Find duplicates in an given array in O(n) time and O(1) extra space.

Objective: Given an array of integers, find out duplicates in it.

Example:

int [] a = {4, 6, 2, 1, 2, 5};
Output: Array has duplicates : 2

int a [] = {1, 6, 5, 2, 3, 3, 2};
Output: Array has duplicates : 2 , 3

Approach:

Naive approach: Use 2 loops. Check each element in the array with all other elements in the array and check if it has the same value.

Time Complexity : O(n^2) Space Complexity: O(1)

Code:

	import java.util.Arrays;

	public class CheckDuplicatesUsingTwoLoops {
	public void hasDuplicates2ForLoops(int [] arrA) {

	for (int i = 0; i < arrA.length; i++) {
	for (int j = i+1; j < arrA.length; j++) {
	if(arrA[i]==arrA[j]){
	System.out.println("Array has duplicates : " + arrA[i]);
	}
	}
	}
	}
	public static void main(String[] args) {
	int a [] = {1, 6, 5, 2, 3, 3, 2};
	new CheckDuplicatesUsingTwoLoops().hasDuplicates2ForLoops(a);
	}
	}

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CheckDuplicatesUsingTwoLoops.java

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Sorting approach: Sort the array, this will bring all the duplicates together if present. Now navigate the array and check the adjacent elements to check for duplicates.

Time Complexity : O(nlogn) Space Complexity: O(n) by using merge sort.

Code:


	import java.util.Arrays;

	public class CheckDuplicatesUsingSorting {
	public void hasDuplicatesUsingSorting(int [] arrA) {
	Arrays.sort(arrA);
	for (int i = 0; i < arrA.length–1; i++) {
	if(arrA[i]==arrA[i+1]){
	System.out.println("Array has duplicates : " + arrA[i]);
	}
	}
	}
	public static void main(String[] args) {
	int a [] = {1, 6, 5, 2, 3, 3, 2};
	new CheckDuplicatesUsingSorting().hasDuplicatesUsingSorting(a);
	}
	}

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CheckDuplicatesUsingSorting.java

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Better Solution : Use Hash map. Store the count of each element of array in a hash table and later check in Hash table if any element has count more than 1. ( Similar approach is used in problem – Find the first non repeating character in a given string

Time Complexity : O(n) and Space Complexity: O(n).

Code:

	import java.util.Arrays;
	import java.util.HashMap;

	public class CheckDuplicatesUsingHashMap {

	public void hasDuplicatesUsingMap(int [] arrA){
	HashMap<Integer, Integer> map = new HashMap<>();
	for (int i = 0; i <arrA.length ; i++) {
	if(map.containsKey(arrA[i])){
	System.out.println("Array has duplicates : " + Math.abs(arrA[i]));
	}else{
	map.put(arrA[i], 1);
	}
	}
	}

	public static void main(String[] args) {
	int a [] = {1, 6, 5, 2, 3, 3, 2};
	new CheckDuplicatesUsingHashMap().hasDuplicatesUsingMap(a);
	}

	}

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CheckDuplicatesUsingHashMap.java

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Better Solution (Conditional) : O(n) time and O(1) extra space.

This solution works only if array has positive integers and all the elements in the array are in range from 0 to n-1 where n is the size of the array.
Navigate the array.
Update the array as for ith index :- arrA[arrA[i]] = arrA[arrA[i]]*-1 (if it already not negative).
If is already negative, we have duplicates, return false.

Note:

The code given below does not handle the case when 0 is present in the array.
To handle 0 in array, while navigating the array, when 0 is encountered, replace it with INT_MIN and if INT_MIN is encountered while traversing, this means 0 is repeated in the array.

Similar approach used in problem : If array has all consecutive numbers.

Code:

	public class CheckDuplicates {
	public void hasDuplicates(int[] arrA) {

	for (int i = 0; i < arrA.length; i++) {
	//check if element is negative, if yes the we have found the duplicate
	if (arrA[Math.abs(arrA[i])] < 0) {
	System.out.println("Array has duplicates : " + Math.abs(arrA[i]));
	} else {
	arrA[Math.abs(arrA[i])] = arrA[Math.abs(arrA[i])] * –1;
	}
	}
	}

	public static void main(String[] args) {
	int a[] = {1, 6, 5, 2, 3, 3, 2};
	new CheckDuplicates().hasDuplicates(a);
	}
	}

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CheckDuplicates.java

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Output:

Array has duplicates : 3
Array has duplicates : 2

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