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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find sum of all element between
// to K1'th and k2'th smallest elements in array
  
// Returns sum between two kth smallest elements of the array
function sumBetweenTwoKth($arr, $n, $k1, $k2)
{
    // Sort the given array
    sort($arr);
  
    // Below code is equivalent to
        $result = 0;
   
        for ($i = $k1; $i < $k2 - 1; $i++)
            $result += $arr[$i];
   
        return $result;
}
  
// Driver program
  
    $arr = array( 20, 8, 22, 4, 12, 10, 14 );
    $k1 = 3;
    $k2 = 6;
    $n = count($arr);;
    echo sumBetweenTwoKth($arr, $n, $k1, $k2);
      
// This code is contributed by mits
?>

chevron_right



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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of above approach
#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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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<<a[0]<<endl;
        ans += a[0];
        a[0] = a[n - 1];
        n--;
        minheapify(a, 0);
    }
  
    System.out.println(ans);
}
}
  
// This code is contributed by mits

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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<<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);
    }
  
    Console.Write(ans);
}
}
  
// This code is contributed by mits

chevron_right



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.

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



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


Be the First to upvote.


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