Issue
As a part of the Java interview question paper I have got following issue to solve. But I am bit wonder whether how can I implement it without any Collection or intermediate Array.
Question:- Count duplicates from int array without using any Collection or another intermediate Array
Input values:- {7,2,6,1,4,7,4,5,4,7,7,3, 1}
Output:- Number of duplicates values: 3
Duplicates values: 7, 4, 1
I have implemented following solution but was not completed one. Any one has some idea? Thanks.
public static void duplicate(int numbers[]) {
for (int i = 0; i < numbers.length; i++) {
boolean duplicate = false;
int j = 0;
while (j < i){
if ((i != j) && numbers[i] == numbers[j]) {
duplicate = true;
}
j++;
}
if (duplicate) {
System.out.print(numbers[i] + " ");
}
}
}
Solution
The easiest way to solve this problem is to sort the array first, and then just walk through the array counting duplicates as you encounter them:
int[] numbers = new int[]{7,2,6,1,4,7,4,5,4,7,7,3,1};
int temp = 0;
// I chose to do a bubble sort of the array,
// but you are free to use any method you wish (e.g. Arrays.sort)
System.out.print("Duplicates values: ");
for (int i=0; i < numbers.length; ++i) {
for (int j=1; j < (numbers.length - i); ++j) {
if (numbers[j-1] > numbers[j]) {
temp = numbers[j-1];
numbers[j-1] = numbers[j];
numbers[j] = temp;
}
}
}
// walk through the sorted array and count duplicates
int numDup = 0, dupCount = 0;
int previous = -1;
for (int i=0; i < numbers.length; ++i) {
if (numbers[i] == previous) {
++numDup;
if (numDup == 1) {
++dupCount;
if (dupCount == 1) {
System.out.print(numbers[i]);
}
else {
System.out.print(", " + numbers[i]);
}
}
}
else {
previous = numbers[i];
numDup = 0;
}
}
System.out.println("\nNumber of duplicates values: " + dupCount);
Output:
Duplicates values: 1, 4, 7
Number of duplicates values: 3
Note that my output order is reverse of what you have, because you need to read through the entire array before you know how many total duplicates you have. Also, I will point out that the only state this solution uses is the input array itself, plus a couple of int
varibles here and there.
This code has been tested in IntelliJ and it works correctly.
Answered By - Tim Biegeleisen