Minimal product subsequence where adjacent elements are separated by a maximum distance of K

Given an array arr[] and an integer K, the task is to find out the minimum product of a subsequence where adjacent elements of the subsequence are separated by a maximum distance of K.

Note: The subsequence should include the first and the last element of the array.

Examples:



Input: arr[] = { 1, 2, 3, 4 }, K = 2
Output: 8
The first element in the subsequence is 1. From 1, we can move to either 2 or 3 (since K = 2). We can move to 3 and then to 4 to have a product of 12. However, we can also move to 2 and then to 4 to have a product of 8. Minimal product subsequence = { 1, 2, 4 }

Input: arr[] = { 2, 3 }, K = 2
Output: 6

Naive Approach: A naive approach is to generate all subsequences of the array and maintain a difference of indices between the adjacent elements and find the minimal product subsequence.

Efficient Approach: An efficient approach is to use dynamic programming. Let dp[i] denote the minimum product of elements till index ‘i’ including arr[i] who are separated by a maximum distance of K. Then dp[i] can be formulated as follows:

dp[i] = arr[i] * min{dp[j]} where j < i and 1 <= i - j <= K.

To calculate dp[i], a window of size K can be maintained and traversed to find the minimum of dp[j] which can then be multiplied to arr[i]. However, this will result in an O(N*K) solution.

To optimize the solution further, values of the product can be stored in an STL set and the minimum value of the product can then be found out in O(log n) time. Since storing products can be a cumbersome task since the product can easily exceed 1018, therefore we will store log values of products since log is a monotonic function and minimization of log values will automatically imply minimization of products.

Below is the implementation of the above approach:

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// C++ implementation of the above approach.
#include <bits/stdc++.h>
  
#define mp make_pair
#define ll long long
using namespace std;
  
const int mod = 1000000007;
const int MAX = 100005;
  
// Function to get the minimum product of subsequence such that
// adjacent elements are separated by a max distance of K
int minimumProductSubsequence(int* arr, int n, int k)
{
    multiset<pair<double, int> > s;
  
    ll dp[MAX];
    double p[MAX];
  
    dp[0] = arr[0];
    p[0] = log(arr[0]);
  
    // multiset will hold pairs
    // pair = (log value of product, dp[j] value)
    // dp[j] = minimum product % mod
    // multiset will be sorted according to log values
    // Therefore, corresponding to the minimum log value
    // dp[j] value can be obtained.
    s.insert(mp(p[0], dp[0]));
  
    // For the first k-sized window.
    for (int i = 1; i < k; i++) {
  
        double l = (s.begin())->first;
        ll min = (s.begin())->second;
  
        // Update log value by adding previous
        // minimum log value
        p[i] = log(arr[i]) + l;
        // Update dp[i]
        dp[i] = (arr[i] * min) % mod;
  
        // Insert it again into the multiset
        // since it is within the k-size window
        s.insert(mp(p[i], dp[i]));
    }
  
    for (int i = k; i < n; i++) {
  
        double l = (s.begin())->first;
        ll min = (s.begin())->second;
  
        p[i] = log(arr[i]) + l;
        dp[i] = (arr[i] * min) % mod;
  
        // Eliminate previous value which falls out
        // of the k-sized window
        multiset<pair<double, int> >::iterator it;
        it = s.find(mp(p[i - k], dp[i - k]));
        s.erase(it);
  
        // Insert newest value to enter in
        // the k-sized window.
        s.insert(mp(p[i], dp[i]));
    }
  
    // dp[n - 1] will have minimum product %
    // mod such that adjacent elements are
    // separated by a max distance K
    return dp[n - 1];
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
  
    cout << minimumProductSubsequence(arr, n, k);
  
    return 0;
}
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Output:
8



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