back(it->second);
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        // T = O(nlogn)
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        sort(ret.begin(), ret.end());
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        return ret;
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    }
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};