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

Given an array of integers and two numbers k1 and k2. Find sum of all elements between given two k1’th and k2’th smallest elements of 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 C++ code uses an interesting method accumulate()

```// C++ program to find sum of all element between
// to K1'th and k2'th smallest elements in array
#include<bits/stdc++.h>
using namespace std;

// Returns sum between two kth smallest element of 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<k2-1; i++)
result += arr[i]; */
return accumulate(arr+k1, arr+k2-1, 0);
}

// Driver program
int main()
{
int arr[] = { 20, 8, 22, 4, 12, 10, 14 } ;
int  k1 = 3 , k2 = 6 ;
int n =  sizeof(arr)/sizeof(arr[0]);
cout << sumBetweenTwoKth(arr, n, k1, k2);
return 0;
}
```

Output:

``` 26
```

Time Complexity: O(n log n)

Method 2 (Using Min Heap)
We can optimize 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 times and sum all extracted elements. (This step takes O ((K2 – k1) * Log n) time)

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

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