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Combine 2 2D arrays that both contain chars with java

I have to write a code as a task for my university and we have to recreate Minesweeper with java and it has to be runned in the command line.
For the matchfield we have to make an array that looks in the end like this picture:
Example how it sould look in the end
And to choose the field we have to use the scanner.
For example if you want to chose field C3, you have to type into the scanner C3.
At the moment im struggleing a little bit with the field.
I had 2 ideas but both didn't work out very well.
in the first try i tried to create everything with 2 for loops and 1 array but my problem was that I couldn't add 2 charrs, so I had the chars 0 to 9 and the charrs A to J.
In the second try I created 3 array, one with the numbers 0 to 9 and anothe array A to J and in the third array i wanted to combine both arrays. And now I'm wondering if this it's possible if I can acctually combine them in the way I want and if it's possible could somebody give me some help?
import java.util.Scanner;
import java.util.Arrays;
public class Minesweeper {
public static void main (String[] args) {
char c = 'A';
char d = '0';
char e = '9';
char f = 'J';
char[][] feldz = new char[11][11];
char[][] feldb = new char[11][11];
char[][] feld = new char[11][11];
for (int i = 0; i < 11; i++) {
for (int j = 0; j < 11; j++) {
if (i == 0 && j == 0) {
feldz[i][j] = ' ';
System.out.print(feldz[i][j] + " |");
}
if (d > e) {
d = '0';
}
if (d <= e && i > 0){
feldz[i][j] = d;
System.out.print(feldz[i][j] + " |");
}
if (i > 0 && j == 10) {
d++;
}
}
System.out.println("");
}
for (int i = 0; i < 11; i++) {
for (int j = 0; j < 11; j++) {
if (i == 0 && j == 0) {
feldb[i][j] = ' ';
System.out.print(feldb[i][j] + " |");
}
if (i > 0 && j == 0){
feldb[i][j] = ' ';
System.out.print(feldb[i][j] + " |");
}
if (c > f) {
c = 'A';
}
if(c <= f && j > 0){
feldb[i][j] = c;
System.out.print(feldb[i][j] + " |");
c++;
}
if (j == 10){
System.out.println("");
}
}
}
}
}

You don't actually need array to print the maze , nested loops is enough for that. 2d Array is only required to store the input. Please try the below code:
int size = 10;
int [][] maze = new int[size][size];
while (true){
System.out.print(' ');
for (int i = 0; i < size; i++) {
System.out.print('|');
System.out.print((char) ('A' + i));
}
for (int i = 0; i < size; i++) {
System.out.println("");
System.out.print(i);
for (int j = 0; j < size; j++) {
System.out.print('|');
if(maze[i][j] > 0) {
System.out.print(maze[i][j]);
} else {
System.out.print(' ');
}
}
}
int row = -1;
int col = -1;
System.out.println("\nEnter CoOrdinates");
Scanner scanner = new Scanner(System.in);
String input = scanner.nextLine();
if(input.length() == 2) {
char charAt = input.charAt(0);
if(charAt >= 'A' && charAt <= 'A'+size-1) {
col = charAt-'A';
}
charAt = input.charAt(1);
if(charAt >= '0' && charAt <= '0'+size-1) {
row = charAt-'0';
}
if(row != -1 && col != -1) {
System.out.println("Enter Action");
input = scanner.nextLine();
int action = Integer.parseInt(input);
maze[row][col] = action;
} else {
System.out.println("Incorrect Input");
}
}
}

Printing only NON-BOUNDARY and CORNER elements of an (n*n) array

I am to write a program that prints ONLY the NON-BOUNDARY AND CORNER elements of an (n*n) array, for my assignment, and this is the main part of the code:
The output I am getting is this:
As you can see, the non-boundary elements (6,7,10,11) are not in their correct positions, which I believe, is because of incorrect printing of tab spaces within the loop. (My code is totally a mess) I would like some help or suggestions to fix this. Thanks!

I generally find that flattening things (the if-conditions in particular), and putting conditions into boolean-returning methods helps. Try something like
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++ {
if (isCorner(i,j,n) || !isEdge(i,j,n)) {
//...
} else {
//...
}
}
System.out.println();
}
where isCorner(i,j,n) and isEdge(i,j,n) are defined something like
public boolean isCorner(int row, int column, int gridSize) {
//...
}

A you got a solution, just missing spaces, I'll add some smart things:
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
boolean visible = (i % (n - 1) == 0) == (j % (n - 1) == 0);
if (visible) {
System.out.printf(" %4d", a[i][j]);
} else {
System.out.print(" ");
}
}
System.out.println();
}
No longer any problems with tabs "\t", though I used spaces here.
Keep it simple, too many cases just cause problems - as you experienced.
The trick here is to consider whether to print or not. Hence I started with
a variable visible.
The border condition
i == 0 || i == n - 1
could also be written with modulo as
i % (n - 1) == 0
If this is "too smart", hard to grasp reading:
boolean iOnBorder = i % (n - 1) == 0;
boolean jOnBorder = j % (n - 1) == 0;
boolean visible = iOnBorder == jOnBorder;
The "X" pattern checks the _equivalence of i-on-border and j-on-border.
For the rest: formatted printf allows padding of a number.

Try this i have optimized your if condition
No need to again check for i == 0 or i == n-1
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(i==0 || i==n-1){
if(j==0 || j==n-1){
System.out.print(a[i][j]);
}
}else{
if(j != 0 && j!= n-1){
System.out.print(a[i][j]);
}
}
System.out.print("\t");
}
System.out.println();
}

Just gave a try in case you might find it helpful.
public static void main(String[] args) throws ParseException {
int[][] ar = new int[4][4];
int[] input = new int[]{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
int pointer=0;
int imin=0,jmin=0,imax=3,jmax=3;
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
ar[i][j]=input[pointer];
pointer++;
}
}
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
if(!((i==imax && j==jmin)||(i==imin && j==jmax)||i==j) && //For skipping the corners
(i == imin || j == jmin || i == imax || j == jmax)){// Not to print the borders
continue;
}
else {
System.out.println(ar[i][j]);
}
}
}
}

Find all positive integers with at least one 7 and 9 under 1,000,000 but how to fix duplicates?

So, for class I have to code a program that determines how many positive integers are
1) Under 1,000,000
2) Have at least one 7 and a 9 in the number
3) Has to be done with the brute-force method.
While the answer is supposed to be 199,262, I keep getting 228530 due to duplicates, can someone take a look to see where I went wrong here? Thanks!
Similar problem but not the same: Java - numbers with at least one 7 and one 9 in its digit
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i<1000000; i++) //Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); //sets a nmber to a string
int length = number2.length() -1; //length goes up to the last character 0-j
for (int j= 0; j <= length; j++) //looking for the first 7 or 9 in string
{
char a = number2.charAt(j); //sets char to the next "letter"
if (a == '7' && oneSeven != true) //if the number is a 7 and there isnt already a seven
{
oneSeven = true; //now there is a seven,
for (int k = j+1; k <= length; k++) //checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9')
{
sevNine = true;
}
}
}
else if (a == '9' && oneNine != true)
{
oneNine = true;
for (int l = j+1; l <= length; l++)
{
char b = number2.charAt(l);
if (b == '7')
{
sevNine = true;
}
}
}
if (sevNine == true)
{
counter++;
sevNine = false;
System.out.println(number2);
}
}
}
System.out.println(counter);

In Java 8 you can try:
public static void main(String[] args) {
final long count = IntStream.rangeClosed(0, 10_00_000)
.filter(i -> String.valueOf(i).contains("7") && String.valueOf(i).contains("9"))
.count();
System.out.println(count);
}

You are not breaking out of the loop once the sevNine is set to true and you increment the counter, so it keeps iterating over each digit on the same number even if it has already included the number... Just add a break statement to exit the for loop iterating over each digit once you increment the counter...
Here's the code.
public static void main(String[] args) {
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i < 1000000; i++) // Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); // sets a nmber to a string
int length = number2.length() - 1; // length goes up to the last character 0-j
for (int j = 0; j <= length; j++) // looking for the first 7 or 9 in string
{
char a = number2.charAt(j); // sets char to the next "letter"
if (a == '7' && oneSeven != true) // if the number is a 7 and there isnt already a seven
{
oneSeven = true; // now there is a seven,
for (int k = j + 1; k <= length; k++) // checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9') {
sevNine = true;
}
}
} else if (a == '9' && oneNine != true) {
oneNine = true;
for (int l = j + 1; l <= length; l++) {
char b = number2.charAt(l);
if (b == '7') {
sevNine = true;
}
}
}
if (sevNine == true) {
counter++;
sevNine = false;
System.out.println(number2);
break;
}
}
}
System.out.println(counter);
}
If you run with the break statement, you should get 199262 as the resulting number.

minimum operations required to make the longest character interval equal to K

I was asked this question in an contest.
Given a string containing only M and L, we can change any "M" to "L" or any "L" to "M". The objective of this function is to calculate the minimum number of changes we have to make in order to achieve the desired longest M-interval length K.
For example, given S = "MLMMLLM" and K = 3, the function should return 1. We can change the letter at position 4 (counting from 0) to obtain "MLMMMLM", in which the longest interval of letters "M" is exactly three characters long.
For another example, given S = "MLMMMLMMMM" and K = 2, the function should return 2. We can, for example, modify the letters at positions 2 and 7 to get the string "MLLMMLMLMM", which satisfies the desired property.
Here's what I have tried till now, but I am not getting correct output:
I am traversing the string and whenever longest char count exceeds K, I'm replacing M with L that point.
public static int solution(String S, int K) {
StringBuilder Str = new StringBuilder(S);
int longest=0;int minCount=0;
for(int i=0;i<Str.length();i++){
char curr=S.charAt(i);
if(curr=='M'){
longest=longest+1;
if(longest>K){
Str.setCharAt(i, 'L');
minCount=minCount+1;
}
}
if(curr=='L')
longest=0;
}
if(longest < K){
longest=0;int indexoflongest=0;minCount=0;
for(int i=0;i<Str.length();i++){
char curr=S.charAt(i);
if(curr=='M'){
longest=longest+1;
indexoflongest=i;
}
if(curr=='L')
longest=0;
}
Str.setCharAt(indexoflongest, 'M');
minCount=minCount+1;
}
return minCount;
}

There are 2 parts to this algorithm as we want to get the longest character interval equal to K.
We already have a interval >= K so now we need to appropriately change some characters so we greedily change every (k + 1) th character and again start counting from 0.
Now if the interval was less than K I will need to run a sliding window over the array. While running this window I am basically considering converting all L's to M's in this window of length K. But this comes with a side effect of increasing the length of the interval as there could be K's outside so this variable (int nec) keeps track of that. So now I have to also consider converting the 2 possible M's outside the (K length) window to L's.
Here's the complete runnable code in C++. Have a good day.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector <int> vi;
typedef pair<int, int> ii;
int change(string s, int k) {
// handling interval >= k
bool flag = false;
int ans = 0;
int cnt = 0;
for(int i=0; i<s.size(); i++) {
if(s[i] == 'M') cnt++;
else cnt = 0;
if(cnt == k) flag = true;
if(cnt > k) s[i] = 'L', ans++, cnt = 0;
}
if(flag) return ans;
// handling max interval < k
// If the interval is too big.
if(k > s.size()) {
cerr << "Can't do it.\n"; exit(0);
}
// Sliding window
cnt = 0;
for(int i=0; i<k; i++) {
if(s[i] == 'L') cnt++;
}
ans = cnt + (s[k] == 'M'); // new edit
int nec = 0; // new edit
for(int i=k; i<s.size(); i++) {
if(s[i-k] == 'L') cnt--;
if(s[i] == 'L') cnt++;
nec = 0;
if(i-k != 0 && s[i-k-1] == 'M')
nec++;
if(i < s.size()-1 && s[i+1] == 'M')
nec++;
ans = min(ans, cnt + nec);
}
return ans;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
string s;
int k;
cin >> s >> k;
int ans = change(s, k);
cout << ans << "\n";
return 0;
}

int
process_data(const char *m, int k)
{
int m_cnt = 0, c_cnt = 0;
char ch;
const char *st = m;
int inc_cnt = -1;
int dec_cnt = -1;
while((ch = *m++) != 0) {
if (m_cnt++ < k) {
c_cnt += ch == 'M' ? 0 : 1;
if ((m_cnt == k) && (
(inc_cnt == -1) || (inc_cnt > c_cnt))) {
inc_cnt = c_cnt;
}
}
else if (ch == 'M') {
if (*st++ == 'M') {
/*
* losing & gaining M carries no change provided
* there is atleast one L in the chunk. (c_cnt != 0)
* Else it implies stretch of Ms
*/
if (c_cnt <= 0) {
int t;
c_cnt--;
/*
* compute min inserts needed to brak the
* stretch to meet max of k.
*/
t = (k - c_cnt) / (k+1);
dec_cnt += t;
}
}
else {
ASSERT(c_cnt > 0, "expect c_cnt(%d) > 0", c_cnt);
ASSERT(inc_cnt != -1, "expect inc_cnt(%d) != -1", inc_cnt);
/* Losing L and gaining M */
if (--c_cnt < inc_cnt) {
inc_cnt = c_cnt;
}
}
}
else {
if (c_cnt <= 0) {
/*
* take this as a first break and restart
* as any further addition of M should not
* happen. Ignore this L
*/
st = m;
c_cnt = 0;
m_cnt = 0;
}
else if (*st++ == 'M') {
/* losing m & gaining l */
c_cnt++;
}
else {
// losing & gaining L; no change
}
}
}
return dec_cnt != -1 ? dec_cnt : inc_cnt;
}

Corrected code:
int
process_data(const char *m, int k)
{
int m_cnt = 0, c_cnt = 0;
char ch;
const char *st = m;
int inc_cnt = -1;
int dec_cnt = -1;
while((ch = *m++) != 0) {
if (m_cnt++ < k) {
c_cnt += ch == 'M' ? 0 : 1;
if ((m_cnt == k) && (
(inc_cnt == -1) || (inc_cnt > c_cnt))) {
inc_cnt = c_cnt;
}
}
else if (ch == 'M') {
if (*st++ == 'M') {
/*
* losing & gaining M carries no change provided
* there is atleast one L in the chunk. (c_cnt != 0)
* Else it implies stretch of Ms
*/
if (c_cnt <= 0) {
c_cnt--;
}
}
else {
ASSERT(c_cnt > 0, "expect c_cnt(%d) > 0", c_cnt);
ASSERT(inc_cnt != -1, "expect inc_cnt(%d) != -1", inc_cnt);
/* Losing L and gaining M */
if (--c_cnt < inc_cnt) {
inc_cnt = c_cnt;
}
}
}
else {
if (c_cnt <= 0) {
/*
* compute min inserts needed to brak the
* stretch to meet max of k.
*/
dec_cnt += (dec_cnt == -1 ? 1 : 0) + ((k - c_cnt) / (k+1));
/*
* take this as a first break and restart
* as any further addition of M should not
* happen. Ignore this L
*/
st = m;
c_cnt = 0;
m_cnt = 0;
}
else if (*st++ == 'M') {
/* losing m & gaining l */
c_cnt++;
}
else {
// losing & gaining L; no change
}
}
}
if (c_cnt <= 0) {
/*
* compute min inserts needed to brak the
* stretch to meet max of k.
*/
dec_cnt += (dec_cnt == -1 ? 1 : 0) + ((k - c_cnt) / (k+1));
}
return dec_cnt != -1 ? dec_cnt : inc_cnt;
}

Cannot read repeating characters

I'm writing a code to read a string and count sets of repeating
public int countRepeatedCharacters()
{
int c = 0;
for (int i = 1; i < word.length() - 1; i++)
{
if (word.charAt(i) == word.charAt(i + 1)) // found a repetition
{
if ( word.charAt(i - 1) != word.charAt(i)) {
c++;
}
}
}
return c;
}
If I try the input
aabbcdaaaabb
I should have 4 sets of repeat decimals
aa | bb | aaaa | bb
and I know I'm not reading the first set aa because my index starts at 1. I tried fixing it around to read zero but then I tr to fix the entire loop to work with the change and I failed, is there any advice as to how to change my index or loop?

Try this code:
public int countRepeatedCharacters(String word)
{
int c = 0;
Character last = null;
bool counted = false;
for (int i = 0; i < word.length(); i++)
{
if (last != null && last.equals(word.charAt(i))) { // same as previous characted
if (!counted) { // if not counted this character yet, count it
c++;
counted = true;
}
}
else { // new char, so update last and reset counted to false
last = word.charAt(i);
counted = false
}
}
return c;
}
Edit - counted aaaa as 4, fixed to count as 1

from what I understood from your question, you want to count number of repeating sets, then this should help.
for (int i = 0; i < word.length()-1; i++){
if (word.charAt(i) == word.charAt(i + 1)){ // found a repetition
if (i==0 || word.charAt(i - 1) != word.charAt(i)) {
c++;
}
}
}

Try this----
public int countRepeatedCharacters()
{
int c = 0,x=0;
boolean charMatched=false;
for (int i = 0; i < word.length(); i++)
{
if(i==word.length()-1)
{
if (word.charAt(i-1) == word.charAt(i))
c++;
break;
}
if (word.charAt(i) == word.charAt(i + 1)) // found a repetition
{
charMatched=true;
continue;
}
if(charMatched==true)
c++;
charMatched=false;
}
return c;
}

Try this method. It counts the sets of repeating charactors.
public static void main(String[] args) {
String word = "aabbcdaaaabbc";
int c = 0;
for (int i = 0; i < word.length()-1; i++) {
// found a repetition
if (word.charAt(i) == word.charAt(i + 1)) {
int k = 0;
while((i + k + 1) < word.length()) {
if(word.charAt(i+k) == word.charAt(i + k + 1)) {
k++;
continue;
}
else {
break;
}
}
c++;
i+=k-1;
}
}
System.out.println(c);
}

You can try something like this:-
public static void main(String str[]) {
String word = "aabbcdaaaabbc";
int c = 1;
for (int i = 0; i < word.length() - 1; i++) {
if (word.charAt(i) == word.charAt(i + 1)) {
c++;
} else {
System.out.println(word.charAt(i)+ " = " +c);
c = 1;
}
}
System.out.println(word.charAt(word.length()-1)+ " = " +c);
}
You can modify this as per your needs, by removing the sysouts and other stuffs.

Using length() -1 is causing you to not consider the last character in your calculations.
This is causing you to lose the last repetitive character.
Finally, I would have done this as follows:
public static int countRepeatedCharacters(String word)
{
boolean withinRepeating = false;
int c = 0;
for (int i = 1; i < word.length(); i++)
{
if (!withinRepeating && (withinRepeating = word.charAt(i) == word.charAt(i - 1)))
c++;
else
withinRepeating = word.charAt(i) == word.charAt(i - 1);
}
return c;
}