You are given an array A of size N. Your task is to find the length of largest dividing sub sequence.A dividing sequence is a sequence a1, a2, …, aN where ai divides aj whenever i < j. For example, 3, 15, 60, 720 is a dividing sequence.
input-
The first line of each test case is N, where N is the size of array.
The second line of each test case contains N space separated integers which is the input for the array.
Output-
the length of largest dividing sub sequence
examples:
Input : arr[] = {2 11 16 12 36 60 71 17 29 144 288 129 432 993}
Output : 5
2 12 36 144 288 is dividing sub sequence of largest size
Input : 1 2 4 8 16
Output : 5
Whole sequence is dividing
This problem is simply a variation of Longest Increasing Subsequence. We can solve this using Dynamic Programming. The idea is to find the longest dividing subsequence ending with every element and finally return maximum of all.
C++
/* Dynamic Programming C++ implementation of lds problem */#include<bits/stdc++.h> using namespace std; /* lds() returns the length of the longest dividing subsequence in arr[] of size n */int lds( int arr[], int n ) { int lds[n]; lds[0] = 1; /* Compute optimized lds values in bottom up manner */ for (int i = 1; i < n; i++ ) { lds[i] = 1; for (int j = 0; j < i; j++ ) if (lds[j] != 0 && arr[i] % arr[j] == 0) lds[i] = max(lds[i], lds[j] + 1); } // Return maximum value in lds[] return *max_element(lds, lds+n);} /* Driver program to test above function */int main() { int arr[] = { 2, 11, 16, 12, 36, 60, 71, 17, 29, 144, 288, 129, 432, 993}; int n = sizeof(arr)/sizeof(arr[0]); printf("Length of lds is %d\n", lds( arr, n ) ); return 0; } |
Java
/* Dynamic Programming Java implementation of lds problem */import java.util.*;import java.lang.*;import java.io.*;class GFG{/* lds() returns the length of the longest dividing subsequence in arr[] of size n */static int lds( Integer arr[], int n ) { Integer lds[]=new Integer[n]; lds[0] = 1; /* Compute optimized lds values in bottom up manner */ for (int i = 1; i < n; i++ ) { lds[i] = 1; for (int j = 0; j < i; j++ ) if (lds[j] != 0 && arr[i] % arr[j] == 0) lds[i] = Math.max(lds[i], lds[j] + 1); } // Return maximum value in lds[] int max=(int)Collections.max(Arrays.asList(lds)); return max;} /* Driver program to test above function */public static void main(String args[]) { Integer arr[] = { 2, 11, 16, 12, 36, 60, 71, 17, 29, 144, 288, 129, 432, 993}; int n =arr.length ; System.out.println("Length of lds is "+lds( arr, n ) ); }} |
Python3
# Dynamic Programming Python3 # implementation of lds problem# lds() returns the length of the longest # dividing subsequence in arr[] of size ndef lds(arr, n): lds = [0 for i in range(n)] lds[0] = 1 # Compute optimized lds values # in bottom up manner for i in range(n): lds[i] = 1 for j in range(i): if (lds[j] != 0 and arr[i] % arr[j] == 0): lds[i] = max(lds[i], lds[j] + 1) return max(lds)# Driver Codearr = [2, 11, 16, 12, 36, 60, 71, 17, 29, 144, 288, 129, 432, 993]print("Length of lds is", lds(arr, len(arr)))# This code is contributed # by Mohit Kumar |
C#
/* Dynamic Programming C# implementation of lds problem */using System;using System.Linq;public class GFG{ /* lds() returns the length of the longest dividing subsequence in arr[] of size n */ static int lds( int []arr, int n ) { int []lds=new int[n]; lds[0] = 1; /* Compute optimized lds values in bottom up manner */ for (int i = 1; i < n; i++ ) { lds[i] = 1; for (int j = 0; j < i; j++ ) if (lds[j] != 0 && arr[i] % arr[j] == 0) lds[i] = Math.Max(lds[i], lds[j] + 1); } // Return maximum value in lds[] int max=lds.Max(); return max; } /* Driver program to test above function */ public static void Main() { int []arr = { 2, 11, 16, 12, 36, 60, 71, 17, 29, 144, 288, 129, 432, 993}; int n =arr.Length ; Console.Write("Length of lds is "+lds( arr, n ) ); }}// This code is contributed by 29AjayKumar |
Javascript
<script>/* Dynamic Programming Javascript implementation of lds problem *//* lds() returns the length of the longest dividing subsequence in arr[] of size n */function lds(arr,n){ let lds = new Array(n); lds[0] = 1; /* Compute optimized lds values in bottom up manner */ for (let i = 1; i < n; i++ ) { lds[i] = 1; for (let j = 0; j < i; j++ ) if (lds[j] != 0 && arr[i] % arr[j] == 0) lds[i] = Math.max(lds[i], lds[j] + 1); } // Return maximum value in lds[] let max=Math.max(...lds); return max;}/* Driver program to test above function */let arr=[2, 11, 16, 12, 36, 60, 71, 17, 29, 144, 288, 129, 432, 993]; let n =arr.length ;document.write("Length of lds is "+lds( arr, n ) );// This code is contributed by unknown2108</script> |
Output:
Length of lds is 5
Time Complexity: O(N*N )
Auxiliary Space: O(N)
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