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

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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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