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Print n smallest elements from given array in their original order
  • Difficulty Level : Easy
  • Last Updated : 01 Mar, 2021

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.

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 <bits/stdc++.h>
using namespace std;
 
// Function to print smallest n numbers
void printSmall(int arr[], int asize, int n)
{
    // Make copy of array
    vector<int> 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[0]);   
    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 

Time Complexity: O(n * log(n))

Auxiliary Space: O(n)
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).




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