Write a C program to find the duplicate element in an array?

Find the duplicate number in array: For example, an array with length 5 contains numbers {1, 2, 3, 4, 4}, the duplicated number is 4. Suppose that the duplicated number in the array is m in given example. The sum of all numbers in the array, denoted as arrSum, should be the result of 1+2+...+(n-1)+m. It is not difficult to get the sum of 1+2+...+(n-1), which is denoted as sum. The duplicated number m will be the difference of arrSum and sum. Following C program demonstrates how do we detect a repeated element in an integer array?.

/* An array of size N contains numbers ranging from 1 to N-1.
   There is exactly one number is duplicated in the array.
   Write a C program to find the duplicated number?
*/
#include <stdio.h>
#define N 5
int main()
{
  int arr[N]; //contains integers from 1 to N-1
  int arrSum = 0; //sum of array elements
  int sum = (N * (N - 1)) / 2; //sum of 1 to N-1
  int i; //loop counter
  printf("Enter %d integers form 1 to %d, with one duplicate:", N, N-1);
  for (i = 0; i < N; i++)
  {
    scanf ("%d", &arr[i]);
    arrSum += arr[i];
  }
 
  printf("\nDuplicate number is: %d\n", arrSum - sum);
}


Get Free Tutorials by Email

About the Author

is the main author for cs-fundamentals.com. He is a software professional (post graduated from BITS-Pilani) and loves writing technical articles on programming and data structures.

Today's Tech News

Facebook to exclude billions from European privacy lawsPosted on Thursday April 19, 2018

A total of 1.5 billion international Facebook users will not be protected under GDPR.

TalkTalk and Vodafone top complaints chart againPosted on Thursday April 19, 2018

TalkTalk, Vodafone and BT drew the most complaints for their services, the regulator says.

Ikea's TaskRabbit app back online after data breachPosted on Thursday April 19, 2018

The odd jobs marketplace has been investigating a "cyber-security incident".

Courtesy BBC News

AD BLOCKER DETECTED!

Advertisements help running this site for free.


To view the content please disable AdBlocker and refresh the page.

×