Related Articles

# Length of longest increasing index dividing subsequence

• Difficulty Level : Expert
• Last Updated : 16 Apr, 2021

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up