Minimum Total Weight Chocolate Hackerrank

Minimum Total Weight Chocolate – Hackerrank Problem Explained

Introduction

The minimum total weight chocolate challenge is a classic variation of the subset‑sum / knapsack family that appears on Hackerrank under titles such as “Chocolate – Minimum Total Weight” or similar practice problems.
So in this problem you are given a list of N chocolates, each with an integer weight w[i]. You also receive a target weight K. Your task is to choose a subset of chocolates whose combined weight is at least K while keeping the total weight as small as possible. If no subset can reach K, you must return ‑1 (or the value indicated by the statement).

At first glance the task looks like a simple “pick the heaviest chocolates until you pass K” greedy idea, but the optimal solution often requires skipping a large chocolate in favor of several smaller ones that together exceed K with less waste. This nuance makes the problem an excellent illustration of dynamic programming (DP) and bitset optimisation techniques that are frequently asked in coding interviews and competitive programming platforms.

Below we walk through the problem statement, develop a rigorous solution, illustrate it with concrete examples, discuss the underlying theory, highlight common pitfalls, and answer frequently asked questions. By the end you will have a complete, production‑ready understanding of how to solve the “minimum total weight chocolate” Hackerrank problem.

Detailed Explanation

Problem Restatement

• Input

  • An integer N – number of chocolates.
  • An array weights[0 … N‑1] – each weights[i] (> 0) is the weight of the i‑th chocolate.
  • An integer K – the required minimum total weight.

• Output

  • The smallest possible sum S such that there exists a subset of weights whose elements add up to S and S ≥ K.
  • If no subset reaches K, output -1.

• Constraints (typical for Hackerrank)

  • 1 ≤ N ≤ 100 (sometimes up to 200)
  • 1 ≤ weights[i] ≤ 1000
  • 1 ≤ K ≤ 10⁵

These limits make a pseudo‑polynomial DP (O(N·K)) feasible, while a naïve exponential search (2^N) would be impossible for the upper bound.

Why a Greedy Approach Fails

A natural greedy heuristic is: sort chocolates in descending order and keep adding them until the sum ≥ K.
Here's the thing — consider weights = [9, 8, 5, 5] and K = 10. Which means greedy picks 9 + 8 = 17 (excess = 7). Still, the subset {5,5} gives sum = 10 (excess = 0), which is strictly better.
Thus, greedy can over‑estimate the answer because it ignores the possibility of replacing a large item with several smaller ones that fit the target more tightly.

The problem is therefore a minimisation version of the subset‑sum problem: find the achievable sum closest to (but not below) K.

Step‑by‑Step Solution Breakdown

We will solve the problem using a 0/1 knapsack DP that tracks which sums are reachable. The DP table dp[s] is a boolean that tells whether a sum s can be formed using some prefix of the chocolates That alone is useful..

1. DP Table Initialization

• Create a boolean array dp of size K + maxWeight, where maxWeight = max(weights).
  • We need to know sums up to K + maxWeight because the optimal answer may overshoot K by at most the weight of a single chocolate (if we overshoot more, we could remove that chocolate and still be ≥ K, contradicting optimality).
  • In practice we can allocate dp of length K + sum(weights) but the tighter bound saves memory.
• Set dp[0] = true (empty subset yields sum 0). All other entries start as false.

2. Transition

For each chocolate weight w:

for s from currentMaxDown to 0:
    if dp[s] is true:
        dp[s + w] = true

We iterate downwards to avoid using the same chocolate more than once (0/1 knapsack property). currentMax tracks the highest sum reachable so far, initially 0 and updated after each chocolate.

3. Finding the Answer

After processing all chocolates, scan dp from K upward:

for s = K to K + maxWeight:
    if dp[s] is true:
        answer = s
        break

If the loop finishes without finding a true entry, answer = -1 Small thing, real impact. That alone is useful..

4. Complexity

• Time: O(N * (K + maxWeight)). With the given constraints this easily runs under a second.
• Space: O(K + maxWeight) booleans (≈ 10⁵ – 10⁶ bits → a

Conclusion
The dynamic programming approach outlined here provides an exact and efficient solution to the problem by systematically exploring all possible subset sums. By focusing on achievable sums up to (K + \text{maxWeight}), the algorithm ensures that the minimal excess is identified without redundant checks. This method is particularly well-suited for the given constraints, as it avoids the exponential complexity of brute-force searches while guaranteeing optimality. The solution is both practical and scalable within the problem’s limits, demonstrating how pseudo-polynomial algorithms can effectively address real-world combinatorial optimization challenges. The bottom line: this approach balances precision and performance, making it a dependable choice for similar subset-sum minimization tasks.