# 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

`// 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 ` |

*chevron_right*

*filter_none*

## 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) ` |

*chevron_right*

*filter_none*

**Output:**

[0, 0, 1, 0, 2]

## Recommended Posts:

- Generate an array of size K which satisfies the given conditions
- Generate original array from an array that store the counts of greater elements on right
- Generate an array B[] from the given array A[] which satisfies the given conditions
- Merge an array of size n into another array of size m+n
- Generate an array using given conditions from a given array
- Generate an Array in which count of even and odd sum sub-arrays are E and O respectively
- Generate elements of the array following given conditions
- Generate an array having Bitwise AND of the previous and the next element
- Generate array with minimum sum which can be deleted in P steps
- Generate two BSTs from the given array such that maximum height among them is minimum
- Generate original array from difference between every two consecutive elements
- Count number of permutation of an Array having no SubArray of size two or more from original Array
- Maximize the size of array by deleting exactly k sub-arrays to make array prime
- Structure Sorting (By Multiple Rules) in C++
- Find the element in the matrix generated by given rules

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.