Skip to content
Related Articles

Related Articles

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

Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 07 Dec, 2022
Improve Article
Save Article

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 an 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 <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.

PHP




<?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
?>

Javascript




<script>
  
// Javascript 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
function sumBetweenTwoKth(arr, k1 , k2)
{
      
    // Sort the given array
    arr.sort(function(a, b){return a - b});
  
    // Below code is equivalent to
    var result = 0;
  
    for(var i = k1; i < k2 - 1; i++)
        result += arr[i];
  
    return result;
}
  
// Driver code
var arr = [ 20, 8, 22, 4, 12, 10, 14 ];
var k1 = 3, k2 = 6;
var n = arr.length;
  
document.write(sumBetweenTwoKth(arr,
                                k1, k2));
  
// This code is contributed by shikhasingrajput 
  
</script>

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 <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;
}

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<<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

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<<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

Javascript




<script>
  
// Javascript implementation of above approach
let n = 7;
  
function minheapify(a, index)
{
    let small = index;
    let l = 2 * index + 1;
    let 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)
    {
        let t = a[small];
        a[small] = a[index];
        a[index] = t;
        minheapify(a, small);
    }
}
  
// Driver code
let i = 0;
let k1 = 3;
let k2 = 6;
  
let a = [ 20, 8, 22, 4, 12, 10, 14 ];
  
let ans = 0;
  
for(i = parseInt(n / 2, 10) - 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);
}
  
document.write(ans);
  
// This code is contributed by vaibhavrabadiya117
  
</script>

Output

26

The 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).

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 Kth Smallest element for the K2 . 
  2. Then just keep an popping the elements until the size of heap is K1, and make sure to add the elements to a variable before popping the elements.

Now the idea revolves around Kth Smallest 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 left out with K elements.
  4.  Then the N-K th Largest Element is Popped and given, which is as same as  K’th 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++




// 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;
long long sumBetweenTwoKth(long long A[], long long N,
                           long long K1, long long K2)
{
    // Using max heap to find K1'th and K2'th smallest
    // elements
    priority_queue<long long> maxH;
  
    // Using this for loop we eliminate the extra elements
    // which are greater than K2'th smallest element as they
    // are not required for us
  
    for (int i = 0; i < N; i++) {
  
        maxH.push(A[i]);
  
        if (maxH.size() > K2) {
            maxH.pop();
        }
    }
    // popping out the K2'th smallest element
    maxH.pop();
  
    long long ans = 0;
    // adding the elements to ans until we reach the K1'th
    // smallest element
    while (maxH.size() > K1) {
  
        ans += maxH.top();
        maxH.pop();
    }
  
    return ans;
}
  
int main()
{
    long long arr[] = { 20, 8, 22, 4, 12, 10, 14 };
    long long k1 = 3, k2 = 6;
    long long 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.*;
public class GFG {
  static long sumBetweenTwoKth(long A[], long N, long K1,
                               long K2)
  {
    // Using max heap to find K1'th and K2'th smallest
    // elements
    PriorityQueue<Long> maxH = new PriorityQueue<>(
      Collections.reverseOrder());
  
    // Using this for loop we eliminate the extra
    // elements which are greater than K2'th smallest
    // element as they are not required for us
  
    for (int i = 0; i < N; i++) {
  
      maxH.add(A[i]);
  
      if (maxH.size() > K2) {
        maxH.remove();
      }
    }
    // poping out the K2'th smallest element
    maxH.remove();
  
    long ans = 0;
    // adding the elements to ans until we reach the
    // K1'th smallest element
    while (maxH.size() > K1) {
  
      ans += maxH.peek();
      maxH.remove();
    }
  
    return ans;
  }
  
  public static void main(String[] args)
  {
    long arr[] = { 20, 8, 22, 4, 12, 10, 14 };
    long k1 = 3, k2 = 6;
    long n = arr.length;
    System.out.println(
      sumBetweenTwoKth(arr, n, k1, k2));
  }
}
// This code is contributed by karandeep1234

Output

26

Time Complexity:

 ( N * log K2 ) + ( (K2-K1) * log (K2-K1) ) + O(N) = O(NLogK2) (Dominant Term)

Reasons:

  • The Traversal O(N) in the function
  • Time Complexity for finding K2’th smallest element is ( N * log K2 )
  • Time Complexity for popping ( K2-K1 ) elements is ( (K2-K1) * log (K2-K1) )
  • 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(K2)

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. 


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!