new questions added

Author: arorarahul

README.md

@@ -196,6 +196,8 @@ time which is a better method.
 - [Find k largest elements from an array](/heaps/question6.c)
 - [Find median in a stream of numbers](/heaps/question7.c)
 - [Given k-sorted lists, Find the minimum range to which at last one number belongs from every list](/heaps/question8.c)
+- [Print out all integers in the form of a^3+b^3 where a and b are integers between 0 and N in sorted order](/heaps/question9.c)
+- [Convert BST to max heap](/heaps/question10.c)
 
 ## Some important concepts to solve algos better
 

heaps/question10.c

@@ -0,0 +1,14 @@
+/*
+Convert BST to max heap
+
+METHOD1:
+Since BST is generally not stored in the array, so we cannot use the data structure to convert directly
+into the max heap as we did in min to max heap or vice versa conversion. Therefore we do a INORDER traversal
+of BST. It will give us numbers in ascending order. Now reversing the array will be descending order which is
+nothing but a MAX HEAP
+Time complexity: O(n)
+Space complexity: O(1)
+
+Note in the method above we could have simply done reverse INORDER TRAVERSAL in order to avoid reversing the 
+array. In that we traverse the RST print and then traverse LST 
+*/

heaps/question8.c

@@ -23,24 +23,26 @@ Space complexity: O(k) //storing the k elements
 #include <stdio.h>
 #include <stdlib.h>
 
-void findInterval(int k, int l1[], int l2[], int l3[], int l4[],int size){
+void findInterval(int arr[], int k, int size){
+
+	
 
 }
 
 int main(){
-	int k, size;
+	int k, size, arr[], elm;
 	printf("enter the value if k\n");
 	printf("enter the size of each array\n");
 	scanf("%d",size);
 	scanf("%d",&k);
 	for(int i=0;i<k;i++){
 		for(int i=0;i<size;i++){
-			
+			printf("enter the %d th element\n");
+			scanf("%d",&elm)
+			arr[i+k*size] = elm;
 		}
 	}
-	
-
-	findInterval(k,l1,l2,l3,l4, size);
 
+	findInterval(arr,k,size);
 	return 0;
 }

heaps/question9.c

@@ -0,0 +1,17 @@
+/*
+Print out all integers in the form of a^3+b^3 where a and b are integers between 0 and N in sorted order
+
+METHOD1:
+Here all integers that lie between 0 and N will, combinations of those will be made with each other and
+a3+b3 will be computed for each and presented in a sorted order. Imagine it has a matrix lets say the
+range is between 0 to 3, then combinations are 00,01,02,03,10,11,12,13 and so on. Therefore 0^3+0^3 and so
+on is computed and each time lets say value is in a two dimensional array. Now the entire column of 
+the array is picked and a min heap is made out of it. Each time min is deleted and new element from the
+row in serial order is inserted and minified. This way we will get all elements in sorted order
+Time complexity: O(n^2)logn //total elements are n^2 inserted and deleting from the min heap takes logn
+time.
+Space complexity: O(n) to store n elements in a heap. matrix wont be taken into consideration because it
+will not be a matrix, its just for understanding purposes. In actual implementation everything can be done
+on runtime
+
+*/

nextquestions.md

@@ -44,5 +44,7 @@ TODO:
 - building min/max heap for duplicate elements
 - Do BST method in question 6 for heaps
 - Do BST method for question 7 in heaps (question7 complete to be done)
+- question8 in heaps to be done still
+- question9 and question 10 to be done still