We Keep Coding

Palindrome Stack

I wrote this method to check to see if a words is a palindrome, but when it is a palindrome it keeps returning false
public boolean isPalindrome(String s){
int i;
int n = s.length();
Stack <Character> stack = new Stack <Character>();
for (i = 0; i < n/2; i++)
stack.push(s.charAt(i));
if (n%2 == 1)
i++;
while(!stack.empty( )) {
char c = stack.pop( );
if (c != s.charAt(i));
return false;
}
return true;
}

I'm not sure why you're not using { } brackets. Try to learn proper Java conventions early.
if (c != s.charAt(i)); // <- this semicolon is your problem
return false;
Is equivalent to:
if (c != s.charAt(i)) {
// Do nothing
}
// Do this no matter what
return false;
Furthermore, the logic on your for-loop may be flawed for similar reasons. Remove your semicolon, and better yet, practice always using brackets:
if (c != s.charAt(i)) {
return false;
}
#jhamon also points out that you never actually increment i in your while loop:
while(!stack.empty( )) {
char c = stack.pop( );
if (c != s.charAt(i)) {
return false;
}
i++;
}

Print the number of unique vowels in a string, Java

I need to find the number of distinct vowels. I came up with the code below but it can't make distinction between same vowels:
public static int count_Vowels(String str) {
str = str.toLowerCase();
int count = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u') {
count++;
}
}
return count;
}

I would start with five variables (one for each vowel) set to 0, iterate the characters in the input and set the corresponding variable to 1 if I find a match, and simply return the accumulated value of said variables. Like,
public static int count_Vowels(String str) {
int a = 0, e = 0, i = 0, o = 0, u = 0;
for (char ch : str.toLowerCase().toCharArray()) {
if (ch == 'a') {
a = 1;
} else if (ch == 'e') {
e = 1;
} else if (ch == 'i') {
i = 1;
} else if (ch == 'o') {
o = 1;
} else if (ch == 'u') {
u = 1;
}
}
return a + e + i + o + u;
}

You could use Set data structure and instead of incrementing the counter just add vowels to the set. At the end you can return just the size of the set.

The problem in your code is that you are not counting the distinct vowels, but all the vowels in the string. A Java-8 way to this:
public static int countDistinctVowels(String str) {
str = str.toLowerCase();
int count = (int) str.chars() // get IntStream of chars
.mapToObj(c -> (char) c) // cast to char
.filter(c -> "aeiou".indexOf(c) > -1) // remove all non-vowels
.distinct() // keep the distinct values
.count(); // count the values
return count;
}
Also use proper Java naming conventions: countDistinctVowels, no count_Distinct_Vowels.

there's definitely an issue here with this counting. at the very least. you should rethink this:
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u')
count++;

You could use the method contains
public static int count_Vowels(String str) {
str = str.toLowerCase();
int count = 0;
count += string.contains("a") ? 1 : 0;
count += string.contains("e") ? 1 : 0;
count += string.contains("i") ? 1 : 0;
count += string.contains("o") ? 1 : 0;
count += string.contains("u") ? 1 : 0;
return count;
}

I made explanations in the comments to the code:
public static int count_Vowels(String str) {
str = str.toLowerCase();
Set<Character> setOfUsedChars = new HashSet<>(); // Here you store used vowels
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u') { // if currently checked character is vowel...
setOfUsedChars.add(str.charAt(i)); // add this vowel to setOfUsedChars
}
}
return setOfUsedChars.size(); // size of this sets is a number of vowels present in input String
}

static void vow(String input){
String output=input.toLowerCase();
int flaga=0,flage=0,flagi=0,flago=0,flagu=0;
for(int i=0;i<input.length();i++) {
if((output.charAt(i))=='a' && flaga==0) {
System.out.print(input.charAt(i)+" ");
flaga++;
}
if(output.charAt(i)=='e' && flage==0) {
System.out.print(input.charAt(i)+" ");
flage++;
}
if(output.charAt(i)=='i' && flagi==0) {
System.out.print(input.charAt(i)+" ");
flagi++;
}
if(output.charAt(i)=='o' && flago==0) {
System.out.print(input.charAt(i)+" ");
flago++;
}
if(output.charAt(i)=='u' && flagu==0) {
System.out.print(input.charAt(i)+" ");
flagu++;
}
}
}

public static void main(String args[]) {
String sentence;
int v=0,c=0,ws=0;
Scanner sc= new Scanner(System.in);
sentence = sc.nextLine();
sc.close();
sentence.toLowerCase();
String res="";
for(int i=0;i<sentence.length();i++) {
if(sentence.charAt(i)=='a'||sentence.charAt(i)=='e'||sentence.charAt(i)=='i'||sentence.charAt(i)=='o'||sentence.charAt(i)=='u') {
if(res.indexOf(sentence.charAt(i))<0) {
res+=sentence.charAt(i);
v++;
}//System.out.println(res.indexOf(sentence.charAt(i)));
}
else if(sentence.charAt(i)==' ')
ws++;
else c++;
}
System.out.println(res);
System.out.println("no of vowels: "+v+"\n"+"no of consonants: "+c+"\n"+"no of
white spaces: "+ws);
}

You can use this Method to Find Count of Distinct vowels.
public static int count_Vowels(String str) {
char[] c = str.toLowerCase().toCharArray();
int Counter=0;
String NewString="";
for(int i=0;i<c.length;i++){
String tempString="";
tempString+=c[i];
if(!NewString.contains(tempString) && (c[i]=='a'||c[i]=='e'||c[i]=='i'||c[i]=='o'||c[i]=='u')){
Counter++;
NewString+=c[i];
}
}
return Counter;
}

Here is a solve for this problem without using objects. It's a crude but great solve for beginners who encounter this problem with limited js experience.
How to count unique vowels is a string;
function hasUniqueFourVowels(str){
let va = 0
let ve = 0
let vi = 0
let vo = 0
let vu = 0
let sum = 0
for(let i = 0; i < str.length; i++){
let char = str[i];
if(char === "i"){
vi = 1
}
if(char === "e"){
ve = 1
}
if(char === "a"){
va = 1
}
if(char === "o"){
vo = 1
}
if(char === "u"){
vu = 1
}
sum = va + vi + vo + ve + vu
if (sum >= 4){
return true
}
}
return false
}

#Test
public void numfindVoweles(){
String s="aeiouaedtesssiou";
char a[]=s.toCharArray();
HashMap<Character,Integer> hp= new HashMap<Character, Integer>();
for(char ch:a){
if(hp.containsKey(ch) && (ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u')){
hp.put(ch,hp.get(ch)+1);
}
else if(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u'){
hp.put(ch,1);
}
}
System.out.println(hp);
}

Analyze a string of characters in java for format A^nB^n

I have a string of characters with As and Bs that I need to analyze for a Language A^nB^n. I can use the following code to work most of the time but when there is a letter that is not an "A" or "B" it may still return true, for example: AABACABAA should not be true, but it says it is. AABB is true; AABBAABB is not true. I have to use stacks and am not allowed to use counting.
public static boolean isL2(String line){
// set up empty stacks
Stack L2Stack = new Stack();
// initialize loop counter
int i = 0;
int n = line.length();
/* Push all 'A's to a_stack */
while ((i < line.length()) && (line.charAt(i) == 'A')) {
char ch = line.charAt(i);
L2Stack.push(ch);
i++;
}
/* Pop an 'A' for each consecutive 'B' */
while ((i < line.length()) && (line.charAt(i) == 'B')) {
if (!L2Stack.empty()){
L2Stack.pop();
i++;
}
else
return false;
}
if (i == n && !L2Stack.empty()){
return false; // more As than Bs
}
if (i != n && L2Stack.empty()){
return false; //more Bs than As
}else
return true;
}

if (i != n && L2Stack.empty()) {
return false; //more Bs than As
}
Should be
if (i != n) {
return false;
}
Since if you haven't finished reading all the characters, you can't return true, regardless of whether or not the stack is empty.
I'm assuming that AAABBBA should return false.
That change would also handle illegal characters.

How to check String using stack

I have some specific task. We have String like "(()[]<>)" or something familiar with this. A question in my interview qustion was how check either String is correct or incorrect. For example: "()[]<>" - true, "([)" - false, "[(])" - false, "([<>])" - true. Thank you guys very much!
I can' t take what's wrong with my code.
Thank a lot guys!!!
Please help!
import java.util.Stack;
public class Test {
public static void main(String[] args) {
String line = "(<>()[])";
Test test = new Test();
boolean res = test.stringChecker(line);
System.out.println(res);
}
public boolean stringChecker(String line){
boolean result = false;
char letter = '\u0000';
char[] arr = line.toCharArray();
Stack<Character> stack = new Stack();
for (int i = 0; i < arr.length; i++) {
if (arr[i] == '(' || arr[i] == '[' || arr[i] == '<') {
stack.push(arr[i]);
}
if(arr[i] == ')' || arr[i] == ']' || arr[i] == '>'){
if(stack.peek() == arr[i]){
result = true;
stack.pop();
}
}
}
return result;
}
}

(0) You are pushing < ( and { but in your peek you are checking for >, ), and }
(1) You are starting with result false and setting it to true on the first successful match. Instead you should start with result true and set it to false on the first failed match.
(2) You should check that the stack is empty when you have run out of characters.
(3) You should check for the stack being empty before you peek.
(4) You might want to check for characters that are not expected.

In addition to #TheodoreNorvell 's explanation here is how an implementation could look like
public boolean stringChecker(String input) {
boolean result = true;
char[] arr = input.toCharArray();
Stack<Character> stack = new Stack<>();
try {
for (int i = 0; result && i < arr.length; i++) {
if (arr[i] == '(' || arr[i] == '[' || arr[i] == '<') {
stack.push(arr[i]);
} else if(arr[i] == ')') {
Character c = stack.pop();
result = c.equals('(');
} else if(arr[i] == ']') {
Character c = stack.pop();
result = c.equals('[');
} else if(arr[i] == '>') {
Character c = stack.pop();
result = c.equals('<');
} else {
// found some char that is not allowed
// here it is not just ignored,
// it invalidates the input
result = false;
}
}
// when the teher is not more chars in the array
// the stack has to be empty
result = result && stack.isEmpty() ;
} catch(EmptyStackException e) {
// found a closing bracket in the array
// but there is nothing on the stack
result = false;
}
return result;
}
#Test
public void stringChecker() {
Assert.assertTrue(stringChecker("[]"));
Assert.assertTrue(stringChecker("[(<>)]"));
Assert.assertFalse(stringChecker("([<>)]"));
Assert.assertFalse(stringChecker(">"));
// invalid char
Assert.assertFalse(stringChecker("<[]e>"));
// stack is not empty
Assert.assertFalse(stringChecker("("));
}
Note that in such a situation a switch-case statement is more elegant than if-else if-else.

Testing parentheses in a equation using stack java

I'm writing a program that will take in an equation and check if all the parentheses line up and it will output if it is good or not.
For Ex: (3+4) is good
((3*8) is NOT Good
I'm not allowed to use java's built in push() pop() methods ext..
I have to make my own which I think I got....I think!
The problem I'm having is in the Test() method.
First I'm not sure how to write the while loop like:
while(there are still characters)
Anyway the output I'm getting is: stack is empty -1
Any help is appreciated. I'm one of the slower program learners and I couldn't be trying any harder. Thanks.
Here's what I got:
public class Stacked {
int top;
char stack[];
int maxLen;
public Stacked(int max) {
top = -1;
maxLen = max;
stack = new char[maxLen];
}
public void push(char item) {
top++;
stack[top] = item;
}
public int pop() {
//x = stack[top];
//top = top - 1;
top--;
return stack[top];
}
public boolean isStackEmpty() {
if(top == -1) {
System.out.println("Stack is empty" + top);
return true;
} else
return false;
}
public void reset() {
top = -1;
}
public void showStack() {
System.out.println(" ");
System.out.println("Stack Contents...");
for(int j = top; j > -1; j--){
System.out.println(stack[j]);
}
System.out.println(" ");
}
public void showStack0toTop() {
System.out.println(" ");
System.out.println("Stack Contents...");
for(int j=0; j>=top; j++){
System.out.println(stack[j]);
}
System.out.println(" ");
}
//}
public boolean test(String p ){
boolean balanced = false;
balanced = false;
//while ( )
for(char i = '('; i < p.length(); i++ ){
push('(');
}
for (char j = ')'; j < p.length(); j++){
pop();
}
if (isStackEmpty()) {
balanced = true;
//return balanced;
}
return balanced;
}
public static void main(String[] args) {
Stacked stacks = new Stacked(100);
String y = new String("(((1+2)*3)");
stacks.test(y);
//System.out.println(stacks.test(y));
}
}
Now I'm getting somewhere. I need to be pointed in the right direction again. Thanks everyone this helped big time. I still have a lot more to do but this is good for now. Eventually I need to create a two more methods: one "infix to postfix" and the other "evaluating postfix" and at the end I'll need to read in answers from a text file instead of putting my own into the main method. Thanks again much appreciated.

Unless you need to actually evaluate the equation, a stack is too complicated a solution here. You simply need a counter:
int openParentheses = 0;
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) == '(') {
openParentheses++;
} else if (p.charAt(i) == ')') {
openParentheses--;
}
//check if there are more closed than open
if (openParentheses < 0) {
return false;
}
}
if (openParentheses == 0) {
return true;
} else {
return false;
}
If you absolutely must use stacks, use this:
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) == '(') {
push('x'); //doesn't matter what character you push on to the stack
} else if (p.charAt(i) == ')') {
pop();
}
//check if there are more closed than open
if (stackIsEmpty()) {
return false;
}
}
if (isStackEmpty()) {
return true;
} else {
return false;
}

I agree with Griff except that you should include another check if you didn't have more closed parentheses than open. (x*y))( is not a valid entry.
int openParentheses = 0;
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) == '(') {
openParentheses++;
} else if (p.charAt(i) == ')') {
openParentheses--;
}
if(openParentheses<0)
return false;
}
if (openParentheses == 0) {
return true;
} else {
return false;
}

You may be required to use a stack, but this could be done with a simple counter. This will show you a how to iterate over the characters of a String:
boolean test(String p) {
int balance = 0;
for (int idx = 0; idx < p.length(); ++idx) {
char ch = p.charAt(idx);
if (ch == '(')
++balance;
else if (ch == ')')
--balance;
if (balance < 0)
return false;
}
return balance == 0;
}
Of course, you could replace the increment and decrement with pushes and pops, respectively, on a stack.

For parsing you can use a for loop over the index and address the character of the string at the certain index.
But you actually do not need a stack, an integer variable openBraces is sufficient:
initialize with 0
for '(' you increment the variable one
for ')' you decrement the variable one
if openBraces is <0, you immediately give an error
if at the end openBraces is not equal to 0, you give an error.
Since you should do your homework yourself, I did not post source code, only explanations ;)

I think you just need this --
for ( int i = 0 ; i < p.length(); i++ ) {
char c = p.charAt(i);
if ( c == '(' )
push('(');
else if ( c == ')' ) {
if ( isStackEmpty() ) {
// Return error here because of unbalanced close paranthesis
}
pop();
}
else {
// do nothing
}
}
You CAN use a stack if you must, but considering how simplistic this is, you just need a counter that you increment and decrement and check for 0 at the end.
If you do use a counter, you should check after every decrement if the value is less than 0. If so, throw an error.
Edited based on Ryan/Dave Ball's comments.

It could be done like this:
String equation = "(2+3))";
Integer counter = 0;
//while(equation)
for(int i=0; i<equation.length();i++)
{
if(equation.charAt(i)=='(')
{
counter++;
}
else
if(equation.charAt(i)==')')
{
counter--;
}
}
if(counter == 0)
{
System.out.println("Is good!!!");
}
else
{
System.out.println("Not good!!!");
}
}