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

<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>

<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>

// JS program to find sum of all element between
// to K1'th and k2'th smallest elements in array

function PriorityQueue () {
    let collection = [];
    this.printCollection = function() {
      (console.log(collection));
    };
    this.enqueue = function(element){
        if (this.isEmpty()){ 
            collection.push(element);
        } else {
            let added = false;
            for (let i=0; i<collection.length; i++){
                 if (element[1] < collection[i][1]){ //checking priorities
                    collection.splice(i,0,element);
                    added = true;
                    break;
                }
            }
            if (!added){
                collection.push(element);
            }
        }
    };
    this.dequeue = function() {
        let value = collection.shift();
        return value[0];
    };
    this.front = function() {
        return collection[0];
    };
    this.size = function() {
        return collection.length; 
    };
    this.isEmpty = function() {
        return (collection.length === 0); 
    };
}

function sumBetweenTwoKth(A, N, K1, K2)
{
    // Using max heap to find K1'th and K2'th smallest
    // elements
    let maxH = new PriorityQueue(); 

    // 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 (let i = 0; i < N; i++) {

        maxH.enqueue(A[i]);

        if (maxH.size() > K2) {
            maxH.dequeue();
        }
    }
    // popping out the K2'th smallest element
    maxH.dequeue();

    let ans = 0;
    // adding the elements to ans until we reach the K1'th
    // smallest element
    while (maxH.size() > K1) {

        ans += maxH.front();
        maxH.dequeue();
    }

    return ans;
}

let arr = [ 20, 8, 22, 4, 12, 10, 14 ];
let k1 = 3, k2 = 6;
let n = arr.length;
console.log(sumBetweenTwoKth(arr, n, k1, k2));

// This code is contributed by akashish__