Given an array arr[] of N integers, the task is to find the longest sub-sequence in the given array such that for all pairs from the sub-sequence (arr[i], arr[j]) where i != j either arr[i] divides arr[j] or vice versa. If no such sub-sequence exists then print -1.
Examples:
Input: arr[] = {2, 4, 6, 1, 3, 11}
Output: 3
Longest valid sub-sequences are {1, 2, 6} and {1, 3, 6}.
Input: arr[] = {21, 22, 6, 4, 13, 7, 332}
Output: 2
Approach: This problem is a simple variation of the longest increasing sub-sequence problem. What changes is the base condition and the trick to reduce the number of computations by sorting the given array.
- First sort the given array so that we only need to check values where arr[i] > arr[j] for i > j.
- Then we move forward using two loops, outer loop runs from 1 to N and the inner loop runs from 0 to i.
- Now in the inner loop we have to find the number of arr[j] where j is from 0 to i – 1 which divides the element arr[i].
- And the recurrence relation will be dp[i] = max(dp[i], 1 + dp[j]).
- We will update the max dp[i] value in a variable named res which will be the final answer.
Below is the implementation of the above approach:
C++
CPP
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the length of the// longest required sub-sequenceint find(int n, int a[]){ // Sort the array sort(a, a + n); // To store the resultant length int res = 1; int dp[n]; // If array contains only one element // then it divides itself dp[0] = 1; for (int i = 1; i < n; i++) { // Every element divides itself dp[i] = 1; // Count for all numbers which // are lesser than a[i] for (int j = 0; j < i; j++) { if (a[i] % a[j] == 0) { // If a[i] % a[j] then update the maximum // subsequence length, // dp[i] = max(dp[i], 1 + dp[j]) // where j is in the range [0, i-1] dp[i] = std::max(dp[i], 1 + dp[j]); } } res = std::max(res, dp[i]); } // If array contains only one element // then i = j which doesn't satisfy the condition return (res == 1) ? -1 : res;}// Driver codeint main(){ int a[] = { 2, 4, 6, 1, 3, 11 }; int n = sizeof(a) / sizeof(int); cout << find(n, a); return 0;} |
Java
// Java implementation of the approachimport java.util.Arrays; import java.io.*;class GFG { // Function to return the length of the// longest required sub-sequencestatic int find(int n, int a[]){ // Sort the array Arrays.sort(a); // To store the resultant length int res = 1; int dp[] = new int[n]; // If array contains only one element // then it divides itself dp[0] = 1; for (int i = 1; i < n; i++) { // Every element divides itself dp[i] = 1; // Count for all numbers which // are lesser than a[i] for (int j = 0; j < i; j++) { if (a[i] % a[j] == 0) { // If a[i] % a[j] then update the maximum // subsequence length, // dp[i] = Math.max(dp[i], 1 + dp[j]) // where j is in the range [0, i-1] dp[i] = Math.max(dp[i], 1 + dp[j]); } } res = Math.max(res, dp[i]); } // If array contains only one element // then i = j which doesn't satisfy the condition return (res == 1) ? -1 : res;}// Driver codepublic static void main (String[] args) { int a[] = { 2, 4, 6, 1, 3, 11 }; int n = a.length; System.out.println (find(n, a));}}// This code is contributed by jit_t |
Python3
# Python3 implementation of the approach # Function to return the length of the # longest required sub-sequence def find(n, a) : # Sort the array a.sort(); # To store the resultant length res = 1; dp = [0]*n; # If array contains only one element # then it divides itself dp[0] = 1; for i in range(1, n) : # Every element divides itself dp[i] = 1; # Count for all numbers which # are lesser than a[i] for j in range(i) : if (a[i] % a[j] == 0) : # If a[i] % a[j] then update the maximum # subsequence length, # dp[i] = max(dp[i], 1 + dp[j]) # where j is in the range [0, i-1] dp[i] = max(dp[i], 1 + dp[j]); res = max(res, dp[i]); # If array contains only one element # then i = j which doesn't satisfy the condition if (res == 1): return -1 else : return res; # Driver code if __name__ == "__main__" : a = [ 2, 4, 6, 1, 3, 11 ]; n = len(a); print(find(n, a)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the length of the// longest required sub-sequencestatic int find(int n, int []a){ // Sort the array Array.Sort(a); // To store the resultant length int res = 1; int []dp = new int[n]; // If array contains only one element // then it divides itself dp[0] = 1; for (int i = 1; i < n; i++) { // Every element divides itself dp[i] = 1; // Count for all numbers which // are lesser than a[i] for (int j = 0; j < i; j++) { if (a[i] % a[j] == 0) { // If a[i] % a[j] then update the maximum // subsequence length, // dp[i] = Math.max(dp[i], 1 + dp[j]) // where j is in the range [0, i-1] dp[i] = Math.Max(dp[i], 1 + dp[j]); } } res = Math.Max(res, dp[i]); } // If array contains only one element // then i = j which doesn't satisfy the condition return (res == 1) ? -1 : res;}// Driver codepublic static void Main () { int []a = { 2, 4, 6, 1, 3, 11 }; int n = a.Length; Console.WriteLine(find(n, a));}}// This code is contributed by anuj_67.. |
javascript
<script>// javascript implementation of the approach// Function to return the length of the// longest required sub-sequencefunction find(n , a){ // Sort the array a.sort(); // To store the resultant length var res = 1; var dp = Array.from({length: n}, (_, i) => 0); // If array contains only one element // then it divides itself dp[0] = 1; for (var i = 1; i < n; i++) { // Every element divides itself dp[i] = 1; // Count for all numbers which // are lesser than a[i] for (var j = 0; j < i; j++) { if (a[i] % a[j] == 0) { // If a[i] % a[j] then update the maximum // subsequence length, // dp[i] = Math.max(dp[i], 1 + dp[j]) // where j is in the range [0, i-1] dp[i] = Math.max(dp[i], 1 + dp[j]); } } res = Math.max(res, dp[i]); } // If array contains only one element // then i = j which doesn't satisfy the condition return (res == 1) ? -1 : res;}// Driver code var a = [ 2, 4, 6, 1, 3, 11 ]; var n = a.length; document.write(find(n, a));// This code is contributed by jit_t// This code is contributed by 29AjayKumar </script> |
Output:
3
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