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Minimum cost to select K strictly increasing elements

  • Difficulty Level : Expert
  • Last Updated : 26 Jul, 2021

Given an array and an integer K. Also given one more array which stores the cost of choosing elements from the first array. The task is to calculate the minimum cost of selecting K strictly increasing elements from the array.
Examples: 
 

Input: N = 4, K = 2
ele[] = {2, 6, 4, 8}
cost[] = {40, 20, 30, 10}
Output: 30
Explanation:
30 is the minimum cost by selecting elements 
6 and 8 from the array with cost 
10 + 20 respectively

Input: N = 11, K = 4
ele = {2, 6, 4, 8, 1, 3, 15, 9, 22, 16, 45}
cost = {40, 20, 30, 10, 50, 10, 20, 30, 40, 20, 10}
Output: 60
Explanation:
60 is the minimum cost by selecting elements 
3, 15, 16, 45 from the array with cost 
10 + 20 + 20 + 10 respectively

 

Approach: 
The given problem can be easily solved using a dynamic programming approach. As the problem asks for increasing elements and then minimum cost then it is clear that we have to move by either selecting ith or not selecting ith element one by one and calculate the minimum cost for each. 
Now, take a 3D DP array which stores our values of minimum cost, where cache[i][prev][cnt] stores the min-cost up to ith element, prev element and count of numbers considered till now.
There are 3 base conditions involved: 
 

  • If k elements are counted return 0.
  • If all elements of array has been traversed return MAX_VALUE.
  • Check if it’s already calculated in dp array.

Now comes the part of either selecting ith element or not selecting ith element: 
 

  • When ith elements is not considered ans = dp(i+1, prev, cnt, s, c)
  • When the ith element is greater than previous element, check if adding its cost makes total cost minimum ans = min(ans, c[i] + dp(i+1, i, cnt+1, s, c))

Below is the implementation of the above approach: 
 



CPP




// C++ program for
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
const int N = 1005;
const int K = 20;
int n, k;
int dp[N + 1][N + 1][K + 1];
 
// Function to calculate
// min cost to choose
// k increasing elements
int minCost(int i, int prev, int cnt,
                    int ele[], int cost[])
{
    // If k elements are
    // counted return 0
    if (cnt == k + 1) {
        return 0;
    }
 
    // If all elements
    // of array has been
    // traversed then
    // return MAX_VALUE
    if (i == n + 1) {
        return 1e5;
    }
 
    // To check if this is
    // already calculated
    int& ans = dp[i][prev][cnt];
    if (ans != -1) {
        return ans;
    }
 
    // When i'th elements
    // is not considered
    ans = minCost(i + 1, prev, cnt, ele, cost);
 
    // When the ith element
    // is greater than previous
    // element check if adding
    // its cost makes total cost minimum
    if (ele[i] > ele[prev]) {
        ans = min(ans, cost[i] + minCost(i + 1,
                           i, cnt + 1, ele, cost));
    }
    return ans;
}
 
// Driver code
int main()
{
 
    memset(dp, -1, sizeof(dp));
    n = 4;
    k = 2;
 
    int ele[n + 1] = { 0, 2, 6, 4, 8 };
 
    int cost[n + 1] = { 0, 40, 20, 30, 10 };
 
    int ans = minCost(1, 0, 1, ele, cost);
 
    if (ans == 1e5) {
        ans = -1;
    }
 
    cout << ans << endl;
 
    return 0;
}

Python3




# Python3 program for
# the above approach
N = 1005;
K = 20;
 
n = 0
k = 0
 
dp = [[[-1 for k in range(K + 1)] for j in range(N + 1)] for i in range(N + 1)]
  
# Function to calculate
# min cost to choose
# k increasing elements
def minCost(i, prev, cnt, ele, cost):
 
    # If k elements are
    # counted return 0
    if (cnt == k + 1):
        return 0;
      
    # If all elements
    # of array has been
    # traversed then
    # return MAX_VALUE
    if (i == n + 1):
        return 100000;
     
    # To check if this is
    # already calculated
    ans = dp[i][prev][cnt];
     
    if (ans != -1):
        return ans;
     
    # When i'th elements
    # is not considered
    ans = minCost(i + 1, prev, cnt, ele, cost);
  
    # When the ith element
    # is greater than previous
    # element check if adding
    # its cost makes total cost minimum
    if (ele[i] > ele[prev]):
        ans = min(ans, cost[i] + minCost(i + 1, i, cnt + 1, ele, cost));
     
    return ans;
 
# Driver code
if __name__=='__main__':
  
    n = 4;
    k = 2;
  
    ele = [ 0, 2, 6, 4, 8 ]
  
    cost = [ 0, 40, 20, 30, 10 ]
  
    ans = minCost(1, 0, 1, ele, cost);
  
    if (ans == 100000):
        ans = -1;
     
    print(ans)
  
# This code is contributed by rutvik_56
Output: 
30

 

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