2D Array - DS

Problem Statement :

```Given a 6*6 2D Array, arr:

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation:

a b c
d
e f g

There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print the maximum hourglass sum. The array will always be 6*6.

Example:

arr=
-9 -9 -9  1 1 1
0 -9  0  4 3 2
-9 -9 -9  1 2 3
0  0  8  6 6 0
0  0  0 -2 0 0
0  0  1  2 4 0

The 16 hourglass sums are:

-63, -34, -9, 12,
-10,   0, 28, 23,
-27, -11, -2, 10,
9,  17, 25, 18

The highest hourglass sum is 28 from the hourglass beginning at row 1, column 2:

0 4 3
1
8 6 6

Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge.

Function Description

Complete the function hourglassSum in the editor below.

hourglassSum has the following parameter(s):

1. int arr[6][6]: an array of integers
Returns
2 .int: the maximum hourglass sum

Input Format:

Each of the 6 lines of inputs arr[i] contains 6 space-separated integers arr[i][j].

Constraints:
-9<=arr[i][j]<=9
0<=i,j<=5

Output Format:

Print the largest (maximum) hourglass sum found in arr.```

Solution :

```                            ```Solution in C :

In C:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

int i,j,k;
int arr[6][6],temp=-9999,a,b;

for(i=0;i<6;i++)
for(j=0;j<6;j++)
scanf("%d",&arr[i][j]);

for(i=0;i<=3;i++)
for(j=0;j<=3;j++)
{
a = arr[i][j]+arr[i][j+1]+arr[i][j+2]+arr[i+1][j+1]+arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2];
if(temp < a)
temp = a ;

}
printf("%d",temp);

return 0;
}

In C++:

#include <cmath>
#include <cstdio>
#include <vector>
#include <climits>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
int a[6][6],s;
int m=INT_MIN;

for(int i=0;i<6;i++)
{
for(int j=0;j<6;j++)
{
cin>>a[i][j];
}
}

for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
s=(a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2]);
if(s>m)
m=s;
}

}
cout<<m;

return 0;
}

In Java:

import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[][] array = new int[6][6];
for (int y = 0; y < 6; y++){
for (int x =0; x<6; x++){
array[x][y] = sc.nextInt();
}
}
int maxHourglass = getHourglass(array, 1,1);
for (int y=1; y<5; y++){
for (int x=1; x<5; x++){
int hourres = getHourglass(array, x, y);
if (hourres > maxHourglass){
maxHourglass = hourres;
}
}
}
System.out.println(maxHourglass);
}

public static int getHourglass(int[][] array, int x, int y) {
return array[x][y] + array[x-1][y-1] + array[x][y-1] + array[x+1][y-1] + array[x-1][y+1]
+ array[x][y+1] + array[x+1][y+1];
}
}

In Python 3:

def convert(x):
res1=[]
for i in range(len(x)):
res1.append(int(x[i]))

return res1

def cal_values(matrix):
a0 = matrix[0]
a1 = matrix[1]
a2 = matrix[2]
res3=[]
for kk in range(4):
temp1 = 0
temp1 = sum(a0[kk:kk+3])
temp1 += a1[kk + 1]
temp1 += sum(a2[kk:kk+3])
res3.append(temp1)

return max(res3)

ss=[]
result=[]
for i in range(6):
temp=input().split()
res2=convert(temp)
ss.append(res2)

for j in range(4):
ee=cal_values(ss[j:j+3])
result.append(ee)

print(max(result))```
```

Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func