Kth Smallest element

//Method 1
/*
Time complexity: O(nlogk)
Auxiliary Space: O(logK)

max_heap will store maximum of k minimum elements of the array
*/
class Solution{
    public:
    int kthSmallest(int arr[], int l, int n, int k) {
            priority_queue<int> pq;
            for(int i = 0; i < k; i++)  pq.push(arr[i]);
            for(int i = k; i <=n; i++){
                if(arr[i] < pq.top()){  //it is because we will keep k minimum elements in max_heap and maximum of that k we will be at top
                    pq.pop();           //that will be our answer ie. kth minimum element
                    pq.push(arr[i]);
                }
            }
            return pq.top();
    }
};

//Method 2
//Quick select
int partition(int arr[],int l,int r){
    int pivot=arr[r];
    int i=l,j=l;
    for(;i<r;i++){
        if(arr[i]<pivot){ swap(arr[i],arr[j]); j++;}
    }
    swap(arr[j],arr[r]);
    return j;
}
int quickselect(int arr[],int l,int r,int k){
    if(l<=r){
        int p=partition(arr,l,r);
        if(k==p) return arr[k];
        if(k<p) return quickselect(arr,l,p-1,k);
        else return quickselect(arr,p+1,r,k);
    }
}
class Solution{
    public:
    // arr : given array
    // l : starting index of the array i.e 0
    // r : ending index of the array i.e size-1
    // k : find kth smallest element and return using this function
    int kthSmallest(int arr[], int l, int r, int k) {
        //find element at k-1 position
        return quickselect(arr,l,r,k-1);
    }
};