# Sum of all elements between k1’th and k2’th smallest elements

Given an array of integers and two numbers k1 and k2. Find the sum of all elements between given two k1’th and k2’th smallest elements of the array. It may be assumed that (1 <= k1 < k2 <= n) and all elements of array are distinct.

Examples :

```Input : arr[] = {20, 8, 22, 4, 12, 10, 14},  k1 = 3,  k2 = 6
Output : 26
3rd smallest element is 10. 6th smallest element
is 20. Sum of all element between k1 & k2 is
12 + 14 = 26

Input : arr[] = {10, 2, 50, 12, 48, 13}, k1 = 2, k2 = 6
Output : 73 ```

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

Method 1 (Sorting)
First sort the given array using a O(n log n) sorting algorithm like Merge Sort, Heap Sort, etc and return the sum of all element between index k1 and k2 in the sorted array.

Below is the implementation of the above idea :

## C++

 `// C++ program to find sum of all element between ` `// to K1'th and k2'th smallest elements in array ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// Returns sum between two kth smallest elements of the array ` `int` `sumBetweenTwoKth(``int` `arr[], ``int` `n, ``int` `k1, ``int` `k2) ` `{ ` `    ``// Sort the given array ` `    ``sort(arr, arr + n); ` ` `  `    ``/* Below code is equivalent to ` `     ``int result = 0; ` `     ``for (int i=k1; i

## Java

 `// Java program to find sum of all element ` `// between to K1'th and k2'th smallest ` `// elements in array ` `import` `java.util.Arrays; ` ` `  `class` `GFG { ` ` `  `    ``// Returns sum between two kth smallest ` `    ``// element of array ` `    ``static` `int` `sumBetweenTwoKth(``int` `arr[], ` `                                ``int` `k1, ``int` `k2) ` `    ``{ ` `        ``// Sort the given array ` `        ``Arrays.sort(arr); ` ` `  `        ``// Below code is equivalent to ` `        ``int` `result = ``0``; ` ` `  `        ``for` `(``int` `i = k1; i < k2 - ``1``; i++) ` `            ``result += arr[i]; ` ` `  `        ``return` `result; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``int` `arr[] = { ``20``, ``8``, ``22``, ``4``, ``12``, ``10``, ``14` `}; ` `        ``int` `k1 = ``3``, k2 = ``6``; ` `        ``int` `n = arr.length; ` ` `  `        ``System.out.print(sumBetweenTwoKth(arr, ` `                                          ``k1, k2)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to find sum of ` `# all element between to K1'th and ` `# k2'th smallest elements in array ` ` `  `# Returns sum between two kth ` `# smallest element of array ` `def` `sumBetweenTwoKth(arr, n, k1, k2): ` ` `  `    ``# Sort the given array ` `    ``arr.sort() ` ` `  `    ``result ``=` `0` `    ``for` `i ``in` `range``(k1, k2``-``1``): ` `        ``result ``+``=` `arr[i]  ` `    ``return` `result ` ` `  `# Driver code ` `arr ``=` `[ ``20``, ``8``, ``22``, ``4``, ``12``, ``10``, ``14` `]  ` `k1 ``=` `3``; k2 ``=` `6` `n ``=` `len``(arr) ` `print``(sumBetweenTwoKth(arr, n, k1, k2)) ` ` `  ` `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# program to find sum of all element ` `// between to K1'th and k2'th smallest ` `// elements in array ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Returns sum between two kth smallest ` `    ``// element of array ` `    ``static` `int` `sumBetweenTwoKth(``int``[] arr, ``int` `n, ` `                                ``int` `k1, ``int` `k2) ` `    ``{ ` `        ``// Sort the given array ` `        ``Array.Sort(arr); ` ` `  `        ``// Below code is equivalent to ` `        ``int` `result = 0; ` ` `  `        ``for` `(``int` `i = k1; i < k2 - 1; i++) ` `            ``result += arr[i]; ` ` `  `        ``return` `result; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 20, 8, 22, 4, 12, 10, 14 }; ` `        ``int` `k1 = 3, k2 = 6; ` `        ``int` `n = arr.Length; ` ` `  `        ``Console.Write(sumBetweenTwoKth(arr, n, k1, k2)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` `

Output:

``` 26
```

Time Complexity: O(n log n)

Method 2 (Using Min Heap)
We can optimize the above solution be using a min heap.
1) Create a min heap of all array elements. (This step takes O(n) time)
2) Do extract minimum k1 times (This step takes O(K1 Log n) time)
3) Do extract minimum k2 – k1 – 1 time and sum all extracted elements. (This step takes O ((K2 – k1) * Log n) time)

Method 2 Implementation

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `n = 7; ` ` `  `void` `minheapify(``int` `a[], ``int` `index) ` `{ ` ` `  `    ``int` `small = index; ` `    ``int` `l = 2 * index + 1; ` `    ``int` `r = 2 * index + 2; ` ` `  `    ``if` `(l < n && a[l] < a[small]) ` `        ``small = l; ` ` `  `    ``if` `(r < n && a[r] < a[small]) ` `        ``small = r; ` ` `  `    ``if` `(small != index) { ` `        ``swap(a[small], a[index]); ` `        ``minheapify(a, small); ` `    ``} ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `i = 0; ` `    ``int` `k1 = 3; ` `    ``int` `k2 = 6; ` ` `  `    ``int` `a[] = { 20, 8, 22, 4, 12, 10, 14 }; ` ` `  `    ``int` `ans = 0; ` ` `  `    ``for` `(i = (n / 2) - 1; i >= 0; i--) { ` `        ``minheapify(a, i); ` `    ``} ` ` `  `    ``// decreasing value by 1 because we want min heapifying k times and it starts ` `    ``// from 0 so we have to decrease it 1 time ` `    ``k1--; ` `    ``k2--; ` ` `  `    ``// Step 1: Do extract minimum k1 times (This step takes O(K1 Log n) time) ` `    ``for` `(i = 0; i <= k1; i++) { ` `        ``// cout<

## Java

 `// Java implementation of above approach ` `class` `GFG ` `{ ` `     `  `static` `int` `n = ``7``; ` ` `  `static` `void` `minheapify(``int` `[]a, ``int` `index) ` `{ ` ` `  `    ``int` `small = index; ` `    ``int` `l = ``2` `* index + ``1``; ` `    ``int` `r = ``2` `* index + ``2``; ` ` `  `    ``if` `(l < n && a[l] < a[small]) ` `        ``small = l; ` ` `  `    ``if` `(r < n && a[r] < a[small]) ` `        ``small = r; ` ` `  `    ``if` `(small != index) ` `    ``{ ` `        ``int` `t = a[small]; ` `        ``a[small] = a[index]; ` `        ``a[index] = t; ` `        ``minheapify(a, small); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `i = ``0``; ` `    ``int` `k1 = ``3``; ` `    ``int` `k2 = ``6``; ` ` `  `    ``int` `[]a = { ``20``, ``8``, ``22``, ``4``, ``12``, ``10``, ``14` `}; ` ` `  `    ``int` `ans = ``0``; ` ` `  `    ``for` `(i = (n / ``2``) - ``1``; i >= ``0``; i--) ` `    ``{ ` `        ``minheapify(a, i); ` `    ``} ` ` `  `    ``// decreasing value by 1 because we want ` `    ``// min heapifying k times and it starts ` `    ``// from 0 so we have to decrease it 1 time ` `    ``k1--; ` `    ``k2--; ` ` `  `    ``// Step 1: Do extract minimum k1 times  ` `    ``// (This step takes O(K1 Log n) time) ` `    ``for` `(i = ``0``; i <= k1; i++) ` `    ``{ ` `        ``a[``0``] = a[n - ``1``]; ` `        ``n--; ` `        ``minheapify(a, ``0``); ` `    ``} ` ` `  `    ``for` `(i = k1 + ``1``; i < k2; i++) ` `    ``{ ` `        ``// cout<

## Python3

 `# Python 3 implementation of above approach  ` `n ``=` `7` ` `  `def` `minheapify(a, index): ` `    ``small ``=` `index ` `    ``l ``=` `2` `*` `index ``+` `1` `    ``r ``=` `2` `*` `index ``+` `2` ` `  `    ``if` `(l < n ``and` `a[l] < a[small]): ` `        ``small ``=` `l ` ` `  `    ``if` `(r < n ``and` `a[r] < a[small]): ` `        ``small ``=` `r ` ` `  `    ``if` `(small !``=` `index): ` `        ``(a[small], a[index]) ``=` `(a[index], a[small]) ` `        ``minheapify(a, small) ` `     `  `# Driver Code ` `i ``=` `0` `k1 ``=` `3` `k2 ``=` `6` ` `  `a ``=` `[ ``20``, ``8``, ``22``, ``4``, ``12``, ``10``, ``14` `]  ` `ans ``=` `0` ` `  `for` `i ``in` `range``((n ``/``/``2``) ``-` `1``, ``-``1``, ``-``1``): ` `    ``minheapify(a, i) ` ` `  `# decreasing value by 1 because we want  ` `# min heapifying k times and it starts  ` `# from 0 so we have to decrease it 1 time  ` `k1 ``-``=` `1` `k2 ``-``=` `1` ` `  `# Step 1: Do extract minimum k1 times  ` `# (This step takes O(K1 Log n) time)  ` `for` `i ``in` `range``(``0``, k1 ``+` `1``): ` `    ``a[``0``] ``=` `a[n ``-` `1``] ` `    ``n ``-``=` `1` `    ``minheapify(a, ``0``) ` ` `  `# Step 2: Do extract minimum k2 – k1 – 1 times and  ` `# sum all extracted elements.  ` `# (This step takes O ((K2 – k1) * Log n) time)*/ ` `for` `i ``in` `range``(k1 ``+` `1``, k2) : ` `    ``ans ``+``=` `a[``0``] ` `    ``a[``0``] ``=` `a[n ``-` `1``]  ` `    ``n ``-``=` `1` `    ``minheapify(a, ``0``)  ` ` `  `print` `(ans) ` ` `  `# This code is contributed  ` `# by Atul_kumar_Shrivastava `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `n = 7; ` ` `  `static` `void` `minheapify(``int` `[]a, ``int` `index) ` `{ ` ` `  `    ``int` `small = index; ` `    ``int` `l = 2 * index + 1; ` `    ``int` `r = 2 * index + 2; ` ` `  `    ``if` `(l < n && a[l] < a[small]) ` `        ``small = l; ` ` `  `    ``if` `(r < n && a[r] < a[small]) ` `        ``small = r; ` ` `  `    ``if` `(small != index) ` `    ``{ ` `        ``int` `t = a[small]; ` `        ``a[small] = a[index]; ` `        ``a[index] = t; ` `        ``minheapify(a, small); ` `    ``} ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `i = 0; ` `    ``int` `k1 = 3; ` `    ``int` `k2 = 6; ` ` `  `    ``int` `[]a = { 20, 8, 22, 4, 12, 10, 14 }; ` ` `  `    ``int` `ans = 0; ` ` `  `    ``for` `(i = (n / 2) - 1; i >= 0; i--) ` `    ``{ ` `        ``minheapify(a, i); ` `    ``} ` ` `  `    ``// decreasing value by 1 because we want ` `    ``// min heapifying k times and it starts ` `    ``// from 0 so we have to decrease it 1 time ` `    ``k1--; ` `    ``k2--; ` ` `  `    ``// Step 1: Do extract minimum k1 times  ` `    ``// (This step takes O(K1 Log n) time) ` `    ``for` `(i = 0; i <= k1; i++) ` `    ``{ ` `        ``// cout<

Output:

``` 26
```

Overall time complexity of this method is O(n + k2 Log n) which is better than sorting based method.

References : https://www.geeksforgeeks.org/heap-sort

This article is contributed by Nishant_Singh (Pintu). 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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

3

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.