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Find K Smallest Numbers After Deleting Given Elements in C++
In this problem, we are given an array arr[] of size n, array del[] of size m, and an integer k. Our task is to find the k smallest numbers after deleting the given elements.
We need to print the first k smallest elements from the array arr[] found after deleting all elements present in the del[] array. If two instances are present in the array delete the first instance.
Let's take an example to understand the problem,
Input : arr[] = {3, 5, 1, 7, 9, 2}, del[] = {1, 9, 3}, k = 2 Output : 2, 5
Explanation −
Array arr[] after deleting the elements : {5, 7, 2} 2 minimum elements are 2, 5.
Solution Approach
A simple solution the problem is by deleting all elements from arr[] that are present in del[]. Then sort the array in ascending order and print first k elements of the array.
Example
Program to illustrate the working of our solution
#include <bits/stdc++.h> using namespace std; void findKminElementDelArray(int arr[], int n, int del[], int m, int k){ for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(arr[j] == del[i]){ arr[j] = INT_MAX; break; } } } sort(arr, arr + n); for (int i = 0; i < k; ++i) { cout<<arr[i]<<" "; } } int main(){ int array[] = { 3, 5, 1, 7, 9, 2 }; int m = sizeof(array) / sizeof(array[0]); int del[] = { 1, 9, 3 }; int n = sizeof(del) / sizeof(del[0]); int k = 2; cout<<k<<" smallest numbers after deleting the elements are "; findKminElementDelArray(array, m, del, n, k); return 0; }
Output
2 smallest numbers after deleting the elements are 2 5
Another approach
Another approach to solve the problem is using hashmap and heap. We will create a min-heap and a hash map. The hashmap will contain all the elements of the array del[]. And then insert elements of the array arr[] that are not present in hash-map to the min-heap. Pop k elements from the heap and then print it.
Example
Program to illustrate the working of our solution
#include <bits/stdc++.h> using namespace std; void findKminElementDelArray(int arr[], int n, int del[], int m, int k){ unordered_map<int, int> deleteElement; for (int i = 0; i < m; ++i) { deleteElement[del[i]]++; } priority_queue<int, vector<int>, greater<int> > minHeap; for (int i = 0; i < n; ++i) { if (deleteElement.find(arr[i]) != deleteElement.end()) { deleteElement[arr[i]]--; if (deleteElement[arr[i]] == 0) deleteElement.erase(arr[i]); } else minHeap.push(arr[i]); } for (int i = 0; i < k; ++i) { cout<<minHeap.top()<<" "; minHeap.pop(); } } int main(){ int array[] = { 3, 5, 1, 7, 9, 2 }; int m = sizeof(array) / sizeof(array[0]); int del[] = { 1, 9, 3 }; int n = sizeof(del) / sizeof(del[0]); int k = 2; cout<<k<<" smallest numbers after deleting the elements are "; findKminElementDelArray(array, m, del, n, k); return 0; }
Output
2 smallest numbers after deleting the elements are 2 5