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Generate an array of size N according to the given rules

  • Last Updated : 26 Mar, 2021

Given a number N, the task is to create an array arr[] of size N, where the value of the element at every index i is filled according to the following rules: 
 

  1. arr[i] = ((i – 1) – k), where k is the index of arr[i – 1] that has appeared second most recently. This rule is applied when arr[i – 1] is present more than once in the array
  2. arr[i] = 0. This rule is applied when arr[i – 1] is present only once or when i = 1.

Examples: 
 

Input: N = 8 
Output: 0 0 1 0 2 0 2 2 
Explanation: 
For i = 0: There is no element in the array arr[]. So, 0 is placed in the first index. arr[] = {0}. 
For i = 1: There is only one element in the array arr[]. The occurrence of arr[i – 1] (= 0) is only one. Therefore, the array is filled according to the rule 2. arr = {0, 0}. 
For i = 2: There are two elements in the array. The second most occurrence of arr[i – 1] = arr[0] = 0. So, arr[2] = 1. arr[] = {0, 0, 1}. 
For i = 3: There is no second occurrence of arr[i – 1] = 1. Therefore, arr[3] = 0. arr[] = {0, 0, 1, 0} 
For i = 4: The second recent occurrence of arr[i – 1] = 0 is 1. Therefore, (i – 1 – k) = (4 – 1 – 1) = 2. arr[] = {0, 0, 1, 0, 2}. 
For i = 5: There is only one occurrence of arr[i – 1] = 2. Therefore, arr[5] = 0. arr[] = {0, 0, 1, 0, 2, 0}. 
For i = 6: The second recent occurrence of arr[i – 1] = 0 is 3. Therefore, (i – 1 – k) = (6 – 1 – 3) = 2. arr[] = {0, 0, 1, 0, 2, 0, 2}. 
For i = 7: The second recent occurrence of arr[i – 1] = 2 is 4. Therefore, (i – 1 – k) = (7 – 1 – 4) = 2. arr[] = {0, 0, 1, 0, 2, 0, 2, 2}
Input: N = 5 
Output: 0 0 1 0 2 
 

 

Approach: The idea is to first create an array filled with zeroes of size N. Then for every iteration, we search if the element has occurred in the near past. If yes, then we follow rule 1. Else, rule 2 is followed to fill the array. 
Below is the implementation of the above approach: 
 

C++




// C++ implementation to generate
// an array of size N by following
// the given rules
#include <bits/stdc++.h>
using namespace std;
 
// Function to search the most recent
// location of element N
// If not present in the array
// it will return -1
int search(int a[], int k, int x)
{
    int j;
     
    for ( j = k - 1; j > -1 ; j--)
    {
        if(a[j] == x)
            return j ;
    }
                 
        return -1 ;
}
 
// Function to generate an array
// of size N by following the given rules
void genArray(int arr[], int N)
{
 
    // Loop to fill the array
    // as per the given rules
    for(int i = 0; i < N - 1; i++)
    {
 
        // Check for the occurrence
        // of arr[i - 1]
        if(search(arr, i, arr[i]) == -1)
                arr[i + 1] = 0 ;
 
        else
            arr[i + 1] = (i-search(arr, i, arr[i])) ;
    }
}
 
// Driver code
int main()
{
    int N = 5 ;
    int size = N + 1 ;
    int a[] = {0, 0, 0, 0, 0};
    genArray(a, N) ;
     
    for (int i = 0; i < N ; i ++)
        cout << a[i] << " " ;
        return 0;
}
 
 
// This code is contributed by shivanisinghss2110

Java




// Java implementation to generate
// an array of size N by following
// the given rules
class GFG
{
 
static int a[];
 
// Function to search the most recent
// location of element N
// If not present in the array
// it will return -1
static int search(int a[],int k, int x)
{
    int j;
     
    for ( j = k - 1; j > -1 ; j--)
    {
            if(a[j] == x)
                return j ;
        }
                 
        return -1 ;
}
 
// Function to generate an array
// of size N by following the given rules
static void genArray(int []arr, int N)
{
 
    // Loop to fill the array
    // as per the given rules
    for(int i = 0; i < N - 1; i++)
    {
 
        // Check for the occurrence
        // of arr[i - 1]
        if(search(arr, i, arr[i]) == -1)
                arr[i + 1] = 0 ;
 
        else
            arr[i + 1] = (i-search(arr, i, arr[i])) ;
    }
}
 
// Driver code
public static void main (String[] args)
{
    int N = 5 ;
    int size = N + 1 ;
    int a[] = new int [N];
    genArray(a, N) ;
     
    for (int i = 0; i < N ; i ++)
        System.out.print(a[i]+" " );
 
}
}
 
// This code is contributed by Yash_R

Python3




# Python implementation to generate
# an array of size N by following
# the given rules
 
 
# Function to search the most recent
# location of element N
# If not present in the array
# it will return -1
def search(a, k, x):
        for j in range(k-1, -1, -1) :
            if(a[j]== x):
                return j
                 
        return -1
 
# Function to generate an array
# of size N by following the given rules
def genArray(arr, N):
 
    # Loop to fill the array
    # as per the given rules
    for i in range(0, N-1, 1):
 
        # Check for the occurrence
        # of arr[i - 1]
        if(search(arr, i, arr[i])==-1):
                arr[i + 1]= 0
 
        else:
            arr[i + 1]=(i-search(arr, i, arr[i]))
             
# Driver code     
if __name__ == "__main__":
    N = 5
    size = N + 1
    a =[0]*N
    genArray(a, N)
     
    print(a)

C#




// C# implementation to generate
// an array of size N by following
// the given rules
 
using System;
 
public class GFG
{
 
static int []a;
 
// Function to search the most recent
// location of element N
// If not present in the array
// it will return -1
static int search(int []a,int k, int x)
{
    int j;
     
    for ( j = k - 1; j > -1 ; j--)
    {
            if(a[j] == x)
                return j ;
        }
                 
        return -1 ;
}
 
// Function to generate an array
// of size N by following the given rules
static void genArray(int []arr, int N)
{
 
    // Loop to fill the array
    // as per the given rules
    for(int i = 0; i < N - 1; i++)
    {
 
        // Check for the occurrence
        // of arr[i - 1]
        if(search(arr, i, arr[i]) == -1)
                arr[i + 1] = 0 ;
 
        else
            arr[i + 1] = (i-search(arr, i, arr[i])) ;
    }
}
 
// Driver code
public static void Main (string[] args)
{
    int N = 5 ;
    int size = N + 1 ;
    int []a = new int [N];
    genArray(a, N) ;
     
    for (int i = 0; i < N ; i ++)
        Console.Write(a[i]+" " );
 
}
}
// This code is contributed by AnkitRai01

Javascript




<script>
 
// Javascript implementation to generate
// an array of size N by following
// the given rules
 
// Function to search the most recent
// location of element N
// If not present in the array
// it will return -1
function search( a, k, x)
{
    var j;
     
    for ( j = k - 1; j > -1 ; j--)
    {
        if(a[j] == x)
            return j ;
    }
                 
        return -1 ;
}
 
// Function to generate an array
// of size N by following the given rules
function genArray(arr, N)
{
 
    // Loop to fill the array
    // as per the given rules
    for(var i = 0; i < N - 1; i++)
    {
 
        // Check for the occurrence
        // of arr[i - 1]
        if(search(arr, i, arr[i]) == -1)
                arr[i + 1] = 0 ;
 
        else
            arr[i + 1] = (i-search(arr, i, arr[i])) ;
    }
}
 
// Driver code
var N = 5 ;
var size = N + 1 ;
var a = [0, 0, 0, 0, 0];
genArray(a, N) ;
document.write("["+a+"]");
 
// This code is contributed by rutvik_56.
</script>
Output: 
[0, 0, 1, 0, 2]

 

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