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.


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 CPP implementation of above approach :

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    // CPP for printing smallest n number in order
    #include <algorithm>
    #include <iostream>
    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;
    }

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    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).



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