Find k closest numbers in an unsorted array
Given an unsorted array and two numbers x and k, find k closest values to x.
Examples:
Input : arr[] = {10, 2, 14, 4, 7, 6}, x = 5, k = 3
Output : 4 6 7
Three closest values of x are 4, 6 and 7.
Input : arr[] = {-10, -50, 20, 17, 80}, x = 20, k = 2
Output : 17, 20
A simple solution is to sort the array. Then apply the method discussed to k closest values in a sorted array.
Time Complexity : O(n Log n)
A better solution is to use Heap Data Structure
1) Make a max heap of differences with first k elements.
2) For every element starting from (k+1)-th element, do following.
…..a) Find difference of current element with x.
…..b) If difference is more than root of heap, ignore current element.
…..c) Else insert the current element to the heap after removing the root.
3) Finally the heap has k closest elements.
C++
// C++ program to find k closest elements #include <bits/stdc++.h> using namespace std; void printKclosest(int arr[], int n, int x, int k) { // Make a max heap of difference with // first k elements. priority_queue<pair<int, int> > pq; for (int i = 0; i < k; i++) pq.push({ abs(arr[i] - x), i }); // Now process remaining elements. for (int i = k; i < n; i++) { int diff = abs(arr[i] - x); // If difference with current // element is more than root, // then ignore it. if (diff > pq.top().first) continue; // Else remove root and insert pq.pop(); pq.push({ diff, i }); } // Print contents of heap. while (pq.empty() == false) { cout << arr[pq.top().second] << " "; pq.pop(); } } // Driver program to test above functions int main() { int arr[] = { -10, -50, 20, 17, 80 }; int x = 20, k = 2; int n = sizeof(arr) / sizeof(arr[0]); printKclosest(arr, n, x, k); return 0; } |
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Java
//Java program to find k closest elements import java.util.Comparator; import java.util.PriorityQueue; class Pair { Integer key; Integer value; public Pair(Integer key, Integer value) { this.key = key; this.value = value; } public Integer getKey() { return key; } public void setKey(Integer key) { this.key = key; } public Integer getValue() { return value; } public void setValue(Integer value) { this.value = value; } } class GFG{ public static void printKclosest(int[] arr, int n, int x, int k) { // Make a max heap. PriorityQueue<Pair> pq = new PriorityQueue<>( new Comparator<Pair>() { public int compare(Pair p1, Pair p2) { return p2.getValue().compareTo( p1.getValue()); } }); // Build heap of difference with // first k elements for(int i = 0; i < k; i++) { pq.offer(new Pair(arr[i], Math.abs(arr[i] - x))); } // Now process remaining elements. for(int i = k; i < n; i++) { int diff = Math.abs(arr[i] - x); // If difference with current // element is more than root, // then ignore it. if(diff > pq.peek().getValue()) continue; // Else remove root and insert pq.poll(); pq.offer(new Pair(arr[i], diff)); } // Print contents of heap. while(!pq.isEmpty()) { System.out.print(pq.poll().getKey() + " "); } } // Driver code public static void main(String[] args) { int arr[] = { -10, -50, 20, 17, 80 }; int x = 20, k = 2; int n = arr.length; printKclosest(arr, n, x, k); } } // This code is contributed by Ashok Borra |
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Python3
# Python3 program to find k closest elements import math import sys from queue import PriorityQueue def printKclosest(arr,n,x,k): # Make a max heap of difference with # first k elements. pq = PriorityQueue() for i in range(k): pq.put((-abs(arr[i]-x),i)) # Now process remaining elements for i in range(k,n): diff = abs(arr[i]-x) p,pi = pq.get() curr = -p # If difference with current # element is more than root, # then put it back. if diff>curr: pq.put((-curr,pi)) continue else: # Else remove root and insert pq.put((-diff,i)) # Print contents of heap. while(not pq.empty()): p,q = pq.get() print("{} ".format(arr[q]),end = "") # Driver program to test above functions if __name__=='__main__': arr = [-10,-50,20,17,80] x,k = 20,2 n = len(arr) printKclosest(arr, n, x, k) # This code is contributed by Vikash Kumar 37 |
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Output:
17 20
Time Complexity : O(n Log k)
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