mailpraveens

// Use Gists to store code you would like to remember later on
console.log(window); // log the "window" object to the console

mailpraveens

static inline BOOL checkIfEmpty(id obj) {
	return thing == nil || 
	([obj respondsToSelector:@selector(length)] && [(NSData *)obj length] == 0) ||
	([obj respondsToSelector:@selector(count)] && [(NSArray *)obj count] == 0);
}

mailpraveens

#include <iostream>
using namespace std;
#include <stdlib.h>     /* qsort */

int compare (const void * a, const void * b) {
    return ( *(int*)a - *(int*)b );
}

int main4()
{

mailpraveens

#include <iostream>
using namespace std;
#include <stdlib.h>     /* qsort */

int compare (const void * a, const void * b) {
  return ( *(int*)a - *(int*)b );
}

int main()
{

mailpraveens

#include<iostream>
#include <list>
using namespace std;

class Graph {
    int V;
    list <int> * adj;
    void DFSUtil(int v, bool visited[]);  // A function used by DFS
    
    

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def calculateLevenSteninDistance(a,b):
    if not a: return len(b)
    if not b: return len(a)
    return min(calculateLevenSteninDistance(a[1:], b[1:])+(a[0] != b[0]), calculateLevenSteninDistance(a[1:], b)+1, calculateLevenSteninDistance(a, b[1:])+1)

mailpraveens

def findDivergePointForOfTwoLinkedLists(list1, list2):
	# Assuming the list class has a method length to give the length of the list
	lengthList1 = list1.length()
	lengthList2 = list2.length()
	differnce = lengthList2 - lengthList1
	if(lengthList2>=lengthList1):
		while(differnce!=0):
			list2 = list2.next
			differnce -= 1

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def findLCA(node,value1,value2):
	if(node == None):
		return None
	#move to right if values are greater
	if(node.value < value1 and node.value < value2):
		return findLCA(node.right,value1, value2)
	#move to left if values are smaller
	if(node.value > value1 and node.value > value2):
		return findLCA(node.left,value1, value2)
	# Whereever it diverges, return the node as its the LCA

mailpraveens

'''
		1
	2		3
  4   5   6    7


 should print 1,2,4,5,6,7,3,1


'''

mailpraveens

#max heap construction
def heapify(items,idx):
	left = leftChild(idx)
	right = rightChild(idx)
	largest = -1
	if(left < len(items) and items[left] > items[idx]):
		largest = left
	else:
		largest = idx