``````/*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];
}
}
}// 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<int>& 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;

}
};
``````