find the duplicate in an array of n integers

1npm i find-array-duplicates
2
3import duplicates from 'find-array-duplicates'
4
5const names = [
6 { 'age': 36, 'name': 'Bob' },
7 { 'age': 40, 'name': 'Harry' },
8 { 'age': 1,  'name': 'Bob' }
9]
10 
11duplicates(names, 'name').single()
12// => { 'age': 36, 'name': 'Bob' }

1// 287. Find the Duplicate Number
2// Medium
3
4// 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.
5
6// Example 1:
7
8// Input: [1,3,4,2,2]
9// Output: 2
10// Example 2:
11
12// Input: [3,1,3,4,2]
13// Output: 3
14// Note:
15
16// You must not modify the array (assume the array is read only).
17// You must use only constant, O(1) extra space.
18// Your runtime complexity should be less than O(n2).
19// There is only one duplicate number in the array, but it could be repeated more than once.
20
21class Solution {
22public:
23    int findDuplicate(vector<int>& nums) {
24        int n=nums.size();
25        int s=nums[0];
26        int f=nums[nums[0]];
27        while(s!=f) {
28            s=nums[s];
29            f=nums[nums[f]];
30        }
31        f=0;
32        while(s!=f) {
33            s=nums[s];
34            f=nums[f];
35        }
36        return s;
37        
38    }
39};

1/*This method is all in one 
2*you can find following things:
3*finding Duplicate elements in array
4*array without duplicate elements
5*number of duplicate elements
6*numbers of pair of dulicate with  repeatation
7*/
8//let given array = [2,3,2,5,3]
9   public static void findDuplicateArray(int [] array)
10   {
11	   int size = array.length;
12     //creating array to hold count frequency of array elements
13	   int [] countFrequency = new int[size];
14     // filling countFrequency with -1 value on every index
15	   for(int i = 0; i < size; i++)
16	   {
17		   countFrequency[i] = -1;//[-1,-1,-1,-1,-1...]
18	   }
19    
20      int count = 1;
21	   for(int i = 0; i < size; i++) 
22	   {
23         //check countFrequency[i] != 0 because 0 means it already counted
24		  if(countFrequency[i] != 0)
25		  {
26		   for(int j = i+1; j < size; j++)  
27		   {
28             //if array[i] == array[j] then increase count value
29			   if(array[i] == array[j])
30			   {
31				   count++;
32                 /*only at first occurence of an element count value 
33                 *will be increased else everywhere it  will be 0 
34                 */
35				   countFrequency[j]= 0;				   
36			   }
37		   }
38		   countFrequency[i] = count;
39	      }
40		  count = 1;
41	   }
42     // array         = [2,3,2,5,3]
43     //countFrequency = [2,2,0,1,0]
44	   System.out.println("array without duplicate elements");
45		    for(int i = 0; i < array.length; i++)
46		    {
47		    	if(countFrequency[i] >= 1)
48		    	System.out.print(array[i] + " ");
49		    }
50		    System.out.println();
51		    
52	   System.out.println("duplicate elements in array");
53		    for(int i = 0; i < array.length; i++)
54		    {
55		    	if(countFrequency[i]/2 >= 1)
56		    	System.out.print(array[i] + " ");
57		    }
58	        System.out.println();
59		    
60	   System.out.println("number of duplicate elements");
61		     count = 0;
62		    for(int i = 0; i < array.length; i++)
63		    {	    	
64		    	if(countFrequency[i]/2 >= 1)
65		            count++;
66		    }
67		    System.out.print(count);
68		    System.out.println();
69		    
70	  System.out.println("numbers of pair of dulicate with  repeatation");
71		     count = 0;
72		    for(int i = 0; i < array.length; i++)
73		    {
74		    	if(countFrequency[i] >= 2)
75		    	{
76		    		int div = countFrequency[i]/2;
77		    		count+=div;
78		    	}
79		    }
80		    System.out.println(count);
81			
82		    int [] array3 = new int [array.length];
83		    for(int i = 0; i < array.length; i++)
84		    {
85		    	for(int j = 0; j < countFrequency[i]; j++)
86		    	{
87		    		array3[i]= array[i];
88		    	}
89		    }
90   }