Reputation: 111
[Java]Find duplicate characters in a string without hashmap and set
I have written a Java program to find duplicate characters in a string without Hashmap and set.
Below is the program,
package practice;
public class Duplicate {
public static void main(String[] args) {
String src= "abcad";
char[] srcChar= src.toLowerCase().toCharArray();
int len=srcChar.length;
int j=0;
boolean flag=false;
char ch;
// System.out.println("Length of the String is "+len1);
// System.out.println("Length of the character array is "+len);
int k=0;
for(int i=0;i<len;i++)
{
// System.out.println("i-----> "+i + " and character is "+srcChar[i]);
for(j=0;j<len;j++)
{
// System.out.println("j-----> "+j + " and character is "+srcChar[j]);
if(srcChar[i]==srcChar[j])
{
k++;
}
}
if(k>1)
{
if(srcChar[i]>1)
{
System.out.println("This character "+srcChar[i]+" has repeated "+k+ " time");
}
else
{
System.out.println("There are no characters repeated in the given string");
}
}
k=0;
}
}
}
Output here is:
This character a has repeated 2 time
This character a has repeated 2 time
Here, I want the output like This character a has repeated 2 time
i.e. not repeating the output twice. Since the character "a" is repeated twice, the output is also repeated twice.
kindly help me to get the output once instead of twice.
Thank you,
Upvotes: 0
Views: 14027
Answers (7)
Reputation: 21
private static void duplicateChar(String str){
char[] arr1 = str.toUpperCase().toCharArray();
int length = str.length();
int count = 1;
String s = "";
char c1 = '\u0000';
for(int i=0;i<length;i++){
count = 1;
for(int j=i+1;j<length;j++){
if(arr1[i] == arr1[j]){
count++;
c1 = arr1[i];
}
if(j == (length-1) && c1 != '\u0000' && !s.contains(String.valueOf(c1))){
s = s+" "+String.valueOf(c1)+" No of times: "+count+"\n";
}
}
}
System.out.println("\nDuplicate char are:\n"+s);
}
Upvotes: 2
Reputation: 11
You can also solve this problem with this code like :
public static void main(String[] args) {
String src = "abcad";
char[] srcChar = src.toLowerCase().toCharArray();
int len = srcChar.length;
int j = 0;
boolean flag = false;
char ch;
// System.out.println("Length of the String is "+len1);
// System.out.println("Length of the character array is "+len);
int k = 0;
for (int i = 0; i < len; i++) {
// System.out.println("i-----> "+i + " and character is "+srcChar[i]);
for (j = 0 + i; j < len; j++) {
// System.out.println("j-----> "+j + " and character is "+srcChar[j]);
if (srcChar[i] == srcChar[j]) {
k++;
}
}
if (k > 1) {
if (srcChar[i] > 1) {
System.out.println("This character " + srcChar[i] + " has repeated " + k + " time");
} else {
System.out.println("There are no characters repeated in the given string");
}
}
k = 0;
}
}
just we need to start the inner loop with j=0+i ; for (j = 0 + i; j < len; j++)
This will you can observe above code;
Upvotes: 0
Reputation: 153
char[] array=value.toCharArray();
int count=0;
char ch;
for(int i=0;i<array.length-1;i++)
{
ch=array[i];
count=1;
if(ch!='#'){
for(int j=i+1;j<array.length;j++)
{
if(ch==array[j]){
count++;
array[j]='#';
}
}
if(count>1)
{
System.out.println("char is " + ch + "count" + count);
}
}
}
Upvotes: 0
Reputation: 1421
class PrintDuplicateCharacter
{
public static void main(String[] args)
{
String str = "HelloJava";
char[] ch = str.toCharArray();
int i=0,j=0;
for(i=0;i<ch.length;i++)
{
int count = 0 ;
for( j = i+1;j<ch.length;j++)
{// 4 6 , 8 , 10
if(ch[i] == ch[j] )
{
count++;
}
}
if(count != 0)
{
System.out.print(str.charAt(i) + " Occured " + count + " time");
}
}
}
}
Upvotes: 2
Reputation: 187
Imports:
import java.util.ArrayList;
import java.util.List;
Code:
public static void main(String args[]) {
String input = "abcad"; // Input value
char[] chars = input.toLowerCase().toCharArray(); // Creates ArrayList
// of all characters
// in the String
List<Character> charR = new ArrayList<>(); // Creates a List used to
// saving the Characters it
// has saved
List<Integer> valR = new ArrayList<>(); // Creates a List that will
// store how many times a
// character is repeated
for (int i = 0; i < chars.length; i++) { // Loop through items in the
// ArrayList
char c = chars[i]; // Create Character value containing the value of
// the item at the "i" index of the ArrayList
if (charR.contains(c)) { // If the List contains item...
for (int i2 = 0; i2 < charR.size(); i2++) { // Loop through its
// items
if (charR.get(i2).equals(c)) { // If you find a match...
valR.set(i2, valR.get(i2) + 1); // Increase repeated
// value by 1
i2 = charR.size(); // Stop loop
} else { // Else...
i2++; // Increase index by 1
}
}
} else { // Else...
charR.add(c); // Add the Character to the List
valR.add(1); // Add the value 1 to the List (Meaning that the
// Character repeated once)
}
}
for (int i = 0; i < charR.size(); i++) { // Loop through all the items
// in the List
System.out.println("'" + charR.get(i) + "' : " + valR.get(i)); // Display
// what
// the
// character
// is
// and
// how
// many
// times
// it
// was
// repeated
}
}
Output:
'a' : 2
'b' : 1
'c' : 1
'd' : 1
Upvotes: 0
Reputation: 315
All you need to fix is to start the second loop from i instead of 0.
for (int i = 0; i < len; i++) {
for (j = i; j < len; j++) {
...
}
...
}
Upvotes: 0
Reputation: 699
You can make a 2 dimensional array, 2 wide, the source strings height. In this array you store a character when it gets replaced and add one to the amount of times it has been replaced.
Something like(I don't know if these counters are correct):
replacements[j][0] = charAt(j);
replacements[j][1] += 1;
You would have to check if the character you are replacing already exists in this array and you can only print elements of the array if they aren't null.
You print this after the original loop.
Upvotes: 0
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