Generate an array of size N according to the given rules
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:
- 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
- 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:
// 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
# 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# 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
<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>
// 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
Output
0 0 1 0 2
Time Complexity: O(N2)
Auxiliary Space: O(1)
Using Dictionary:
To generate the array of size N while adhering to the given rules with improved efficiency, we can utilize an approach that avoids searching for elements in the array for each iteration. Instead, we can maintain a dictionary to store the most recent index of each element encountered. This approach reduces the time complexity significantly compared to the provided solution by eliminating the need for repeated linear searches.
- We initialize a dictionary to store the most recent index of each element encountered in the array.
- We start filling the array with zeroes.
- For each iteration, we check if the current element has occurred previously.
- If yes, we calculate the difference between the current index and the second most recent occurrence of the element.
- If not, we continue filling the array with zeroes.
- Finally, we return the generated array.
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
vector<int> genArray(int N)
{
vector<int> arr(N);
unordered_map<int, int>
recentIndex; // Map to store most recent index of
// elements
for (int i = 1; i < N; i++) {
if (recentIndex.find(arr[i - 1])
!= recentIndex.end()
&& recentIndex[arr[i - 1]] != i - 1) {
arr[i] = (i - 1) - recentIndex[arr[i - 1]];
}
recentIndex[arr[i - 1]] = i - 1;
}
return arr;
}
// Driver code
int main()
{
int N = 5;
vector<int> result = genArray(N);
for (int num : result) {
cout << num << " ";
}
cout << endl;
return 0;
}
import java.util.HashMap;
import java.util.Map;
public class GenArray {
public static int[] genArray(int N)
{
int[] arr = new int[N];
Map<Integer, Integer> recentIndex
= new HashMap<>(); // Dictionary to store most
// recent index of elements
for (int i = 1; i < N; i++) {
if (recentIndex.containsKey(arr[i - 1])
&& recentIndex.get(arr[i - 1]) != i - 1) {
arr[i]
= (i - 1) - recentIndex.get(arr[i - 1]);
}
recentIndex.put(arr[i - 1], i - 1);
}
return arr;
}
// Driver code
public static void main(String[] args)
{
int N = 5;
int[] result = genArray(N);
for (int num : result) {
System.out.print(num + " ");
}
}
}
// This code is contributed by shivamgupta0987654321
def genArray(N):
arr = [0] * N
recent_index = {} # Dictionary to store most recent index of elements
for i in range(1, N):
if arr[i - 1] in recent_index and recent_index[arr[i - 1]] != i - 1:
arr[i] = i - 1 - recent_index[arr[i - 1]]
recent_index[arr[i - 1]] = i - 1
return arr
# Driver code
if __name__ == "__main__":
N = 5
result = genArray(N)
print(result)
function genArray(N) {
let arr = new Array(N).fill(0);
let recentIndex = {}; // Object to store most recent index of elements
for (let i = 1; i < N; i++) {
if (recentIndex[arr[i - 1]] !== undefined && recentIndex[arr[i - 1]] !== i - 1) {
arr[i] = i - 1 - recentIndex[arr[i - 1]];
}
recentIndex[arr[i - 1]] = i - 1;
}
return arr;
}
// Driver code
let N = 5;
let result = genArray(N);
console.log(result);
Output
[0, 0, 1, 0, 2]
Time Complexity: O(N)
Space Complexity: O(N)
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