for (i = 1; i < 3; i++)
{
for (j = 0; j < 2; j++)
{
System.out.print(a[i][j] + " ");
}
System.out.println();
}
The above code is not able to print numbers which are below the left diagonal. For a 3x3 matrix my code is printing:
1 2 3
4 5 6
7 8 9
OUTPUT :
4 5
7 8
Desired output:
4
7 8
You could add an if statment like this :
for (i = 0; i < 3; i++) {
for (j = 0; j < 2; j++) {
if (i>j) {
System.out.print(a[i][j] + " ");
}
}
System.out.println();
}
Or better you could do :
for(int i=0;i<a.length;i++) {
for(int j=0;i>j;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
A simple way to achieve this depending on the matrix size is
final int matrixSize = 7; // your matrix size
for (int i = 0; i < matrixSize; ++ i) {
for (int j = 0; j < i; ++ j) System.out.print(a[i][j] + " ");
System.out.print("\n");
}
I want to print the following pattern in java:
a+1357+1
b+246+2
a+13+3
b+2+4
following is my code, but with this i can only print odd no. or only even no.s
public static void main(String[] args) {
int rows = 7;
for(int i = rows; i >= 1; i=i-2) {
for(int j = 1; j <= i; j=j+2) {
System.out.print(j + " ");
}
System.out.println();
}
}
DEMO
var rows = 4;
for (var i = 4; i > 0; i--) {
for (var j = 1; j <= i; j++) {
document.write((i % 2) + (2 * j) - 1 + " ");
}
document.write('<br>');
}
public static void main(String[] args) {
int rows = 4;
for(int i = rows; i > 0; i--) {
for(int j = 1; j <= i; j++) {
System.out.print((i%2)+(2*j)-1 + " ");
}
System.out.println();
}
}
You need to make a pattern for it. here you can use (i%2)+(2*j)-1
With only a few updates of your code (but not very readable):
int rows = 7;
for (int i = rows; i >= 1; i = i - 2) {
System.out.print((((i + 1) % 4) == 0 ? "a" : "b") + " + ");
for (int j = 1; j <= i; j = j + 2) {
System.out.print((j + ((i + 2) % 4) / 2));
}
System.out.println(" + " + (10 - i) / 2);
}
But instead of using my code, I suggest you write down exactly how the "pattern" is defined and write new code based on your specification. These loops are not optimal.
Am confused about getting the "sub diagonal" for 2D array, I can get the columns , rows and major diagonal, but i don't know how to get sub diagonal, here is what i'v done so far:
int size = input.nextInt();
int[][] list = initiateArr(size);
for (int i = 0; i < list.length; i++) {
for (int j = i ; j < i + 1; j++) {
System.out.print(list[i][j] + " ");
}
}
My attempt for "sub diagonal":
for (int i = 0; i < list.length; i++) {
for (int j = (list.length / 2) + 1; j > list.length - (i + 1) ; j--) {
System.out.print(list[i][j] + " ");
System.out.println("j = " + j);
}
}
How to get sub diagonal?
I can get the major diagonal right as described in the output 1111, the sub diagonal should be 1010.
Here is the answer. Let me know if you have any questions.
public static void main(String args[]) {
int[][] list = {{1,2,3}, {4,5,6},{7,8,9}};
int matrixSize = 3;
System.out.println("Subdiagonal");
for( int i = 0; i < matrixSize ; i ++){
System.out.print( list[i][matrixSize - i -1] + " ");
}
System.out.println("");
System.out.println("Major ");
for( int i = 0; i < matrixSize ; i ++){
System.out.print( list[i][i] + " ");
}
}
I am working on a problem where I've to print the largest sum among all the hourglasses in the array. You can find the details about the problem here-
What I tried:
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for (int arr_i = 0; arr_i < 6; arr_i++) {
for (int arr_j = 0; arr_j < 6; arr_j++) {
arr[arr_i][arr_j] = in.nextInt();
}
}
int sum = 0;
int tmp_sum = 0;
for (int arr_i = 0; arr_i < 4; arr_i++) {
for (int arr_j = 0; arr_j < 4; arr_j++) {
if (arr[arr_i][arr_j] > 0) {
sum = sum + (arr[arr_i][arr_j]) + (arr[arr_i][arr_j + 1]) + (arr[arr_i][arr_j + 2]);
sum = sum + (arr[arr_i + 1][arr_j + 1]);
sum = sum + (arr[arr_i + 2][arr_j]) + (arr[arr_i + 2][arr_j + 1]) + (arr[arr_i + 2][arr_j + 2]);
if (tmp_sum < sum) {
tmp_sum = sum;
}
sum = 0;
}
}
}
System.out.println(tmp_sum);
}
}
Input:
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 9 2 -4 -4 0
0 0 0 -2 0 0
0 0 -1 -2 -4 0
Output:
12
Expected Output:
13
Screenshot:
I don't know where I'm doing wrong. I cannot understand why the expected output is 13. According to the description given in the problem it should be 10. Is this a wrong question or my understanding about this is wrong?
Remove the if (arr[arr_i][arr_j] > 0) statement. It prevents finding the answer at row 1, column 0, because that cell is 0.
Comments for other improvements to your code:
What if the best hourglass sum is -4? You should initialize tmp_sum to Integer.MIN_VALUE. And name it maxSum, to better describe it's purpose.
You shouldn't define sum outside the loop. Declare it when it is first assigned, then you don't have to reset it to 0 afterwards.
Your iterators should be just i and j. Those are standard names for integer iterators, and keeps code ... cleaner.
If you prefer longer names, use row and col, since that is what they represent.
You don't need parenthesis around the array lookups.
For clarity, I formatted the code below to show the hourglass shape in the array lookups.
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for (int i = 0; i < 6; i++){
for (int j = 0; j < 6; j++){
arr[i][j] = in.nextInt();
}
}
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int sum = 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 (maxSum < sum) {
maxSum = sum;
}
}
}
System.out.println(maxSum);
This was my solution. I wrapped an if statement around the code that calculates the sum, that makes sure we don't go out of bounds.
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
int max = Integer.MIN_VALUE;
int tempMax = 0;
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
arr[i][j] = in.nextInt();
}
}
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
if (i + 2 < 6 && j + 2 < 6) {
tempMax += arr[i][j] + arr[i][j + 1] + arr[i][j + 2];
tempMax += arr[i + 1][j + 1];
tempMax += arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2];
if (max < tempMax) {
max = tempMax;
}
tempMax = 0;
}
}
}
System.out.println(max);
}
Here's the simple and easy to understand C# equivalent code for your hourglass problem.
class Class1
{
static int[][] CreateHourGlassForIndex(int p, int q, int[][] arr)
{
int[][] hourGlass = new int[3][];
int x = 0, y = 0;
for (int i = p; i <= p + 2; i++)
{
hourGlass[x] = new int[3];
int[] temp = new int[3];
int k = 0;
for (int j = q; j <= q + 2; j++)
{
temp[k] = arr[i][j];
k++;
}
hourGlass[x] = temp;
x++;
}
return hourGlass;
}
static int findSumOfEachHourGlass(int[][] arr)
{
int sum = 0;
for (int i = 0; i < arr.Length; i++)
{
for (int j = 0; j < arr.Length; j++)
{
if (!((i == 1 && j == 0) || (i == 1 && j == 2)))
sum += arr[i][j];
}
}
return sum;
}
static void Main(string[] args)
{
int[][] arr = new int[6][];
for (int arr_i = 0; arr_i < 6; arr_i++)
{
string[] arr_temp = Console.ReadLine().Split(' ');
arr[arr_i] = Array.ConvertAll(arr_temp, Int32.Parse);
}
int[] sum = new int[16];
int k = 0;
for (int i = 0; i < 4; i++)
{
for (int j = 0; j < 4; j++)
{
int[][] hourGlass = CreateHourGlassForIndex(i, j, arr);
sum[k] = findSumOfEachHourGlass(hourGlass);
k++;
}
}
//max in sum array
Console.WriteLine(sum.Max());
}
}
Happy Coding.
Thanks,
Ankit Bajpai
You can try this code:
I think this will be easy to understand for beginners.
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for(int arr_i=0; arr_i < 6; arr_i++){
for(int arr_j=0; arr_j < 6; arr_j++){
arr[arr_i][arr_j] = in.nextInt();
}
}
int sum = 0;
int sum2 = 0;
int sum3 = 0;
int x = 0;
int max = Integer.MIN_VALUE;
for(int i = 0; i < 4; i++){
for(int j = 0; j < 4; j++){
for(int k = 0; k < 3; k++){
sum += arr[i][j+k]; //top elements of hour glass
sum2 += arr[i+2][j+k]; //bottom elements of hour glass
sum3 = arr[i+1][j+1]; //middle elements of hour glass
x = sum + sum2 + sum3; //add all elements of hour glass
}
if(max < x){
max = x;
}
sum = 0;
sum2 = 0;
sum3 = 0;
x = 0;
}
}
System.out.println(max);
}
}
Here is another easy option, hope it helps:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a[][] = new int[6][6];
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
a[i][j] = in.nextInt();
}
}
int hg = Integer.MIN_VALUE, sum;
for(int i=0; i<4; i++){
for(int j=0; j<4; j++){
sum = 0;
sum = sum + a[i][j] + a[i][j+1] + a[i][j+2];
sum = sum + a[i+1][j+1];
sum = sum + a[i+2][j] + a[i+2][j+1] + a[i+2][j+2];
if(sum>hg)
hg = sum;
}
}
System.out.println(hg);
in.close();
}
}
there is another opetion in case of -(minus) and zero output we can use shorted ser Treeset for the same . below is the sameple code
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for(int i=0; i < 6; i++){
for(int j=0; j < 6; j++){
arr[i][j] = in.nextInt();
}
}
int sum=0;int output=0;
Set<Integer> set=new TreeSet<Integer>();
for(int k=0;k<4;k++ )
{
for(int y=0;y<4;y++)
{
sum=arr[k][y]+arr[k][y+1]+arr[k][y+2]+arr[k+1][y+1]+arr[k+2][y]+arr[k+2][y+1]+arr[k+2][y+2]; set.add(sum);
}
}
int p=0;
for(int u:set)
{
p++;
if(p==set.size())
output=u;
}
System.out.println(output);
}
}
Solved in PHP, may be helpful.
<?php
$handle = fopen ("php://stdin","r");
$input = [];
while(!feof($handle))
{
$temp = fgets($handle);
$input[] = explode(" ",$temp);
}
$maxSum = PHP_INT_MIN;
for($i=0; $i<4; $i++)
{
for($j=0; $j<4; $j++)
{
$sum = $input[$i][$j] + $input[$i][$j + 1] + $input[$i][$j + 2]
+ $input[$i + 1][$j + 1] +
$input[$i + 2][$j] + $input[$i + 2][$j + 1] + $input[$i + 2][$j + 2];
if($sum > $maxSum)
{
$maxSum = $sum;
}
}
}
echo $maxSum;
?>
Passes all test cases
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner read = new Scanner(System.in);
int rowSize = 6;
int colSize = 6;
int[][] array = new int[rowSize][colSize];
for(int row = 0; row < rowSize; row++) {
for(int col = 0; col < colSize; col++) {
array[row][col] = read.nextInt();
}
}
read.close();
int max = Integer.MIN_VALUE;
for(int row = 0; row < 4; row++) {
for(int col = 0; col < 4; col++) {
int sum = calculateHourglassSum(array, row, col);
if(sum > max) {
max = sum;
}
}
}
System.out.println(max);
}
private static int calculateHourglassSum(int[][] array, int rowIndex, int colIndex) {
int sum = 0;
for(int row = rowIndex; row < rowIndex + 3; row++) {
for(int col = colIndex; col < colIndex + 3; col++) {
if(row == rowIndex + 1 && col != colIndex + 1) {
continue;
}
sum += array[row][col];
}
}
return sum;
}
}
function galssSum(array) {
let maxGlass = 0;
if (array[0].length == 3) {
maxGlass = 1;
} else if (array[0].length > 3) {
maxGlass = array.length - 2;
}
let maxValue = -100000;
for (let i = 0; i < maxGlass; i++) {
for (let j = 0; j < maxGlass; j++) {
let a = array[i][j] + array[i][j + 1] + array[i][j + 2];
let b = array[i + 1][j + 1];
let c = array[i + 2][j] + array[i + 2][j + 1] + array[i + 2][j + 2];
let sum = a + b + c;
if (maxValue<sum) {
maxValue = sum;
}
}
}
return maxValue;
}
console.log(galssSum([[1, 1, 1, 0, 0, 0], [0, 1, 0, 0, 0, 0], [1, 1, 1, 0, 0, 0], [0, 0, 2, 4, 4, 0], [0, 0, 0, 2, 0, 0], [0, 0, 1, 2, 4, 0]]));
int hourglassSum(vector<vector<int>> vec) {
int res = 0;
int size = ((vec[0].size())-2) * ((vec.size())-2);
//cout<<size<<endl;
vector<int> res_vec(size);
int j = 0;
int itr =0 ;
int cnt = 0;
int mid = 0;
int l =0;
while((l+2) < vec.size())
{
while((j+2) < vec.size())
{
for(int i =j ;i<j+3; i+=2)
{
//cout<<i<<" :";
for(int k=l;k<l+3;k++)
{
//cout<<k<<" ";
res_vec[itr] += vec[i][k];
}
//cout<<endl;
}
res_vec[itr] += vec[j+1][l+1];
//cout<<endl;
itr++;
j++;
}
l++;
j=0;
}
int max=res_vec[0];
for(int i =1;i<res_vec.size();i++)
{
if(max < res_vec[i])
{
max = res_vec[i];
}
//cout<<res_vec[i]<< " ";
}
res = max;
//cout<<endl;
return res;
}
// Complete the hourglassSum function below.
static int hourglassSum(int[][] arr) {
int max = Integer.MIN_VALUE;
for (int i = 0; i < arr.length - 2; i++) {
for (int j = 0; j < arr.length - 2; j++) {
int hourGlassSum = (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]);
max = Math.max(hourGlassSum,max);
}
}
return max;
}
public static int hourglassSum(List<List<Integer>> arr) {
// Write your code here
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int sum = arr.get(i).get(j) +arr.get(i).get(j+1) +
arr.get(i).get(j+2)+arr.get(i+1).get(j+1)+
arr.get(i+2).get(j)+arr.get(i+2).get(j+1)+arr.get(i+2).get(j+2);
if (maxSum < sum) {
maxSum = sum;
}
}
}
return maxSum;
}
}
Iterative way,Passing all test cases in hackerank web
public static int hourglassSum(List<List<Integer>> arr) {
// Write your code here
int rowsCount=arr.size();
int colCount=arr.get(0).size();
Integer max=Integer.MIN_VALUE;
Integer subSum=0;
for(int r=0; (r+3)<=rowsCount; r++)
{
for(int c=0; (c+3)<=colCount; c++)
{
subSum= hourglassSubSum(arr,r,c);
System.out.println("r,c,subSum "+r+" "+c+" "+" "+subSum);
if(subSum>max)
{
max=subSum;
}
}
}
return max;
}
public static int hourglassSubSum(List<List<Integer>> hourglassArray,
int rowIndex,int colIndex) {
// Write your code here
Integer subSum=0;
for(int i=rowIndex;i<(rowIndex+3);i++)
{
for(int j=colIndex;j<(colIndex+3);j++)
{
if(i==(rowIndex+1) && (j==colIndex || j==colIndex+2))
{
continue;
}
subSum=subSum+hourglassArray.get(i).get(j);
}
}
return subSum;
}
Solution for actual "2D Array - DS" challenge from HackerRank https://www.hackerrank.com/challenges/2d-array
public static int hourglassSum(List<List<Integer>> arr) {
int maxSum = Integer.MIN_VALUE;
for (int col=0; col <= 3; col++) {
for (int row=0; row <= 3; row++) {
int sum = calcHourglass(arr, col, row);
maxSum = Math.max(sum, maxSum);
}
}
return maxSum;
}
private static int calcHourglass(List<List<Integer>> arr, int col, int row) {
int sum = 0;
for (int i=0; i < 3; i++) {
sum += arr.get(row).get(col+i); // the top of the hourglass
sum += arr.get(row+2).get(col+i); // the bottom of the hourglass
}
sum += arr.get(row+1).get(col+1); // the center
return sum;
}
import java.io.*;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int low = -9,high = 5;
int lh = low * high;
int sum = 0, i, j;
int max = 0;
int a[][] = new int[6][6];
for (i = 0; i < 6; i++) {
for (j = 0; j < 6; j++) {
a[i][j] = in.nextInt();
}
}
for (i = 0; i < 4; i++) {
for (j = 0; j < 4; j++) {
sum = (a[i][j] + a[i][j+1] + a[i][j+2]);
sum = sum + a[i+1][j+1];
sum = sum + (a[i+2][j] + a[i+2][j+1] + a[i+2][j+2]);
if (sum > lh) lh = sum;
}
}
System.out.print(lh);
}
}
Here you go..
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a[][] = new int[6][6];
int max = 0;
for (int i = 0; i < 6; i++) {
for (int j = 0; j < 6; j++) {
a[i][j] = in.nextInt();
}
}
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int sum = 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 (sum > max || (i == 0 && j == 0)) {
max = sum;
}
}
}
System.out.println(max);
}
I have a tutorial in class today with bubble sort and I've got an error I don't know how to fix.
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 8
at BubbleSorter.main(BubbleSorter.java:24)
It is not assessed but I would like to get through it to continue on.
Thank you. Below is my entire code.
public class BubbleSorter {
public static void main(String[] args)
{
int i;
int array[] = { 12, 9, 4, 99, 120, 1, 3, 10 };
System.out.println("Array Values before the sort:\n");
for (i = 0; i < array.length; i++)
System.out.print(array[i] + " ");
System.out.println();
System.out.println();
bubble_srt(array, array.length);
System.out.print("Array Values after the sort:\n");
for (i = 0; i < array.length; i++)
;
System.out.print(array[i] + " ");
System.out.println();
System.out.println("PAUSE");
}
private static void bubble_srt(int[] array, int length) {
int i, j, t = 0;
for (i = 0; i < length; i++) {
for (j = 1; j < (length - 1); j++) {
if (array[j - 1] > array[j]) {
t = array[j - 1];
array[j - 1] = array[j];
array[j] = t;
}
}
}
}
}
You have a small mistake:
This:
for (i = 0; i<array.length; i++);
System.out.print(array[i] + " ");
should be:
// v - Semicolon removed
for (i = 0; i<array.length; i++)
System.out.print(array[i] + " ");
Also change
for (j = 1; j < (length - 1); j++) {
to
for (j = 1; j < length; j++) {
You left out the last element of the Array!
Output
Array Values before the sort:
12 9 4 99 120 1 3 10
Array Values after the sort:
1 3 4 9 10 12 99 120
PAUSE
You need to change 2 lines
for (i = 0; i < array.length; i++)
;
System.out.print(array[i] + " ");
to
for (i = 0; i < array.length; i++)
System.out.print(array[i] + " ");
and
for (j = 1; j < (length - 1); j++) {
to
for (j = 1; j < (length); j++) {
On line 18 you need to get rid of the ; or put in brackets {