Print n smallest elements from given array in their original order

We are given an array of m-elements, we need to find n smallest elements from the array but they must be in the same order as they are in given array.

Examples:

Input : arr[] = {4, 2, 6, 1, 5},
n = 3
Output : 4 2 1
Explanation :
1, 2 and 4 are 3 smallest numbers and
4 2 1 is their order in given array.

Input : arr[] = {4, 12, 16, 21, 25},
n = 3
Output : 4 12 16
Explanation :
4, 12 and 16 are 3 smallest numbers and
4 12 16 is their order in given array.

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Make a copy of original array and then sort copy array. After sorting the copy array, save all n smallest numbers. Further for each element in original array, check whether it is in n-smallest number or not if it present in n-smallest array then print it otherwise move forward.

• Make copy_arr[]
• sort(copy_arr)
• For all elements in arr[] -
• Find arr[i] in n-smallest element of copy_arr
• If found then print the element
• Below is the implementation of above approach :

C++

 // CPP for printing smallest n number in order #include #include using namespace std;    // Function to print smallest n numbers void printSmall(int arr[], int asize, int n) {     // Make copy of array     vector copy_arr(arr, arr + asize);        // Sort copy array     sort(copy_arr.begin(), copy_arr.begin() + asize);        // For each arr[i] find whether     // it is a part of n-smallest     // with binary search     for (int i = 0; i < asize; ++i)         if (binary_search(copy_arr.begin(),                  copy_arr.begin() + n, arr[i]))             cout << arr[i] << " "; }    // Driver program int main() {     int arr[] = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };     int asize = sizeof(arr) / sizeof(arr);         int n = 5;     printSmall(arr, asize, n);     return 0; }

Java

 // Java for printing smallest n number in order import java.util.*;    class GFG  {       // Function to print smallest n numbers static void printSmall(int arr[], int asize, int n) {     // Make copy of array     int []copy_arr = Arrays.copyOf(arr,asize);        // Sort copy array     Arrays.sort(copy_arr);        // For each arr[i] find whether     // it is a part of n-smallest     // with binary search     for (int i = 0; i < asize; ++i)     {         if (Arrays.binarySearch(copy_arr,0,n, arr[i])>-1)             System.out.print(arr[i] + " ");     } }    // Driver code public static void main(String[] args)  {     int arr[] = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };     int asize = arr.length;      int n = 5;     printSmall(arr, asize, n); } }    // This code is contributed by Princi Singh

Python3

 # Python3 for printing smallest n number in order    # Function for binary_search def binary_search(arr, low, high, ele):     while low < high:         mid = (low + high) // 2         if arr[mid] == ele:             return mid         elif arr[mid] > ele:             high = mid         else:             low = mid + 1     return -1    # Function to print smallest n numbers def printSmall(arr, asize, n):        # Make copy of array     copy_arr = arr.copy()        # Sort copy array     copy_arr.sort()        # For each arr[i] find whether     # it is a part of n-smallest     # with binary search     for i in range(asize):         if binary_search(copy_arr, low = 0,                           high = n, ele = arr[i]) > -1:             print(arr[i], end = " ")    # Driver Code if __name__ == "__main__":     arr = [1, 5, 8, 9, 6, 7, 3, 4, 2, 0]     asize = len(arr)     n = 5     printSmall(arr, asize, n)    # This code is conributed by # sanjeev2552

C#

 // C# for printing smallest n number in order using System;         class GFG  {       // Function to print smallest n numbers static void printSmall(int []arr, int asize, int n) {     // Make copy of array     int []copy_arr = new int[asize];     Array.Copy(arr, copy_arr, asize);        // Sort copy array     Array.Sort(copy_arr);        // For each arr[i] find whether     // it is a part of n-smallest     // with binary search     for (int i = 0; i < asize; ++i)     {         if (Array.BinarySearch(copy_arr, 0, n, arr[i])>-1)             Console.Write(arr[i] + " ");     } }    // Driver code public static void Main(String[] args)  {     int []arr = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };     int asize = arr.Length;      int n = 5;     printSmall(arr, asize, n); } }    // This code has been contributed by 29AjayKumar

Output :

1 3 4 2 0

For making a copy of array we need space complexity of O(n) and then for sorting we will need complexity of order O(n log n). Further for each element in arr[] we are performing searching in copy_arr[], which will result O(n) for linear search but we can improve it by applying binary search and hence our overall time complexity will be O(n log n).

My Personal Notes arrow_drop_up Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.