# Length of longest increasing index dividing subsequence

Given an array arr[] of size N, the task is to find the longest increasing sub-sequence such that index of any element is divisible by index of previous element (LIIDS). The following are the necessary conditions for the LIIDS:
If i, j are two indices in the given array. Then:

• i < j
• j % i = 0
• arr[i] < arr[j]

Examples:

Input: arr[] = {1, 2, 3, 7, 9, 10}
Output:
Explanation:
The subsequence = {1, 2, 7}
Indices of the elements of sub-sequence = {1, 2, 4}
Indices condition:
2 is divisible by 1
4 is divisible by 2
OR
Another possible sub-sequence = {1, 3, 10}
Indices of elements of sub-sequence = {1, 3, 6}
Indices condition:
3 is divisible by 1
6 is divisible by 3
Input: arr[] = {7, 1, 3, 4, 6}
Output:
Explanation:
The sub-sequence = {1, 4}
Indices of elements of sub-sequence = {2, 4}
Indices condition:
4 is divisible by 2

Approach: The idea is to use the concept of Dynamic programming to solve this problem.

• Create a dp[] array first and initialize the array with 1. This is because, the minimum length of the dividing subsequence is 1.
• Now, for every index ‘i’ in the array, increment the count of the values at the indices at all its multiples.
• Finally, when the above step is performed for all the values, the maximum value present in the array is the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the length of` `// the longest increasing sub-sequence` `// such that the index of the element is` `// divisible by index of previous element`   `#include ` `using` `namespace` `std;`   `// Function to find the length of` `// the longest increasing sub-sequence` `// such that the index of the element is` `// divisible by index of previous element` `int` `LIIDS(``int` `arr[], ``int` `N)` `{` `    ``// Initialize the dp[] array with 1 as a` `    ``// single element will be of 1 length` `    ``int` `dp[N + 1];`   `    ``int` `ans = 0;` `    ``for` `(``int` `i = 1; i <= N; i++) {` `        ``dp[i] = 1;` `    ``}`   `    ``// Traverse the given array` `    ``for` `(``int` `i = 1; i <= N; i++) {`   `        ``// If the index is divisible by` `        ``// the previous index` `        ``for` `(``int` `j = i + i; j <= N; j += i) {`   `            ``// if increasing` `            ``// subsequence identified` `            ``if` `(arr[j] > arr[i]) {` `                ``dp[j] = max(dp[j], dp[i] + 1);` `            ``}` `        ``}`   `        ``// Longest length is stored` `        ``ans = max(ans, dp[i]);` `    ``}` `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `arr[] = { 1, 2, 3, 7, 9, 10 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``cout << LIIDS(arr, N);` `    ``return` `0;` `}`

## Java

 `// Java program to find the length of` `// the longest increasing sub-sequence` `// such that the index of the element is` `// divisible by index of previous element` `import` `java.util.*;`   `class` `GFG{`   `// Function to find the length of` `// the longest increasing sub-sequence` `// such that the index of the element is` `// divisible by index of previous element` `static` `int` `LIIDS(``int` `arr[], ``int` `N)` `{`   `    ``// Initialize the dp[] array with 1 as a` `    ``// single element will be of 1 length` `    ``int``[] dp = ``new` `int``[N + ``1``];` `    ``int` `ans = ``0``;` `    `  `    ``for``(``int` `i = ``1``; i <= N; i++)` `    ``{` `       ``dp[i] = ``1``;` `    ``}`   `    ``// Traverse the given array` `    ``for``(``int` `i = ``1``; i <= N; i++)` `    ``{` `        `  `       ``// If the index is divisible by` `       ``// the previous index` `       ``for``(``int` `j = i + i; j <= N; j += i)` `       ``{` `           `  `          ``// If increasing` `          ``// subsequence identified` `          ``if` `(j < arr.length && arr[j] > arr[i])` `          ``{` `              ``dp[j] = Math.max(dp[j], dp[i] + ``1``);` `          ``}` `       ``}` `       `  `       ``// Longest length is stored` `       ``ans = Math.max(ans, dp[i]);` `    ``}` `    ``return` `ans;` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``7``, ``9``, ``10` `};` `    ``int` `N = arr.length;`   `    ``System.out.println(LIIDS(arr, N));` `}` `}`   `// This code is contributed by AbhiThakur`

## Python3

 `# Python3 program to find the length of ` `# the longest increasing sub-sequence ` `# such that the index of the element is ` `# divisible by index of previous element `   `# Function to find the length of ` `# the longest increasing sub-sequence ` `# such that the index of the element is ` `# divisible by index of previous element ` `def` `LIIDS(arr, N):` `    `  `    ``# Initialize the dp[] array with 1 as a ` `    ``# single element will be of 1 length ` `    ``dp ``=` `[]` `    ``for` `i ``in` `range``(``1``, N ``+` `1``):` `        ``dp.append(``1``)` `    `  `    ``ans ``=` `0` `    `  `    ``# Traverse the given array ` `    ``for` `i ``in` `range``(``1``, N ``+` `1``):` `        `  `        ``# If the index is divisible by ` `        ``# the previous index ` `        ``j ``=` `i ``+` `i` `        ``while` `j <``=` `N:` `            `  `            ``# If increasing ` `            ``# subsequence identified ` `            ``if` `j < N ``and` `i < N ``and` `arr[j] > arr[i]:` `                ``dp[j] ``=` `max``(dp[j], dp[i] ``+` `1``)` `            `  `            ``j ``+``=` `i` `            `  `        ``# Longest length is stored ` `        ``if` `i < N:` `            ``ans ``=` `max``(ans, dp[i])` `        `  `    ``return` `ans`   `# Driver code ` `arr ``=` `[ ``1``, ``2``, ``3``, ``7``, ``9``, ``10` `]`   `N ``=` `len``(arr)`   `print``(LIIDS(arr, N))`   `# This code is contributed by ishayadav181`

## C#

 `// C# program to find the length of` `// the longest increasing sub-sequence` `// such that the index of the element is` `// divisible by index of previous element` `using` `System; `   `class` `GFG{`   `// Function to find the length of` `// the longest increasing sub-sequence` `// such that the index of the element is` `// divisible by index of previous element` `static` `int` `LIIDS(``int``[] arr, ``int` `N)` `{`   `    ``// Initialize the dp[] array with 1 as a` `    ``// single element will be of 1 length` `    ``int``[] dp = ``new` `int``[N + 1];` `    ``int` `ans = 0;` `    `  `    ``for``(``int` `i = 1; i <= N; i++)` `    ``{` `        ``dp[i] = 1;` `    ``}`   `    ``// Traverse the given array` `    ``for``(``int` `i = 1; i <= N; i++)` `    ``{` `        `  `        ``// If the index is divisible by` `        ``// the previous index` `        ``for``(``int` `j = i + i; j <= N; j += i)` `        ``{` `            `  `            ``// If increasing` `            ``// subsequence identified` `            ``if` `(j < arr.Length && arr[j] > arr[i])` `            ``{` `                ``dp[j] = Math.Max(dp[j], dp[i] + 1);` `            ``}` `        ``}` `        `  `        ``// Longest length is stored` `        ``ans = Math.Max(ans, dp[i]);` `    ``}` `    ``return` `ans;` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int``[] arr = ``new` `int``[] { 1, 2, 3, 7, 9, 10 };` `    ``int` `N = arr.Length;`   `    ``Console.WriteLine(LIIDS(arr, N));` `}` `}`   `// This code is contributed by sanjoy_62`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N log(N)), where N is the length of the array.

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