Given an array arr[] of N positive integers, the task is to find the largest strictly increasing subsequence of arr[] such that the indices of the selected elements in arr[], and the selected elements are multiples of each other individually.
Note: Consider 1 based indexing for the array arr[].
Examples:
Input: arr[] = {1, 4, 2, 3, 6, 4, 9}
Output: 3
Explanation:
We can choose index 1, 3, 6 and values are 1, 2, 4:
Here every greater index is divisible by smaller index and every greater index value is greater than the smaller index value.
Input: arr[] = {5, 3, 4, 6}
Output: 2
Explanation:
We can choose index 1 and 3 and values are 3 and 6:
Here, every greater index is divisible by smaller index and every greater index value is greater than the smaller index value.
Naive Approach:
The naive approach is to simply generate all possible subsequence and for each subsequence check two conditions:
- first check if elements are in strictly increasing order and
- secondly check if the index of the selected elements in arr[] is multiple of each other.
Out of all possible subsequence which satisfies the given two conditions select the largest subsequence.
Time Complexity: O( N * 2N )
Auxiliary Space: O( N )
Efficient Approach:
We can optimize the code by using Dynamic Programming by avoiding redundant calculation of the repeated sub problem by caching its result.
- Create an array dp[] of size equal to the size of arr[], where dp[i] represents size of largest subsequence till i-th index which satisfies the given conditions.
- Initialize array dp[] with 0.
- Now, iterate the array arr[] from the end.
- For each index find the indices j which divide the current index and check if the value at current index is greater than the element at index j.
- If it is then update dp[j] as:
dp[j] = max(dp[current] + 1, dp[j])
Finally, traverse the array dp[] and print the maximum value.
Below is the implementation of the efficient approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;
// Function that print maximum length// of arrayvoid maxLength(int arr[], int n)
{ // dp[] array to store the
// maximum length
vector<int> dp(n, 1);
for (int i = n - 1; i > 1; i--) {
// Find all divisors of i
for (int j = 1;
j <= sqrt(i); j++) {
if (i % j == 0) {
int s = i / j;
if (s == j) {
// If the current value
// is greater than the
// divisor's value
if (arr[i] > arr[s]) {
dp[s] = max(dp[i] + 1,
dp[s]);
}
}
else {
// If current value
// is greater
// than the divisor's value
// and s is not equal
// to current index
if (s != i
&& arr[i] > arr[s])
dp[s] = max(dp[i] + 1,
dp[s]);
// Condition if current
// value is greater
// than the divisor's value
if (arr[i] > arr[j]) {
dp[j] = max(dp[i] + 1,
dp[j]);
}
}
}
}
}
int max = 0;
// Computing the greatest value
for (int i = 1; i < n; i++) {
if (dp[i] > max)
max = dp[i];
}
// Printing maximum length of array
cout << max << "\n";
}// Driver Codeint main()
{ // Given array arr[]
int arr[] = { 0, 1, 4, 2, 3, 6, 4, 9 };
int size = sizeof(arr) / sizeof(arr[0]);
// Function Call
maxLength(arr, size);
return 0;
} |
Java
// Java program for the above approach import java.util.*;
import java.io.*;
class GFG{
// Function that print maximum length// of arraystatic void maxLength(int arr[], int n)
{ // dp[] array to store the
// maximum length
int dp[] = new int[n];
for(int i = 1; i < n; i++)
{
dp[i] = 1;
}
for(int i = n - 1; i > 1; i--)
{
// Find all divisors of i
for(int j = 1;
j <= Math.sqrt(i); j++)
{
if (i % j == 0)
{
int s = i / j;
if (s == j)
{
// If the current value
// is greater than the
// divisor's value
if (arr[i] > arr[s])
{
dp[s] = Math.max(dp[i] + 1,
dp[s]);
}
}
else
{
// If current value is greater
// than the divisor's value
// and s is not equal
// to current index
if (s != i && arr[i] > arr[s])
dp[s] = Math.max(dp[i] + 1,
dp[s]);
// Condition if current
// value is greater
// than the divisor's value
if (arr[i] > arr[j])
{
dp[j] = Math.max(dp[i] + 1,
dp[j]);
}
}
}
}
}
int max = 0;
// Computing the greatest value
for(int i = 1; i < n; i++)
{
if (dp[i] > max)
max = dp[i];
}
// Printing maximum length of array
System.out.println(max);
}// Driver Codepublic static void main(String[] args)
{ // Given array arr[]
int arr[] = { 0, 1, 4, 2, 3, 6, 4, 9 };
int size = arr.length;
// Function call
maxLength(arr, size);
}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approachfrom math import *
# Function that print maximum length # of arraydef maxLength (arr, n):
# dp[] array to store the
# maximum length
dp = [1] * n
for i in range(n - 1, 1, -1):
# Find all divisors of i
for j in range(1, int(sqrt(i)) + 1):
if (i % j == 0):
s = i // j
if (s == j):
# If the current value
# is greater than the
# divisor's value
if (arr[i] > arr[s]):
dp[s] = max(dp[i] + 1, dp[s])
else:
# If current value
# is greater
# than the divisor's value
# and s is not equal
# to current index
if (s != i and arr[i] > arr[s]):
dp[s] = max(dp[i] + 1, dp[s])
# Condition if current
# value is greater
# than the divisor's value
if (arr[i] > arr[j]):
dp[j] = max(dp[i] + 1, dp[j])
Max = 0
# Computing the greatest value
for i in range(1, n):
if (dp[i] > Max):
Max = dp[i]
# Printing maximum length of array
print(Max)
# Driver Codeif __name__ == '__main__':
# Given array arr[]
arr = [ 0, 1, 4, 2, 3, 6, 4, 9]
size = len(arr)
# Function call
maxLength(arr, size)
# This code is contributed by himanshu77 |
C#
// C# program for the above approach using System;
class GFG{
// Function that print maximum length// of arraystatic void maxLength(int[] arr, int n)
{ // dp[] array to store the
// maximum length
int[] dp = new int[n];
for(int i = 1; i < n; i++)
{
dp[i] = 1;
}
for(int i = n - 1; i > 1; i--)
{
// Find all divisors of i
for(int j = 1;
j <= Math.Sqrt(i); j++)
{
if (i % j == 0)
{
int s = i / j;
if (s == j)
{
// If the current value
// is greater than the
// divisor's value
if (arr[i] > arr[s])
{
dp[s] = Math.Max(dp[i] + 1,
dp[s]);
}
}
else
{
// If current value is greater
// than the divisor's value
// and s is not equal
// to current index
if (s != i && arr[i] > arr[s])
dp[s] = Math.Max(dp[i] + 1,
dp[s]);
// Condition if current
// value is greater
// than the divisor's value
if (arr[i] > arr[j])
{
dp[j] = Math.Max(dp[i] + 1,
dp[j]);
}
}
}
}
}
int max = 0;
// Computing the greatest value
for(int i = 1; i < n; i++)
{
if (dp[i] > max)
max = dp[i];
}
// Printing maximum length of array
Console.WriteLine(max);
}// Driver Codepublic static void Main()
{ // Given array arr[]
int[] arr = new int[] { 0, 1, 4, 2,
3, 6, 4, 9 };
int size = arr.Length;
// Function call
maxLength(arr, size);
}}// This code is contributed by sanjoy_62 |
Javascript
<script>// Javascript Program to implement// the above approach// Function that print maximum length// of arrayfunction maxLength(arr, n)
{ // dp[] array to store the
// maximum length
let dp = [];
for(let i = 1; i < n; i++)
{
dp[i] = 1;
}
for(let i = n - 1; i > 1; i--)
{
// Find all divisors of i
for(let j = 1;
j <= Math.sqrt(i); j++)
{
if (i % j == 0)
{
let s = i / j;
if (s == j)
{
// If the current value
// is greater than the
// divisor's value
if (arr[i] > arr[s])
{
dp[s] = Math.max(dp[i] + 1,
dp[s]);
}
}
else
{
// If current value is greater
// than the divisor's value
// and s is not equal
// to current index
if (s != i && arr[i] > arr[s])
dp[s] = Math.max(dp[i] + 1,
dp[s]);
// Condition if current
// value is greater
// than the divisor's value
if (arr[i] > arr[j])
{
dp[j] = Math.max(dp[i] + 1,
dp[j]);
}
}
}
}
}
let max = 0;
// Computing the greatest value
for(let i = 1; i < n; i++)
{
if (dp[i] > max)
max = dp[i];
}
// Printing maximum length of array
document.write(max);
}// Driver Code // Given array arr[]
let arr = [ 0, 1, 4, 2, 3, 6, 4, 9 ];
let size = arr.length;
// Function call
maxLength(arr, size);
// This code is contributed by chinmoy1997pal.
</script> |
Output:
3
Time Complexity: O(N*(sqrt(N)) Since, for each index of the array, we calculate its all divisor, this takes O(sqrt(N))
Auxiliary Space: O(N)