Kth smallest Element in unsorted Array

(Referrence:geeksforgeeks, Kth largest Element in Array)

This was a common algorithm problem appearing in interviews.

There is four basic solutions.

Solution 1--Sort first

A simple solution are to sort the given array using a O (n log n) sorting algorithm like Merge sort,heap sort, etc and retur n the element at index k-1 in the sorted array. Time complexity of this solution is O (n log n).

Java Arrays.sort ()

1  Public class solution{2      Public int findkthsmallest (intint  k) {3        arrays.sort (nums); 4         return Nums[k]; 5     }6 }

Solution 2--Construct Min Heap

A simple optomization are to create a Min Heap of the given n elements and call Extractmin () K times.

To build a heap, time complexity is O (n). So total time complexity is O (n + k log n).

Java Priority Queue

Using priorityqueue (collection<? extends e> c), we can construct a heap from array or other object in linear time.

By Defaule, it would create a min-heap.

Example

1  Public intGenerateheap (int[] nums) {2         intLength =nums.length;3integer[] NewArray =NewInteger[length];4          for(inti = 0; i < length; i++)5Newarray[i] =(Integer) nums[i];6Priorityqueue<integer> PQ =NewPriorityqueue<integer>(Arrays.aslist (NewArray));7}

Comparator Example

Comparator cmp = colletions.reverseorder ();

Solution 3--Use Max Heap

1. Build a max-heap of size K. Put Nums[0] to nums[k-1] to heap.

2. For each element after nums[k-1], the compare it with root of the heap.

A. If current >= root, move on.

B. If current < Root, remove root and put current into heap.

3. Return Root.

Time Complexity is O ((n-k) log k).

(Java:priorityqueue)

(codes)

1  Public classSolution {2      Public intFindkthsmallest (int[] Nums,intk) {3         //Construct A max heap of size K4         intLength =nums.length;5Priorityqueue<integer> PQ =NewPriorityqueue<integer>(k, Collections.reverseorder ());6          for(inti = 0; I < K; i++)7 Pq.add (Nums[i]);8          for(inti = k; i < length; i++) {9             intCurrent =Nums[i];Ten             intRoot =Pq.peek (); One             if(Current <root) { A                 //Remove Head - Pq.poll (); -                 //ADD new Node the Pq.add (current); -             } -         } -         returnPq.peek (); +     } -}

Solution 4--Quick Select

 Public classSolution {Private voidSwapint[] Nums,intIndex1,intindex2) {        intTMP =NUMS[INDEX1]; NUMS[INDEX1]=Nums[index2]; NUMS[INDEX2]=tmp; }    //Pick last element as pivot//Place all smaller elements before pivot//Place all bigger elements after pivot    Private intPartitionint[] Nums,intStartintend) {        intPivot =Nums[end]; intCurrentsmaller = start-1;  for(inti = start; I < end; i++) {            //If Current element <= Pivot, put it to right position            if(Nums[i] <=pivot) {Currentsmaller++;            Swap (Nums, I, Currentsmaller); }        }        //Put pivot to right positioncurrentsmaller++;        Swap (Nums, end, Currentsmaller); returnCurrentsmaller; }     Public intQuickselect (int[] Nums,intStartintEndintk) {intpos =partition (Nums, start, end)if(pos = = K-1)            returnNums[pos]; if(Pos < K-1)            returnQuickselect (Nums, pos + 1, end, K-(Pos-start + 1)); Else            returnQuickselect (Nums, start, pos-1, K); }}

The worst case time complexity of this method was O (N2), but it works with O (n) on average.

Kth smallest Element in unsorted Array