Dynamic Programming Solution for Finding Minimum Elements Summing to Given Value

If you're working with an array where elements can repeat, and you need to find the minimum number of elements that sum up to a specific value, this article provides a detailed explanation and implementation using dynamic programming. We will break down the problem, explain the solution, and provide a Python implementation.

Problem Definition

The problem statement is as follows: Given an array and a target sum, find the minimum number of elements from the array that can sum up to the target. Elements in the array can be repeated.

Steps to Solve the Problem

1. Define the Problem

Let's denote the array as arr with length n and the target sum as target. The goal is to find the minimum number of elements needed to achieve the sum target.

2. Create a DP Array

Initialize a dynamic programming (DP) array, dp, where dp[i] represents the minimum number of elements needed to sum up to the value i. Initially, set dp[0] to 0 since no elements are required to form a sum of 0. Set all other dp[i] values to a large number (e.g., infinity) to represent an impossible sum.

3. Fill the DP Array

- Iterate through each element in the array.

- For each element, iterate through all possible sums from the element value up to target.

- Use the recurrence relation to update the dp array:

dp[j] min(dp[j], dp[j - arr[i]] 1)

This means that if you can form the sum j - arr[i] using dp[j - arr[i]] elements, you can form j by adding one more element arr[i].

4. Get the Result

After filling the dp array, the value at dp[target] will give you the minimum number of elements needed to sum to the target. If dp[target] is still infinity, it means no possible arrangement of elements can sum to the target.

Example Code

Python Implementation

Here’s a Python implementation of the above approach:

def min_elements(arr, target):
    # Initialize dp array
    dp  [float('inf')] * (target   1)
    dp[0]  0  # Base case
    # Fill the dp array
    for num in arr:
        for j in range(num, target   1):
            dp[j]  min(dp[j], dp[j - num]   1)
    # Return the result
    return dp[target] if dp[target] ! float('inf') else -1

Example Usage

Consider the following example:

arr  [1, 2, 3]
target  7
print(min_elements(arr, target))  # Output: 3, e.g. 3   2   2  7

Explanation of the Code

The dp array is initialized to a size of target 1 to accommodate all sums from 0 to target.

The outer loop iterates through each number in the array, and the inner loop updates the dp array for all sums that can include the current number.

Finally, the function returns the minimum number of elements needed to form the target sum or -1 if it's not possible.

This approach ensures that you efficiently find the minimum number of elements needed to achieve the desired sum, taking advantage of the ability to use elements multiple times.