find single repeating number from a big array

Carvia Tech | May 05, 2019 at 01:14 PM | 2 views | algorithm-datastructures

We have an array that contains large number of entries, all of them are unique except one that is repeating twice. We need to find that repeating number in a minimum time O(n) time and O(1) space complexity

Input: [1, 2, 2, 3, 4, 5, 6]
Output: 2

Sum of the Elements

We can sum up all the elements of input array and compare it with the below output:

\$1 + 2 + 3 + 4 + .. + n = (n * (n + 1)) / 2\$

The difference between actual array sum and math sum will give us the single duplicate number.

In Kotlin, we can write the below program to find the single duplicate number:

Kotlin: print single duplicate number
val array = arrayOf(1, 2, 3, 4, 5, 6, 7, 8, 9, 2)
val sum = array.sum()
val mathSum = ((array.size -1) * array.size) / 2
val duplicateNumber = sum - mathSum
Program output


Top articles in this category:
  1. Citibank Java developer interview questions
  2. BlackRock Top Java Interview Questions: Investment Banking Domain
  3. Cracking core java interviews - question bank
  4. Morgan Stanley Java Interview Questions
  5. Top 50 Multi-threading Java Interview Questions for Investment Banking Domain
  6. Sapient Global Market Java Interview Questions and Coding Exercise
  7. UBS Top 10 Java Interview Questions

Find more on this topic:
Java Interviews image
Java Interviews

Interview - Product Companies, eCommerce Companies, Investment Banking, Healthcare Industry, Service Companies and Startups.

Last updated 1 month ago

Recommended books for interview preparation:
You may also be interested in..
Generic placeholder image
ebook PDF - Cracking Spring Microservices Interviews for Java Developers
You may also be interested in..
Generic placeholder image
ebook PDF - Cracking Java Interviews v3.5 by Munish Chandel

This website uses cookies to ensure you get the best experience on our website. more info