Create kth-largest-smallest.md

possible solutions

Author: CodingWallah

kth-largest-smallest.md

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+### [Back2Home](https://github.com/CodingWallah/ArraysCodingQuestions/main/README.md) | [Go2Video](#)
+
+1) min heap = uses for largest ?
+
+2) max heap = uses for smallest ?
+
+### WHEN max and when min HEAPS and WHY ?
+
+ Note- k is small = min; | k is large = MAX;
+```java
+
+//sol1 = using sorting or trick
+// Time complexity = O(NlogN)
+class Solution {
+ public int findKthLargest(int[] nums, int k) {
+ Arrays.sort(nums);
+ return nums[nums.length-k];
+ }
+}
+
+```
+
+```java
+
+//MIN HEAP APPROACH = Best Solution O(N + NlogK) time complexity
+PriorityQueue<Integer> minheap = new PriorityQueue<Integer>();
+ for(int i=0;i<k;i++)
+ minheap.add(nums[i]);
+ 
+ 
+ for(int i=k;i<nums.length;i++){
+ if(nums[i]>minheap.peek()){
+ minheap.poll();
+ minheap.add(nums[i]);
+ }
+ }
+return minheap.peek();
+```
+```java
+
+//MIN HEAP EASY & CONCISE IMPLEMENTATION
+PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); 
+ for(int i : nums){
+ pq.add(i);
+ if(pq.size()>k){
+ pq.remove();
+ }
+ }
+ 
+ return pq.remove();
+
+```
+
+
+```java
+//MAX HEAP APPROACH = Best Solution O(N + NlogK) time complexity
+PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(Collections.reverseOrder()); 
+ for(int i=0;i<nums.length-k+1;i++)
+ maxHeap.add(nums[i]);
+ 
+ for(int i=nums.length-k+1;i<nums.length;i++){
+ if(nums[i]<maxHeap.peek()){
+ maxHeap.poll();
+ maxHeap.add(nums[i]);
+ }
+ }
+
+ return maxHeap.peek(); 
+ }
+ 
+```
+
+```java
+//MAX HEAP EASY & CONCISE IMPLEMENTATION
+PriorityQueue<Integer> pq = new PriorityQueue<Integer>(Collections.reverseOrder()); 
+ for(int i : nums){
+ pq.add(i);
+ 
+ for(int i=1;i<k;i++)
+ pq.poll();
+ 
+ return pq.peek(); 
+ 
+```