Question

Given an array A of n integers, describe an O(n log n) algorithm to decide if the elements of A are all distinct. Why does your algorithm run in O(n log n) time?

Answer 1

#include<iostream>

using namespace std;

/*Merges two subarrays of arr[].

// First subarray is arr[l..m]

*/ Second subarray is arr[m+1..r]

void merge(int arr[], int l, int m, int r)

{

    int n1 = m - l + 1;

    int n2 = r - m;

    // Create temp arrays

    int L[n1], R[n2];

    // Copy data to temp arrays L[] and R[]

    for(int i = 0; i < n1; i++)

        L[i] = arr[l + i];

    for(int j = 0; j < n2; j++)

        R[j] = arr[m + 1 + j];

    // Merge the temp arrays back into arr[l..r]

    

    // Initial index of first subarray

    int i = 0;

    

    // Initial index of second subarray

    int j = 0;

    

    // Initial index of merged subarray

    int k = l;

    

    while (i < n1 && j < n2)

    {

        if (L[i] <= R[j])

        {

            arr[k] = L[i];

            i++;

        }

        else

        {

            arr[k] = R[j];

            j++;

        }

        k++;

    }

    // Copy the remaining elements of

    // L[], if there are any

    while (i < n1)

    {

        arr[k] = L[i];

        i++;

        k++;

    }

    // Copy the remaining elements of

    // R[], if there are any

    while (j < n2)

    {

        arr[k] = R[j];

        j++;

        k++;

    }

}

// l is for left index and r is

// right index of the sub-array

// of arr to be sorted */

void mergeSort(int arr[], int l, int r)

{

    if (l < r)

    {

        

        // Same as (l+r)/2, but avoids

        // overflow for large l and h

        int m = l + (r - l) / 2;

        // Sort first and second halves

        mergeSort(arr, l, m);

        mergeSort(arr, m + 1, r);

        merge(arr, l, m, r);

    }

}

/*Here you have sorted an array using Merge Sort which will sort your array in O(nlogn) time.

and the using liner search you can compare each element if there is a common element adjacent element then you can return false else return true.

*/

int main()

{

    int arr[] = { 12, 11, 13, 5, 6, 7 };

    int arr_size = sizeof(arr) / sizeof(arr[0]);

    cout << "Given array is \n";

    printArray(arr, arr_size);

    mergeSort(arr, 0, arr_size - 1);

for(int i=0;i<arr_size-1;i++)

if(arr[i]==arr[i+1])

cout<<"Array does not contain distinct element \n ";

if(i==arr_size)

court<<"Distinct Element\n";

return 0;

}