The initial occurrence of a duplicated element within the array is identified at index 4, which corresponds to the element (2) that is duplicated from index 1. Therefore, the duplicate elements within the provided array are 2, 3, and 8.
ALGORITHM:
- STEP 1: START
- STEP 2: INITIALIZE arr= {1, 2, 3, 4, 2, 7, 8, 8, 3}.
- STEP 3: length = sizeof(arr)/sizeof(arr[0])
- STEP 4: PRINT "Duplicate elements in given array:"
- STEP 5: SET i=0. REPEAT STEP 6 to STEP 9 UNTIL i<length
- STEP 6: SET j=i+1. REPEAT STEP 7 and STEP 8 UNTIL i<length
- STEP 7: if(arr[i] == arr[j]) PRINT arr[j]
- STEP 8: j=j+1
- STEP 9: i+1.
- STEP 10: RETURN 0.
- STEP 11: END
PROGRAM:
Example
#include <stdio.h>
int main()
{
//Initialize array
int arr[] = {1, 2, 3, 4, 2, 7, 8, 8, 3};
//Calculate length of array arr
int length = sizeof(arr)/sizeof(arr[0]);
printf("Duplicate elements in given array: \n");
//Searches for duplicate element
for(int i = 0; i < length; i++) {
for(int j = i + 1; j < length; j++) {
if(arr[i] == arr[j])
printf("%d\n", arr[j]);
}
}
return 0;
}
Output:
Output
Duplicate elements in given array:
2
3
8