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

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 <bits/stdc++.h> 
  
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<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; 
} 

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. 

Output:

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

#include <bits/stdc++.h> 
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<<a[0]<<endl; 
        a[0] = a[n - 1]; 
        n--; 
        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 = k1 + 1; i < k2; i++) { 
        // cout<<a[0]<<endl; 
        ans += a[0]; 
        a[0] = a[n - 1]; 
        n--; 
        minheapify(a, 0); 
    } 
  
    cout << ans; 
  
    return 0; 
} 

Output:

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

References : https://tutorialspoint.dev/slugresolver/heap-sort

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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

C++

Java

Python3

C#

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