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How to eliminate a nested for loop to get O(n)?

In the code below I have a double for loop resulting in a time complexity of O^2 in method getResponse(). This code prompts the user for a 10 integer sequence string and an uppercase sensitive pin. It then converts the pin to numbers on a phone pad ie. [ABC] --> 2, [DEF] --> 3. Lastly a response array is generated with each digit of the new phone pin corresponding to indexes of sequence. So input "0123456789","HAM", response = "426"
import java.util.Scanner;
public class Test {
public static final int SEQ_DIGITS = 10;
public static final String ERR_SEQ = "Invalid sequence";
public static final String ERR_PIN = "Invalid PIN";
public static int letterToPhone(char c) {
int phoneNumber = 0;
if (Character.toString(c).matches("[ABC]")) {
phoneNumber = 2;
} else if (Character.toString(c).matches("[DEF]")) {
phoneNumber = 3;
} else if (Character.toString(c).matches("[GHI]")) {
phoneNumber = 4;
} else if (Character.toString(c).matches("[JKL]")) {
phoneNumber = 5;
} else if (Character.toString(c).matches("[MNO]")) {
phoneNumber = 6;
} else if (Character.toString(c).matches("[PQRS]")) {
phoneNumber = 7;
} else if (Character.toString(c).matches("[TUV]")) {
phoneNumber = 8;
} else if (Character.toString(c).matches("[WXYZ]")) {
phoneNumber = 9;
}
return phoneNumber;
}
public static int[] getResponse(String pin, int[] values) {
int[] response = new int[pin.length()];
for(int i = 0; i < pin.length(); i++) {
for (int j = 0; j < values.length; j++) {
int x = letterToPhone(pin.charAt(i));
if(x == j) {
response[i] = values[j];
}
}
}
return response;
}
public static boolean stringIsLengthK(String s, int k) {
boolean isLength = false;
if (s.length() == k) {
isLength = true;
}
return isLength;
}
public static boolean allDigits(String s) {
boolean isDigit = true;
for (int i = 0; i < s.length(); i++) {
if (!(Character.isDigit(s.charAt(i)))) {
isDigit = false;
break;
}
}
return isDigit;
}
public static boolean allUppercaseLetters(String s) {
boolean isUpper = true;
for (int i = 0; i < s.length(); i++) {
if (!(Character.isUpperCase(s.charAt(i)))) {
isUpper = false;
break;
}
}
return isUpper;
}
public static int[] digitStringToIntArray(String s) {
int[] arrayS = new int[s.length()];
for(int i = 0; i < arrayS.length; i++) {
for(int j = 0; j < SEQ_DIGITS; j++) {
if (((int) s.charAt(i) - 48) == j) {
arrayS[i] = j;
}
}
}
return arrayS;
}
public static int countValues(int value, int[] values) {
int count = 0;
for(int i = 0; i < values.length; i++) {
if(value == values[i]) {
count++;
}
}
return count;
}
public static int numPossible(int[] response, int[] values) {
int product = 1;
int[] count = new int[response.length];
for (int i = 0; i < count.length; i++) {
count[i] = countValues(response[i], values);
}
for(int i=0; i<response.length; i++){
product = product * count[i];
}
return product;
}
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
System.out.printf("Enter value sequence: ");
final String seq = in.nextLine();
System.out.printf("Enter PIN: ");
final String pin = in.nextLine();
if (!(allUppercaseLetters(pin))) {
throw new AssertionError(ERR_PIN);
} else if (!(allDigits(seq)) || !(stringIsLengthK(seq, SEQ_DIGITS))) {
throw new AssertionError(ERR_SEQ);
}
int[] seqArray = new int[SEQ_DIGITS];
seqArray = digitStringToIntArray(seq);
int[] response = new int[SEQ_DIGITS];
response = getResponse(pin, seqArray);
System.out.printf("Response: ");
for (int i = 0; i < response.length; i++) {
System.out.printf("%d", response[i]);
}
System.out.printf("%n");
numPossible(response, seqArray);
} catch (Error e) {
System.out.println(e.getMessage());
}
}
}
I want to be to able to accommodate larger sequence numbers without a scaling of n^2. Is there a way to change the for loop to instead compare the int x = letterToPhone(pin.charAt(i)); value in getResponse() to a range of integers such as "[0-9]"

One easy optimization of constant factors is to move the call to letterToPhone() out of the inner loop.
And yes, you can compare the x value to a range, eliminating the need for the inner loop.
for(int i = 0; i < pin.length(); i++) {
int x = letterToPhone(pin.charAt(i));
if ( (0 <= x) && (x < values.length)) {
response[i] = values[x];
}
}
Another optimization of constant factors would be to replace all the function calls in letterToPhone() with a switch statement. The compiler may choose to optimize that into a table lookup.

Why else statement is executing although if statement is true?

import java.util.Scanner;
class candidate {
public String name;
public int count;
public candidate(String name) {
super();
this.name = name;
}
}
public class DayScholar {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
candidate[] candidates = new candidate[3];
candidates[0] = new candidate("vikas");
candidates[1] = new candidate("ganesh");
candidates[2] = new candidate("teja");
System.out.print("No. of voters : ");
int voters = in.nextInt();
in.nextLine();
for (int i = 0; i < voters; i++) {
System.out.print("vote : ");
String name = in.nextLine().toLowerCase();
for (int j = 0; j < 3; j++) {
Here is the code, although if the statement is true else is also executing. How to check the condition
if (name.equals(candidates[j].name)) {
candidates[j].count++;
} else { **//problem here**
System.out.println("N");
break;
}
}
}
int highest = 0;
String winner = "";
for (int i = 0; i < 3; i++) {
if (candidates[i].count > highest) {
highest = candidates[i].count;
winner = candidates[i].name;
} else if (candidates[i].count == highest) {
winner += ("\n" + candidates[i].name);
}
}
System.out.println(winner);
}
}

Assuming the user enters a valid name, the following loop will increment the count field on the candidate with the matching name, and print N for the other 2 candidates.
for (int j = 0; j < 3; j++) {
if (name.equals(candidates[j].name)) {
candidates[j].count++;
} else {
System.out.println("N");
break;
}
}
To fix, you need the loop to just set the index of the matching candidate, then do the increment or printing after the loop:
int matchingIndex = -1; // -1 = not found
for (int j = 0; j < 3; j++) {
if (name.equals(candidates[j].name)) {
matchingIndex = j;
break;
}
}
if (matchingIndex == -1) {
System.out.println("N");
} else {
candidates[matchingIndex].count++;
}

How to count capital letters with compareTo() method?

Is there a way to count capital letters in a string using the method compareTo()? This is my current code so far, I don't know what to add in the if statement.
import java.util.*;
public class countcapitalletters
{
public static void main(String args[])
{
Scanner scan = new Scanner(System.in);
System.out.print("enter a string");
String input = scan.nextLine();
int count = 0;
for(int i=0; i<=input.length()-1; i++)
{
if(input.substring(i,i+1)
{
count = count+1;
}
}
System.out.println(count);
}
}
i dont know what to add to my if statement.

You can write a method as follows:
public int countUpperCase(String input) {
int count = 0;
for (int i = 0; i < input.length(); i++) {
String currentChar = input.substring(i, i + 1);
if (currentChar.compareTo("A") >= 0 && currentChar.compareTo("Z") <= 0) {
count = count + 1;
}
}
return count;
}

output the smallest integer in an array, and the index of it

I've been coding this program but I've got a little bit stuck and would like some advice. This is what I've got so far:
import java.util.Scanner;
public class SmallestInArray
{
public static void main(String[] args)
{
int[] array = new int[10];
input(array);
output(array);
}
public static void input(int[] array)
{
Scanner kybd = new Scanner(System.in);
System.out.println("Enter 10 integers: ");
for (int i = 0; i < array.length; i++) {
array[i] = kybd.nextInt();
}
}
public static int findSmallest(int[] array, int first)
{
int smallestPos = first;
for (int i = first + 1; i < array.length; i++) {
if (array[i] < array[smallestPos]) {
smallestPos = i;
}
}
return smallestPos;
}
public static void output(int[] array)
{
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + " ");
}
System.out.println();
}
}
Everything is fine other than the findSmallest method, as I'd like to output the smallest value and the index of it, but I'm not quite sure what to pass as the parameters in the main method?

Please find the refactored code which gets you both the value and index of smallest element in the array.
import java.util.Scanner;
public class SmallestInArray
{
int index_of_smallest_element;
public static void main(String[] args)
{
int[] array = new int[10];
input(array);
SmallestInArray smallestInArray = new SmallestInArray();
System.out.printf("Smallest Value:%d corresponding Index:%d\n",smallestInArray.findSmallest(array), smallestInArray.index_of_smallest_element);
output(array);
}
public static void input(int[] array)
{
System.out.println("Enter 10 integers: ");
try (Scanner kybd = new Scanner(System.in))
{
for (int i = 0; i < array.length; i++)
{
array[i] = kybd.nextInt();
}
}
}
public int findSmallest(int[] array)
{
int smallestValue = array[0];
index_of_smallest_element = 0;
for (int i = 1; i < array.length; i++) {
if (smallestValue > array[i]) // it doesn't accounts for duplicate values
{
smallestValue = array[i];
index_of_smallest_element = i;
}
}
return smallestValue;
}
public static void output(int[] array)
{
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + " ");
}
}}
Hope this helps

Try this
public static void main(String[] args)
{
int[] array = new int[10];
input(array);
output(array);
int smallestPos = findSmallest(array, 0 /* P.S. this parameter seem to be useless */);
int smallestVal = array[smallestPos];
// output the two
}

How can I avoid repetition of the same number?

This is what I want :
Let the user enter as many numbers as they want until a non number is entered (you may
assume there will be less than 100 numbers). Find the most frequently entered number. (If
there are more than one, print all of them.)
Example output:
Input: 5
Input: 4
Input: 9
Input: 9
Input: 4
Input: 1
Input: a
Most common: 4, 9
I have got to the point in my code where I have managed to find out which are the most common numbers. However, I don't want to print out the same number over and over again; example from above: Most common: 4, 9, 9, 4
What needs to be done?
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String[] input = new String[100];
System.out.print("Input: ");
input[0] = in.readLine();
int size = 0;
for (int i = 1; i < 100 && isNumeric(input[i-1]); i++) {
System.out.print("Input: ");
input[i] = in.readLine();
size = size + 1;
}
/*for (int i = 0; i < size; i++) { //testing
System.out.println(input[i]);
}*/
int numOccur;
int[] occur = new int[size];
for(int i = 0; i < size; i++) {
numOccur = 0;
for (int j = 0; j < size; j++) {
if(input[i].equals(input[j])) {
numOccur = numOccur + 1;
}
}
occur[i] = numOccur;
//System.out.println(numOccur); //testing
}
int maxOccur = 0;
for(int i = 0; i < size; i++) {
if(occur[i] > maxOccur) {
maxOccur = occur[i];
}
}
//System.out.println(maxOccur); //testing
for (int i = 0; i < size && !numFound; i++) {
if(occur[i] == maxOccur) {
System.out.println(input[i]);
}
}
}
//checks if s is an in, true if it is an int
public static boolean isNumeric (String s) {
try {
Integer.parseInt(s);
return true; //parse was successful
} catch (NumberFormatException nfe) {
return false;
}
}
Found the solution!
String[] mostCommon = new String[size];
int numMostCommon = 0;
boolean numFound = false;
for (int i = 0; i < size; i++) {
int isDifferent = 0;
if (occur[i] == maxOccur) {
for (int j = 0; j < size; j++) {
if (!(input[i].equals(mostCommon[j]))) {
isDifferent = isDifferent + 1;
}
}
if (isDifferent == size) {
mostCommon[numMostCommon] = input[i];
numMostCommon = numMostCommon + 1;
}
}
}
for (int i = 0; i < numMostCommon - 1; i++) {
System.out.print("Most common: " + mostCommon[i] + ", ");
}
System.out.println(mostCommon[numMostCommon - 1]);

you could use the hash table for this to store the frequenceis as the limit is very less i.e. less than 100.
pseudo code would be like:
vector<int> hash(101)
cin>>input
if(isnumeric(input))
hash[input]++
else{
max=max_element(hash.begin(),hash.end());
for(int i=0;i<100;i++)
if(hash[i]==max)
print i
}

Set<Integer> uniqueMaxOccur = new HashSet<Integer>();
for (int i = 0; i < size ; i++) {
if(occur[i] == maxOccur) {
//System.out.println(input[i]);
uniqueMaxOccur.add(input[i]);
}
}
and display the values in the set

You can use a Set and store the values already printed.

What about something like this?
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
Map<string,int> numberLookup = new HashMap<string,int>();
Boolean doContinue = true;
while (doContinue)
{
System.out.print("Input: ");
String input = in.readLine();
if (isNumeric(input))
{
if (!numberLookup.containsKey(input))
numberLookup.put(input,1);
else
numberLookup.put(input, numberLookup.get(input) + 1);
}
else
doContinue = false;
}
maxOccur = numberLookup.values().max();
System.out.print("These numbers were all entered " + maxOccur + " times:");
Iterator it = numberLookup.entrySet().iterator();
while (it.hasNext())
{
(Map.Entry)it.next();
System.out.println(pairs.getKey());
}
}
Sorry, I'm a C# person and don't have a Java compiler on me, so this might need some tweaking.