[Algorithm] 5. Kth Largest Element

Description

Find K-th largest element in an array.

Example

In array [9,3,2,4,8], the 3rd largest element is 4.

In array [1,2,3,4,5], the 1st largest element is 5, 2nd largest element is 4, 3rd largest element is 3 and etc.

Challenge

O(n) time, O(1) extra memory.

Notice

You can swap elements in the array

Answer

    /**
     * @param n: An integer
     * @param nums: An array
     * @return: the Kth largest element
     */
    int kthLargestElement(int n, vector<int> &nums) {
        // find out the kth largest number in an array.
        int temp=0;
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<nums.size()-i; j++)
            {
                // Swap the larger number to the end.
                if(nums[j] >= nums[j+1])
                {
                    temp = nums[j];
                    nums[j] = nums[j+1];
                    nums[j+1] = temp;
                }
            }
        }
        
        return nums[nums.size()-n];
    }

Better Solution

    /**
     * @param n: An integer
     * @param nums: An array
     * @return: the Kth largest element
     */
     
    int quick_sort(vector<int> &nums, int l, int r)
    {
        int key=nums[l];
        while(l<r)
        {
            while( (nums[r] >= key) && (l<r) )  r--;
            swap(nums[l], nums[r]);
            
            while( (nums[l] <= key) && (l<r) )  l++;
            swap(nums[l], nums[r]);
        }
        
        return l;
    }
    
    int kthLargestElement(int n, vector<int> &nums) {
        // Using qsort method to find the kth largest number.
        if(n>nums.size() || n==0 || nums.size()==0)   return -1;
        
        int left=0, right=nums.size()-1, k=nums.size()-n, pivot=0;
        
        while(true){
            pivot = quick_sort(nums, left, right);
            if(pivot==k){
                return nums[k];
            }else if(pivot<k){
                left=pivot+1;
                right=nums.size()-1;
            }else{
                left=0;
                right=pivot-1;
            }
        }
    }

 

Best Solution

    /**
     * @param n: An integer
     * @param nums: An array
     * @return: the Kth largest element
     */
    int kthLargestElement(int n, vector<int> &nums) {
        // Using priority queue to solve it.
        if(n>nums.size() || n==0 || nums.size()==0)   return -1;
        
        priority_queue<int, vector<int>, greater<int>> pri_queue;
        for(int i=0; i<n; i++)
        {
            pri_queue.push(nums[i]);
        }
        for(int i=n; i<nums.size(); i++)
        {
            if( nums[i] > pri_queue.top() )
            {
                pri_queue.pop();
                pri_queue.push(nums[i]);
            }
        }
        
        return pri_queue.top();
    }

Tips

It is a good way to use STL container(priority queue) to solve the problem.