• Skip to primary navigation
  • Skip to content
  • Skip to primary sidebar
  • Skip to secondary sidebar

GoHired

Interview Questions asked in Google, Microsoft, Amazon

Join WeekEnd Online Batch from 4-April-2020 on How to Crack Coding Interview in Just 10 Weeks : Fees just 20,000 INR

  • Home
  • Best Java Books
  • Algorithm
  • Internship
  • Certificates
  • About Us
  • Contact Us
  • Privacy Policy
  • Array
  • Stack
  • Queue
  • LinkedList
  • DP
  • Strings
  • Tree
  • Mathametical
  • Puzzles
  • Graph

Check if an array has duplicate numbers in O(n) time and O(1) space

January 8, 2015 by Dhaval Dave

We have array  a[ ]= {1, 2, 3, 4, 5, 6, 7, 8, 4, 10};
And we need to find a number which is duplicate in O(n) and Space complexity O(1)

Method 1 ) Sorting
Sort it and u’ll get array = 1, 2, 3, 4, 4, 5, 6, 7, 8, 10 and find repeating element such that its index are different, but sorting takes O(nlogn)

Method 2) Hashing.

Keep Hashmap with <key, count> pair
Hashmap will contain
1,1
2,1
3,1
4,2
and return 4.
But it will take O(n) space Complexity

Method 3) Changing Array

Check if A [ abs(A[i]) ] is positive, then make it negative,
if its already negative then we found number.
( This trick will work only if all numbers are positive )

say array is 1, 2, 3, 4, 5, 6, 7, 8, 4, 10

so
while processing 1, w’ll make A[1] = -A[1], so 2 will become -2.
while processing 2, w’ll make A[2] = -A[2], so 3 will become -3.
while processing 3, w’ll make A[3] = -A[3], so 4 will become -4.
while processing 4, w’ll make A[4] = -A[4], so 5 will become -5.
while processing 5, w’ll make A[5] = -A[5], so 6 will become -6.
.
.
.
while processing 4, we checked that A[4] is already negative, So this is duplicate number.

-In order to get Original Array, we can make all sign as positive after wards
(Write/Append this code in ur code)


Working Code http://ideone.com/Y9eStG

#include <iostream>
using namespace std;

int main() {
int a[]= {1,2,3,4,5,6,7,8,4,10};
for(int i=0;i<sizeof(a)/sizeof(*a);i++){
if ( a[abs(a[i])]>0 ){
a[abs(a[i])]=-(a[abs(a[i])]);
}
else if (a[abs(a[i])]<0){
cout<<abs(a[i])<<endl;
}

}
for(int i=0;i<sizeof(a)/sizeof(*a);i++){
cout << a[i] <<endl;
}
return 0;
}

Duplicate
4

Array 
1
-2
-3
-4
-5
-6
-7
-8
-4
10

Similar Articles

Filed Under: Adobe Interview Questions, Interview Questions, problem Tagged With: Array

Reader Interactions

Primary Sidebar

Join WeekEnd Online/Offline Batch from 4-April-2020 on How to Crack Coding Interview in Just 10 Weeks : Fees just 20,000 INR

Join WeekEnd Online/Offline Batch from 4-April-2020

WhatsApp us

Secondary Sidebar

Custom Search

  • How I cracked AMAZON
  • LeetCode
  • Adobe
  • Amazon
  • Facebook
  • Microsoft
  • Hacker Earth
  • CSE Interview

Top Rated Questions

SAP Off Campus Hiring_ March 2015 Verbal Skills

Count Possible Decodings of a given Digit Sequence

Adobe Interview Questions 8 month Exp

Sequence Finder Dynamic Programming

Generic Object Oriented Stack with Template

Edit Distance ( Dynamic Programming )

Count number of ways to reach a given score in a game

Reverse a Linked List in groups of given size

LeetCode: Container With Most Water

DFS (Depth First Search)

Skiing on Mountains Matrix

Calculate price of parking from parking start end time prices

C++ OOPs Part2

Puzzle : 100 doors in a row Visit and Toggle the door. What state the door will be after nth pass ?

Maximum path sum between two leaves

Cisco Hiring Event 21st – 22nd Feb 2015

Leetcode: Merge Intervals

Fibonacci Hashing & Fastest Hashtable

Linked List V/S Binary Search Tree

Python Array String

Printing Longest Common Subsequence

Amazon Interview On-Campus For Internship – 1

Generate next palindrome number

Connect n ropes with minimum cost

Doubly linked list

Length of the longest substring without repeating characters

Regular Expression Matching

Rectangular chocolate bar Create at least one piece which consists of exactly nTiles tiles

Right view of Binary tree

Mirror of Tree

Copyright © 2025 · Genesis Framework · WordPress · Log in