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Matrix manipulation: logic not fetching correct answer for higher order NXN matrix data

I came across below problem related to Matrix Manipulation.
problem statement
There is a NxN matrix,divided into N * N cells. Each cell has a predefined value. Which would be given as an input. Iteration has to happen K number of times which is also given in the test input. We have to make sure that we pick the optimum/min value of rows/columns at each iteration. Final output is the cumulative sum of optimum value saved at the end of each iteration.
Steps 1. Sum up the individual row and column and find the min sum of rows and columns, (it could be a row or a column, just need the minimum row or a column)
Step 2. Store the sum found above separately
Step 3.
Increment elements of the min. sum row or column. by 1
Repeat steps 1,2,3 from 1 to Kth value
add the sum at each iteration(specified in step2)
output is the sum obtained on on the Kth iteration.
Sample data
2 4
1 3
2 4
Output data
22
I was able to write a code (in java) and tested the same for some sample test cases. The output worked fine. The code works fine for sample data matrix of lower order, say, 2x2,4x4,even till 44x40 (that has less iteration). However, when the matrix size is increased to 100X100 (complex iteration), I see the expected output output values differ at 10s and hundreds place of the digit from the actual output and its random. Since I am not able to find a correct pattern of output vs input. Now, it is taking a toll on me to really debugging 500th loop to identify the issue. Is there any better way or approach to solve such problem related to huge matrix manipulation. Has anyone come across issues similar to this and solved it.
I am mainly interested in knowing the correct approach to solve given matrix problem. What Data structure to use in java. At present, I am using primitive DS and arrays int[] or long[] to solve this problem. Appreciate any help in this regard.

Which data structure?
What you need here is a data structure which allows you to efficiently query and update the minimum sum line. The most commonly used for this is a heap https://en.wikipedia.org/wiki/Heap_(data_structure).
For your purposes it's probably best to just implement the simplest kind, an array-based binary heap:
See here: https://en.wikipedia.org/wiki/Binary_heap
And here: http://courses.cs.washington.edu/courses/cse373/11wi/homework/5/BinaryHeap.java
..for implementation details.
Procedure:
Initialize your heap to size M + N where M, N are the number of rows and columns.
Before the loop, pre-compute the sum of each row and column, and add them as objects to the heap. Also add two arrays A, B which store the row and columon objects separately.
Now heapify the heap array with respect to the line sum attribute. This ensures the heap follows the criterion of the binary heap structure (parent always > children). Read the sources to find out more about how to implement this (quite easy for a fixed array)
For each iteration, look at the first element in the heap array. This is always the one with the smallest line sum. If this is a row object, then increment the sum attribute by N (no. of columns), and increment each object in B (list of columns) by 1. Do the same if it's a column.
After this, always heapify before the next iteration.
At the end, just return the first element's attribute.
Time complexity:
The original naive solution (looping through all columns and rows every time) is .
Using a heap, the heapify operation at each step is (for a binary heap).
This means the total complexity is , FAR smaller. The max term is to compensate for the fact that at each iteration it may be either rows or columns which are incremented.
As a side note, there are other heap structure types which have even better time complexity than the binary heap, e.g. binomial trees, Fibonacci heaps etc. These however are far more complicated, and have higher constant-factor overheads as a result. Thus for your project I feel they are not necessary, as many of them need phenomenal data set sizes to justify for the constant factor overhead.
Besides, they all support the same external operations as the binary heap, as defined by the Abstract Data Structure of Heap.
(heapify is an internal operation specific to the binary heap structure. Quite a few of the other ones are theoretically superior as they do this operation implicitly and "lazily")

O(KN + N*N) Solution:
You can just work with sum of columns and rows, and not store or manipulate them directly.
First sum all the columns and rows, in a 2*N array, first row being sum of columns, a[0][0] is sum of first column, a[0][1] is sum of second column, and second row is sum of rows, a[1][0] sum of first row, etc...
Then do the following for iterating:
Find min in array a .
Add it to the answer.
Add N to the min of row or column selected.
If the min is row add one to all cols and if it is a column add one to all rows.
If needed any further explanation, don't hesitate to comment.

I am doing like this for solving the above problem...
void matrixManipulation() throws IOException {
int N = Reader.nextInt();
int[][] matrix = new int[N][N];
int K = Reader.nextInt();
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
matrix[i][j] = Reader.nextInt();
}
}
// System.out.println("********Inital position**********");
// for (int i = 0; i < N; i++) {
// for (int j = 0; j < N; j++) {
// System.out.print(matrix[i][j]);
// }
// System.out.println();
// }
// System.out.println("********Inital position**********");
CalculateSum calculateSum = new CalculateSum();
int[] row = new int[N];
int[] row_clone = new int[N];
int[] col = new int[N];
int[] col_clone = new int[N];
int test =0;
for (int kk = 0; kk < K; kk++) {
row = calculateSum.calculateRowSum(matrix, N);
row_clone = row.clone();
/* just sort it either Arrarys sort or any other ---starts here*/
// for (int i = 1; i < row.length; i++) {
// row_orignial[i] = row[i];
// }
// Arrays.sort(row);
Node root1 = insert(null, row[0], 0, row.length);
for (int i = 1; i < row.length; i++) {
insert(root1, row[i], 0, row.length);
}
sortArrayInOrderTrvsl(root1, row, 0);
/* just sort it either Arrarys sort or any other ---ends here*/
col = calculateSum.calculateColumnSum(matrix, N);
col_clone = col.clone();
/* just sort it either Arrarys sort or any other ---starts here*/
// for (int i = 1; i < col.length; i++) {
// col_orignial[i] = col[i];
// }
// Arrays.sort(col);
Node root2 = insert(null, col[0], 0, col.length);
for (int i = 1; i < row.length; i++) {
insert(root2, col[i], 0, col.length);
}
sortArrayInOrderTrvsl(root2, col, 0);
/* just sort it either Arrary.sort or any other---ends here */
int pick = 0;
boolean rowflag = false;
int rowNumber = 0;
int colNumber = 0;
if (row[0] < col[0]) {
pick = row[0];// value
rowflag = true;
for (int i = 0; i < N; i++) {
if (pick == row_clone[i])
rowNumber = i;
}
} else if (row[0] > col[0]) {
pick = col[0];// value
rowflag = false;
for (int i = 0; i < N; i++) {
if (pick == col_clone[i])
colNumber = i;
}
} else if(row[0] == col[0]){
pick = col[0];
rowflag = false;
for (int i = 0; i < N; i++) {
if (pick == col_clone[i])
colNumber = i;
}
}
test= test + pick;
if (rowflag) {
matrix = rowUpdate(matrix, N, rowNumber);
} else {
matrix = columnUpdate(matrix, N, colNumber);
}
System.out.println(test);
// System.out.println("********Update Count"+kk+" position**********");
// for (int i = 0; i < N; i++) {
// for (int j = 0; j < N; j++) {
// System.out.print(matrix[i][j]);
// }System.out.println();
// }
// System.out.println("********Update Count"+kk+" position**********");
}
// System.out.println("********Final position**********");
// for (int i = 0; i < N; i++) {
// for (int j = 0; j < N; j++) {
// System.out.print(matrix[i][j]);
// }System.out.println();
// }
// System.out.println("********Final position**********");
// System.out.println(test);
}

Sort all Columns independently form a 2d array java

I need to have the columns organized in increasing order. Right now I have the following done but, it is sorting the rows and not columns.Any help would be nice, ive been working on this all day. Thanks.
public static double[][] sortColumns(double[][] m) {
double[][] sortedArray = new double[m.length][m.length];
for (int i = 0; i < m.length; i++) {
double[] temp = new double[m.length];
for (int j = 0; j < m.length; j++) {
temp[j] = m[j][i];
}
Arrays.sort(temp);
for (int j = 0; j < temp.length; j++) {
sortedArray[j][i] = temp[j];
}
}
return sortedArray;
}

If you change
temp[j] = m[j][i];
to
temp[j] = m[i][j];
and
sortedArray[j][i] = temp[j];
to
sortedArray[i][j] = temp[j];
then your existing algorithm will work fine. It just means you'll be copying columns to your "temporary sorting area" instead of rows.

In your current solution, you are just mistaking on indexes, just like David Wallace tells you in his answer. I propose you a different answer, enumerating the possible solutions of this problem.
You have at least 4 solutions :
instead of storing your data like you are currently doing it, use the transponate of your matrix
implement yourself an efficient sorting algorithm that takes a bi-dimensional array and the index of a column in argument
at each turn of you loop, fill an array with the current column, sort it, and copy it back (if you don't care about using some additional memory, do it). That is what you are currently trying to do
transponate your matrix, sort its lines, transponate it back (if you don't want to use too much memory, use this)
I prefer the last solution, which code is :
public static double[][] sortColumns(double[][] m) {
double[][] sortedArray = new double[m.length][m.length];
// compute the transponate of m
for (int i=0 ; i<m.length ; i++)
for (int j=0 ; j<m[i].length ; j++)
sortedArray[j][i] = m[i][j];
// sort the lines of the transponate
for (int i=0; i<sortedArray.length; i++)
Arrays.sort(sortedArray[i]);
// transponate back the result of the sorting
for (int i=0 ; i<sortedArray.length ; i++)
for (int j=i+1 ; j<sortedArray[i].length ; j++) {
double tmp = sortedArray[i][j];
sortedArray[i][j] = sortedArray[j][i];
sortedArray[j][i] = tmp;
}
return sortedArray;
}

When I look at your code I see the following line:
double[][] sortedArray = new double[m.length][m.length];
it doesn't look right to me.
you need to find length and breath of the array so i would do something like this:
length = m.length;
breath = m[0].length;
if i m not sure of all rows have same no of elements i may do that check by a for loop and initialize with the max.. wud lead to memory wastage but thats another demon to tame :)
next when we write m[x][y] x represents the rows and y represents the columns so when ur doing :
for (int j = 0; j < m.length; j++) {
temp[j] = m[j][i];
}
Arrays.sort(temp);
you are fetching all the values from a column i, assigning it to temp array and sorting the column.
hope that helps

Algorithm to modify a matrix and set columns and rows to zero

This is technically a code challenge.
I was asked an interesting question at an interview and am hoping for some insight as the best answer I could come up with was O(2n^2) - n-squared category, but still pretty much brute force.
Let's say you have a matrix that's M by N size ( an array of arrays (int[][]) )
1 2 4 3 1
0 5 3 7 7
5 8 9 2 8
6 7 0 8 9
If a cell contains a Zero, then set that entire row and column to zero.
Making the result:
0 2 0 3 1
0 0 0 0 0
0 8 0 2 8
0 0 0 0 0
What is the fastest and/or best way to do this?
My own answer is to iterate the entire array of arrays, keep track of rows and columns to zero out, and then zero them out.
public void zeroOut(int[][] myArray){
ArrayList<Integer> rowsToZero = new....
ArrayList<Integer> columnsToZero = new....
for(int i=0; i<myArray.length; i++){ // record which rows and columns will be zeroed
for(int j=0; j<myArray[i].length; i++){
if(myArray[i][j] == 0){
if(!rowsToZero.contains(i)) rowsToZero.add(i);
if(!columnsToZero.contains(j)) columnsToZero.add(j);
}
}
}
for(int row : rows){ // now zero the rows
myArray[row] = int[myArray.length];
}
for(int i=0; i<myArray.length; i++){
for(int column: columns){ // now zero the columns
myArray[i][column] = 0;
}
}
}
Is there a better algorithm? Is there a better data-structure to represent this matrix?

you can do this by taking two int but the only condition is the no of rows and cols should less than or equal to 32. You can do the same with greater than 32 but you have to take array of ints.
So the logic is :
take two ints i.e. row and col
traverse the matrix if matrix[i][j] = 0 than set the corresponding bits in the row and col
after traversal traverse again and set the matrix[i][j] = 0 if corresponding bit of either row or column is set.
The time complexity is same O(N^2) but it is memory efficient. Please find my code below .
Check whether the array[row][col] == 0 if 0 than set the corresponding bit in r and c.
int r = 0, c = 0;
for (int row = 0; row < 5; row++) {
for (int col = 0; col < 7; col++) {
if (array[row][col] == 0) {
r = r | (1<<row);
c = c | (1<<col);
}
}
}
Now if either of the bit is set than make the cell to 0.
for (int row = 0; row < 5; row++) {
for (int col = 0; col <7; col++) {
if (((c&(1<<col))!=0) || ((r&(1<<row))!=0)) {
array[row][col] = 0;
}
}
}

What about splitting the matrices into equal smaller matrice parts and calculate the deteriminant, so that you can predict that there is a zero within this matric part.
And then only use the brute force mechanism to this preselected matrices to determin
in wich row or column the zero is.
The determinant is just a suggestion maybe you can use some other kind of linear algebraic
algorithms and rules to predict a zero value
UPDATE:
if you use Quicksort concept to organize temporay every row. Then you have just to loop until the first none zero element occurs.
You need to remember during sorting process which column index was associated with the 0
Means
1 2 6 0 3
Quciksort (
0 1 2 4 6
When you remember the column index you now directly know which row to fill with 0 and which column,
Average of Quicksort ist O(n log n) Worstcase n * n
Maybe this already improves the overall complexisitiy.

Seems no one really came up with a significantly faster/better algorithm so far, so this one seems to be it. Thanks for your input everyone.
public void zeroOut(int[][] myArray){
ArrayList<Integer> rowsToZero = new....
ArrayList<Integer> columnsToZero = new....
for(int i=0; i<myArray.length; i++){ // record which rows and columns will be zeroed
for(int j=0; j<myArray[i].length; i++){
if(myArray[i][j] == 0){
if(!rowsToZero.contains(i)) rowsToZero.add(i);
if(!columnsToZero.contains(j)) columnsToZero.add(j);
}
}
}
for(int row : rows){ // now zero the rows
myArray[row] = int[myArray.length];
}
for(int i=0; i<myArray.length; i++){
for(int column: columns){ // now zero the columns
myArray[i][column] = 0;
}
}
}

I just came across this question and have developed a solution for it.
I am hoping I can get some feedback for the code about how it's better/worse and it's runtime.
Pretty new to all this :)
public static void zeroMatrix(int[][] arr1)
{
ArrayList<Integer> coord = new ArrayList<>();
int row = arr1.length;
int column = arr1[0].length;
for(int i=0; i < row; i++)
{
for(int j=0; j < column; j++)
{
if(arr1[i][j]==0)
{
coord.add((10*i) + j);
}
}
}
for(int n : coord)
{
int j=n%10;
int i=n/10; int k=0;
int l=0;
while(k<row)
{
arr1[k][j]=0;
k++;
}
while(l<column)
{
arr1[i][l]=0;
l++;
}
}
}

I have used HashMap.See if this can help in any way
import java.util.*;
class ZeroMatrix
{
public static void main(String args[])
{
int mat[][]=new int[][]{{0,2,3,4},{1,2,3,4},{1,2,3,4},{1,2,3,4},{1,2,3,0}};
HashMap<Integer,Integer> ht=new HashMap<Integer,Integer>();
for(int i=0;i<5;i++)
{
for(int j=0;j<4;j++)
{
if(mat[i][j]==0)
ht.put(i,j);
}
}
//Set the Respected Rows and colums to Zeros
Set set=ht.entrySet();
Iterator itr=set.iterator();
while(itr.hasNext())
{
Map.Entry m=(Map.Entry)itr.next();
int i=(Integer)m.getKey();
int k=(Integer)m.getValue();
for(int j=0;j<4;j++)
{
mat[i][j]=0;
}
for(int j=0;j<5;j++)
{
mat[j][k]=0;
}
}
//Printing the Resultant Zero Matrix
for(int i=0;i<5;i++)
{
for(int j=0;j<4;j++)
{
System.out.print(mat[i][j]);
}
System.out.println();
}
}
}

Can't figure out ArrayIndexOutofBoundsException error

Thanks in advance for your feedback. I can't make sense of an ArrayIndexOutofBoundsException error in the snippet of code below. Is it obvious to any of you?
public void run() {
int x = 10;
int y = 10;
double[][] Lx = new double[x][y];
double[][] Ly = new double[x][y];
for (int i=0; i<Lx.length; i++) {
for (int j=0; j<Lx[0].length; i++) {
Lx[j][i] = 2*j+i-1;
Ly[j][i] = Math.sqrt(3)*(i-1);
}
}
}

Your j for loop is incorrect in incrementing i, not j. Change
for (int j=0; j<Lx[0].length; i++) {
to
for (int j=0; j<Lx[0].length; j++) {
In addition, your array access indices are backwards. Change
Lx[j][i]
to
Lx[i][j]
and likewise with Ly. Otherwise you'll get a flipped 2D array.

With addition to #rgettman's answer
As in here your x and y is same so you will not have any issue with Lx[0].length. But it is good to use Lx[i].length.
Lx[0].length will always give the length or number of columns present in the 0th row. But Lx[i].length will give the length or number of columns present in the ith row.
So in case your x and y differs you will not again get ArrayIndexOutBoundException.

It should be j++ not i++ in the nested for-loop.
It's better not to use i and j for the index because they look too similar.
i and k are much more distinct than i and j

Logic Solving Algorithm for Sudoku (Java)

I decided to write a logic solving algorithm for my Sudoku application. What I wrote works for a limited amount of grid values, but then the recursion stops way too soon.
What my methods do:
addToThirdDimension(): A three dimensional array stores any possible values that can be put into the grid value at logicGrid[x][y]. This method refreshes the three dimensional array. It does this by testing values 1-9 in every grid index, and if it's valid, it adds that number to the array. If not, it sets that value to zero.
checkValues(): Checks how many possibilities are left in the three dimensional grid. It goes through the logicGrid and returns the number of non-zero values are in the grid.
checkSingleValue(int row, int col): Checks logicGrid[row][col] to see if there is one and only one value left in there (If there is one value left, it is the only possibility for the grid element at [row, col]). It returns the amount of non-zero values that are in that grid location.
getSingleValue(int row, int col): Returns the single number that's left in logicGrid[row][col]
immutableValues: A two dimensional boolean array that stores whether or not a specific grid element is immutable or not. If it is immutable, the solve method should not touch it.
public boolean solveWithLogic(){
addToThirdDimension();
if(checkValues() == 0){
return true;
}
for(int row = 0; row < 9; row++){
for(int col = 0; col < 9; col++){
if(!immutableValues[row][col]){
if(checkSingleValue(row, col) == 1){
sGrid[row][col] = getSingleValue(row, col);
setValues[row][col] = true;
addToThirdDimension();
}
}
}
}
if(checkValues() != 0){
solveWithLogic();
} else{
return true;
}
return false;
}
I cannot see where I am going wrong. After a certain number of tries, checkValues returns 0 even though there should be more possibilities. Here is the code for addToThirdDimension() as I am sure that if something is wrong, it is here.
sGrid is the main two-dimensional integer array that stores the values for the puzzle.
public void addToThirdDimension(){
logicGrid = new int[9][9][9];
for(int x = 0; x < 9; x++){
for(int y = 0; y < 9; y++){
for(int z = 0; z < 9; z++){
logicGrid[x][y][z] = z + 1;
}
}
}
int[][] temp1 = sGrid;
for(int row = 0; row < 9; row++){
for(int col = 0; col < 9; col++){
if(setValues[row][col]){
for(int i = 0; i < 9; i++){
logicGrid[row][col][i] = 0;
}
} else{
for(int i = 1; i <= 9; i++){
temp1[row][col] = i;
if(!isColumnValid(col, temp1) && !isRowValid(row, temp1) &&
!isQuadrantValid(row, col, temp1){
logicGrid[row][col][i-1] = 0;
}
}
}
temp1[row][col] = sGrid[row][col];
}
}
}
The code isn't too efficient at the moment. I want to get it working before I start minimizing solve times.

The first thing I would do is create a SudukoCell object that stores your possible values in it. Then create a SudukoBoard with a 2d array of SudukoCells. Also give it an array of SudukoAreas. One area for rows, one area for cols, and one area for blocks.
Add your suduko cells appropriately.
This will help you consolidate your legwork and prevent silly mistakes.
then every time you solve a number, you can go to the cells in each of its areas and remove the number you solved from them.