I have an integer array crr_array and I want to count elements, which occur repeatedly. First, I read the size of the array and initialize it with numbers read from the console. In the array new_array, I store the elements that are repeated. The array times stores the number of consecutive occurrences of an element. Then, I try to search for the repeating sequences and print them in a specific format. However, it does not work.
// Get integer array size
Scanner input = new Scanner(System.in);
System.out.println("Enter array size: ");
int size = input.nextInt();
int[] crr_array = new int[size];
int[] new_array= new int[size];
int[] times = new int[size];
// Read integers from the console
System.out.println("Enter array elements: ");
for (int i = 0; i < crr_array.length; i++) {
crr_array[i] = input.nextInt();
times[i] = 1;
}
// Search for repeated elements
for (int j = 0; j < crr_array.length; j++) {
for (int i = j; i < crr_array.length; i++) {
if (crr_array[j] == crr_array[i] && j != i) {
new_array[i] = crr_array[i];
times[i]++;
}
}
}
//Printing output
for (int i = 0; i < new_array.length; i++) {
System.out.println("\t" + crr_array[i] + "\t" + new_array[i] + "\t" + times[i]);
}
I want the output to look like this:
There are <count_of_repeated_element_sequences> repeated numbers
<repeated_element>: <count> times
...
For example:
There are 3 repeated numbers:
22: 2 times
4: 3 times
1: 2 times
How can I find the repeated elements and their counts? How can I print them as shown above?
This question is related to
java
private static void getRepeatedNumbers() {
int [] numArray = {2,5,3,8,1,2,8,3,3,1,5,7,8,12,134};
Set<Integer> nums = new HashSet<Integer>();
for (int i =0; i<numArray.length; i++) {
if(nums.contains(numArray[i]))
continue;
int count =1;
for (int j = i+1; j < numArray.length; j++) {
if(numArray[i] == numArray[j]) {
count++;
}
}
System.out.println("The "+numArray[i]+ " is repeated "+count+" times.");
nums.add(numArray[i]);
}
}
public class DuplicationNoInArray {
/**
* @param args
* the command line arguments
*/
public static void main(String[] args) throws Exception {
int[] arr = { 1, 2, 3, 4, 5, 1, 2, 8 };
int[] result = new int[10];
int counter = 0, count = 0;
for (int i = 0; i < arr.length; i++) {
boolean isDistinct = false;
for (int j = 0; j < i; j++) {
if (arr[i] == arr[j]) {
isDistinct = true;
break;
}
}
if (!isDistinct) {
result[counter++] = arr[i];
}
}
for (int i = 0; i < counter; i++) {
count = 0;
for (int j = 0; j < arr.length; j++) {
if (result[i] == arr[j]) {
count++;
}
}
System.out.println(result[i] + " = " + count);
}
}
}
public class FindRepeatedNumbers
{
public static void main(String[] args)
{
int num[]={1,3,2,4,1,2,4,6,7,5};
Arrays.sort(num);
for(int j=1;j<num.length;j++)
{
if(num[j]==num[j-1])
{
System.out.println(num[j]);
}
}
}
}
public static void duplicatesInteger(int arr[]){
Arrays.sort(arr);
int count=0;
Set s=new HashSet();
for(int i=0;i<=arr.length-1;i++){
for(int j=i+1;j<=arr.length-1;j++){
if(arr[i]==arr[j] && s.add(arr[i])){
count=count+1;
}
}
System.out.println(count);
}
}
public static void main(String[] args) {
Scanner input=new Scanner(System.in);
int[] numbers=new int[5];
String x=null;
System.out.print("enter the number 10:"+"/n");
for(int i=0;i<5;i++){
numbers[i] = input.nextInt();
}
System.out.print("Numbers : count"+"\n");
int count=1;
Arrays.sort(numbers);
for(int z=0;z<5;z++){
for(int j=0;j<z;j++){
if(numbers[z]==numbers[j] & j!=z){
count=count+1;
}
}
System.out.print(numbers[z]+" - "+count+"\n");
count=1;
}
public class ArrayDuplicate {
private static Scanner sc;
static int totalCount = 0;
public static void main(String[] args) {
int n, num;
sc = new Scanner(System.in);
System.out.print("Enter the size of array: ");
n =sc.nextInt();
int[] a = new int[n];
for(int i=0;i<n;i++){
System.out.print("Enter the element at position "+i+": ");
num = sc.nextInt();
a[enter image description here][1][i]=num;
}
System.out.print("Elements in array are: ");
for(int i=0;i<a.length;i++)
System.out.print(a[i]+" ");
System.out.println();
duplicate(a);
System.out.println("There are "+totalCount+" repeated numbers:");
}
public static void duplicate(int[] a){
int j = 0,count, recount, temp;
for(int i=0; i<a.length;i++){
count = 0;
recount = 0;
j=i+1;
while(j<a.length){
if(a[i]==a[j])
count++;
j++;
}
if(count>0){
temp = a[i];
for(int x=0;x<i;x++){
if(a[x]==temp)
recount++;
}
if(recount==0){
totalCount++;
System.out.println(+a[i]+" : "+count+" times");
}
}
}
}
}
You have to use or read about associative arrays, or maps,..etc. Storing the the number of occurrences of the repeated elements in array, and holding another array for the repeated elements themselves, don't make much sense.
Your problem in your code is in the inner loop
for (int j = i + 1; j < x.length; j++) {
if (x[i] == x[j]) {
y[i] = x[i];
times[i]++;
}
}
with O(n log(n))
int[] arr1; // your given array
int[] arr2 = new int[arr1.length];
Arrays.sort(arr1);
for (int i = 0; i < arr1.length; i++) {
arr2[i]++;
if (i+1 < arr1.length)
{
if (arr1[i] == arr1[i + 1]) {
arr2[i]++;
i++;
}
}
}
for (int i = 0; i < arr1.length; i++) {
if(arr2[i]>0)
System.out.println(arr1[i] + ":" + arr2[i]);
}
package jaa.stu.com.wordgame;
/**
* Created by AnandG on 3/14/2016.
*/
public final class NumberMath {
public static boolean isContainDistinct(int[] arr) {
boolean isDistinct = true;
for (int i = 0; i < arr.length; i++)
{
for (int j = 0; j < arr.length; j++) {
if (arr[i] == arr[j] && i!=j) {
isDistinct = false;
break;
}
}
}
return isDistinct;
}
public static boolean isContainDistinct(float[] arr) {
boolean isDistinct = true;
for (int i = 0; i < arr.length; i++)
{
for (int j = 0; j < arr.length; j++) {
if (arr[i] == arr[j] && i!=j) {
isDistinct = false;
break;
}
}
}
return isDistinct;
}
public static boolean isContainDistinct(char[] arr) {
boolean isDistinct = true;
for (int i = 0; i < arr.length; i++)
{
for (int j = 0; j < arr.length; j++) {
if (arr[i] == arr[j] && i!=j) {
isDistinct = false;
break;
}
}
}
return isDistinct;
}
public static boolean isContainDistinct(String[] arr) {
boolean isDistinct = true;
for (int i = 0; i < arr.length; i++)
{
for (int j = 0; j < arr.length; j++) {
if (arr[i] == arr[j] && i!=j) {
isDistinct = false;
break;
}
}
}
return isDistinct;
}
public static int[] NumberofRepeat(int[] arr) {
int[] repCount= new int[arr.length];
for (int i = 0; i < arr.length; i++)
{
for (int j = 0; j < arr.length; j++) {
if (arr[i] == arr[j] ) {
repCount[i]+=1;
}
}
}
return repCount;
}
}
call by NumberMath.isContainDistinct(array) for find is it contains repeat or not
call by int[] repeat=NumberMath.NumberofRepeat(array) for find repeat count. Each location contains how many repeat corresponding value of array...
for (int i = 0; i < x.length; i++) {
for (int j = i + 1; j < x.length; j++) {
if (x[i] == x[j]) {
y[i] = x[i];
times[i]++;
}
}
}
If you have values in a short set of possible values then you can use something like Counting Sort
If not you have to use another data structure like a Dictionary, in java a Map
int[] array
Map<Integer, Integer>
where Key = array value for example array[i] and value = a counter
Example:
int[] array = new int [50];
Map<Integer,Integer> counterMap = new HashMap<>();
//fill the array
for(int i=0;i<array.length;i++){
if(counterMap.containsKey(array[i])){
counterMap.put(array[i], counterMap.get(array[i])+1 );
}else{
counterMap.put(array[i], 1);
}
}
package com.core_java;
import java.util.Arrays;
import java.util.Scanner;
public class Sim {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter array size: ");
int size = input.nextInt();
int[] array = new int[size];
// Read integers from the console
System.out.println("Enter array elements: ");
for (int i = 0; i < array.length; i++) {
array[i] = input.nextInt();
}
Sim s = new Sim();
s.find(array);
}
public void find(int[] arr) {
int count = 1;
Arrays.sort(arr);
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
if (count > 1) {
System.out.println();
System.out.println("repeated element in array " + arr[i] + ": " + count + " time(s)");
i = i + count - 1;
}
count = 1;
}
}
}
Source: Stackoverflow.com