Given a list of integers, the task is to generate another list with N minimum elements.
Examples:
Input: {23, 1, 6, 2, 90}
Output: {1, 2}
Input: {34, 21, 56, 42, 89, 90, -1}
Output: {-1, 21}
Approach: The idea is to use the concept used in Find the smallest and second smallest elements in an array.
- First, find the smallest number in the given list.
- Then add that current minimum element to another list that will contain the N minimum elements.
- Remove the current minimum element from the given list.
- Continue with the same process until the N minimum elements are found.
Below is the implementation of the above approach:
C++
// C++ program to find N // minimum elements #include <bits/stdc++.h> using namespace std; void Nminelements(vector<int>list1, int N) { vector<int> final_list; int n = list1.size(); for (int i = 0; i < N; i++) { int min1 = 9999999; for (int j = 0; j < n; j++) { if (list1[j] < min1) min1 = list1[j]; } // remove minimum element // from list so that it do // not come in calculation again list1.erase(remove(list1.begin(), list1.end(), min1), list1.end()); final_list.push_back(min1); } for(int i = 0; i < final_list.size(); i++) cout << final_list[i] << " "; } // Driver code int main() { vector<int> list1 = {4, 78, 34, 10, 8, 21, 11, 231}; int N = 2; Nminelements(list1, N); } // This code is contributed by Subhadeep |
Python3
# Python program to find N minimum elements def Nminelements(list1, N): final_list =[]; for i in range(0, N): min1 = 9999999; for j in range(len(list1)): if list1[j]<min1: min1 = list1[j]; # remove minimum element from list so # that it do not come in calculation again list1.remove(min1); final_list.append(min1) print(final_list) # Driver code list1 = [4, 78, 34, 10, 8, 21, 11, 231]; N = 2; Nminelements(list1, N) |
4 8
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