Example 1: java array check duplicates
duplicates = false;
for(i = 0; i < zipcodeList.length; i++) {
for(j = i + 1; k < zipcodeList.length; j++) {
if(j != i && zipcodeList[j] == zipcodeList[i]) {
duplicates = true;
}
}
}
Example 2: java find duplicates in array
package dto;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
/**
* Java Program to find duplicate elements in an array. There are two straight
* forward solution of this problem first, brute force way and second by using
* HashSet data structure. A third solution, similar to second one is by using
* hash table data structure e.g. HashMap to store count of each element and
* print element with count 1.
*
* @author java67
*/
public class DuplicatesInArray{
public static void main(String args[]) {
String[] names = { "Java", "JavaScript", "Python", "C", "Ruby", "Java" };
// First solution : finding duplicates using brute force method
System.out.println("Finding duplicate elements in array using brute force method");
for (int i = 0; i < names.length; i++) {
for (int j = i + 1; j < names.length; j++) {
if (names[i].equals(names[j]) ) {
// got the duplicate element
}
}
}
// Second solution : use HashSet data structure to find duplicates
System.out.println("Duplicate elements from array using HashSet data structure");
Set store = new HashSet<>();
for (String name : names) {
if (store.add(name) == false) {
System.out.println("found a duplicate element in array : "
+ name);
}
}
// Third solution : using Hash table data structure to find duplicates
System.out.println("Duplicate elements from array using hash table");
Map nameAndCount = new HashMap<>();
// build hash table with count
for (String name : names) {
Integer count = nameAndCount.get(name);
if (count == null) {
nameAndCount.put(name, 1);
} else {
nameAndCount.put(name, ++count);
}
}
// Print duplicate elements from array in Java
Set> entrySet = nameAndCount.entrySet();
for (Entry entry : entrySet) {
if (entry.getValue() > 1) {
System.out.println("Duplicate element from array : "
+ entry.getKey());
}
}
}
}
Output :
Finding duplicate elements in array using brute force method
Duplicate elements from array using HashSet data structure
found a duplicate element in array : Java
Duplicate elements from array using hash table
Duplicate element from array : Java
Example 3: find repeated elements in array java
for (String name : names) {
if (set.add(name) == false) {
// your duplicate element
}
}
Example 4: Find the duplicate in an array of N integers.
// 287. Find the Duplicate Number
// Medium
// Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
// Example 1:
// Input: [1,3,4,2,2]
// Output: 2
// Example 2:
// Input: [3,1,3,4,2]
// Output: 3
// Note:
// You must not modify the array (assume the array is read only).
// You must use only constant, O(1) extra space.
// Your runtime complexity should be less than O(n2).
// There is only one duplicate number in the array, but it could be repeated more than once.
class Solution {
public:
int findDuplicate(vector& nums) {
int n=nums.size();
int s=nums[0];
int f=nums[nums[0]];
while(s!=f) {
s=nums[s];
f=nums[nums[f]];
}
f=0;
while(s!=f) {
s=nums[s];
f=nums[f];
}
return s;
}
};
Example 5: find duplicate elements in array in java
/*This method is all in one
*you can find following things:
*finding Duplicate elements in array
*array without duplicate elements
*number of duplicate elements
*numbers of pair of dulicate with repeatation
*/
//let given array = [2,3,2,5,3]
public static void findDuplicateArray(int [] array)
{
int size = array.length;
//creating array to hold count frequency of array elements
int [] countFrequency = new int[size];
// filling countFrequency with -1 value on every index
for(int i = 0; i < size; i++)
{
countFrequency[i] = -1;//[-1,-1,-1,-1,-1...]
}
int count = 1;
for(int i = 0; i < size; i++)
{
//check countFrequency[i] != 0 because 0 means it already counted
if(countFrequency[i] != 0)
{
for(int j = i+1; j < size; j++)
{
//if array[i] == array[j] then increase count value
if(array[i] == array[j])
{
count++;
/*only at first occurence of an element count value
*will be increased else everywhere it will be 0
*/
countFrequency[j]= 0;
}
}
countFrequency[i] = count;
}
count = 1;
}
// array = [2,3,2,5,3]
//countFrequency = [2,2,0,1,0]
System.out.println("array without duplicate elements");
for(int i = 0; i < array.length; i++)
{
if(countFrequency[i] >= 1)
System.out.print(array[i] + " ");
}
System.out.println();
System.out.println("duplicate elements in array");
for(int i = 0; i < array.length; i++)
{
if(countFrequency[i]/2 >= 1)
System.out.print(array[i] + " ");
}
System.out.println();
System.out.println("number of duplicate elements");
count = 0;
for(int i = 0; i < array.length; i++)
{
if(countFrequency[i]/2 >= 1)
count++;
}
System.out.print(count);
System.out.println();
System.out.println("numbers of pair of dulicate with repeatation");
count = 0;
for(int i = 0; i < array.length; i++)
{
if(countFrequency[i] >= 2)
{
int div = countFrequency[i]/2;
count+=div;
}
}
System.out.println(count);
int [] array3 = new int [array.length];
for(int i = 0; i < array.length; i++)
{
for(int j = 0; j < countFrequency[i]; j++)
{
array3[i]= array[i];
}
}
}
Example 6: java find duplicates in array
// Uses a set, which does not allow duplicates
for (String name : names)
{
if (set.add(name) == false)
{
// print name your duplicate element
}
}