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

• Difficulty Level : Easy
• Last Updated : 20 Jul, 2022

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 ```

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.

Implementation:

## 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

 ``

## Javascript

 ``

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)

Time Complexity Analysis:

• By doing a simple analysis, we can observe that time complexity of step3 [ Determining step for overall time complexity ] can reach to O(nlogn) also.
• Take a look at the following description:
• Time Complexity of step3 is:  O((k2-k1)*log(n)) .
• In worst case, (k2-k1) would be almost O(n) [ Assume situation when k1=0  and k2=len(arr)-1 ]
• When O(k2-k1) =O(n) then overall complexity will be O(n* Log n ) .
• but in most of the cases…it will be lesser than O(n Log n) that is equal to sorting approach described above.

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<

## Javascript

 ``

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

### Method 3 : (Using Max Heap – most optimized )

The Below Idea uses the Max Heap Strategy to find the solution.

Algorithm:

1. The idea is to find the KthSmallest element for both the K1 and K2.
2. Then just traverse the array and sum the elemennts Less than K1 and More than K2 Value.

Now the idea revolves around KthSmallest Finding:

1.  The CRUX over here is that, we are storing the K smallest elements in the MAX Heap
2.  So while every push, if the size goes over K, then we pop the Maximum value.
3.  This way after whole traversal. we are leftout with K elements.
4.  Then the N-K th Largest Element is Poped and given, which is as same as  KTh Smallest element.

So by this manner we can write a functional code with using the C++ STL Priority_Queue, we get the most time and space optimized solution.

## C++

 `#include ``using` `namespace` `std;`  `//O(NlogK) Time to find Kth Smallest Element in Array``long` `long` `KthSmallest(``long` `long` `A[], ``long` `long` `N, ``long` `long` `K){``  ``priority_queue<``long` `long``>maxH; ``// MAX Heap` `  ``for``(``int` `i=0; iK){``      ``//O(log K)``      ``maxH.pop(); ``//Re-heapify happens``    ``}``  ``}` `  ``return` `maxH.top();``}``    ` `long` `long` `sumBetweenTwoKth( ``long` `long` `A[], ``long` `long` `N, ``long` `long` `K1, ``long` `long` `K2){``  ``long` `long` `K1val = KthSmallest(A,N,K1);``  ``long` `long` `K2val = KthSmallest(A,N,K2);` `  ``//Now just traverse and sum up all vals between these above vals``  ``long` `long` `sum = 0;``  ``for``(``int` `i=0; iK1val && A[i]

Output

`26`

Time Complexity:

O(N+ NLogK) = O(NLogK) (Dominant Term)

Reasons:

• The Traversal O(N) in the function.
• The K1th Smallest and K2 Smallest – O(N*LogK)
• As 1 Insertion takes O(LogK) where K is the size of Heap.
• As 1 Deletion takes  O(LogK) where K is the size of Heap.

Extra Space Complexity:

O(K)

Reasons:

• As we use Heap / Priority Queue and we only store at max K elements, not more than that.

The above Method-3 Idea, Algorithm, and Code are contributed by Balakrishnan R (rbkraj000 – GFG ID). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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