# 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 :

 `// 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<``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 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 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# 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).

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