Answers for "duplicate find in java"

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
     }
}

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<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;
        
    }
};

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];
		    	}
		    }
   }