Maximum Coins From K Consecutive Bags

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ArraysSliding WindowsGreedy Algorithms

There are an infinite amount of bags on a number line, one bag for each coordinate. Some of these bags contain coins.

You are given a 2D array coins, where coins[i] = [l_i, r_i, c_i] denotes that every bag from l_i to r_i contains c_i coins.

The segments that coins contain are non-overlapping.

You are also given an integer k.

Return the maximum amount of coins you can obtain by collecting k consecutive bags.

Example 1:

Input: coins = [[8,10,1],[1,3,2],[5,6,4]], k = 4

Output: 10

Explanation:

Selecting bags at positions [3, 4, 5, 6] gives the maximum number of coins: 2 + 0 + 4 + 4 = 10.

Example 2:

Input: coins = [[1,10,3]], k = 2

Output: 6

Explanation:

Selecting bags at positions [1, 2] gives the maximum number of coins: 3 + 3 = 6.

Constraints:

1 <= coins.length <= 10⁵
1 <= k <= 10⁹
coins[i] == [l_i, r_i, c_i]
1 <= l_i <= r_i <= 10⁹
1 <= c_i <= 1000
The given segments are non-overlapping.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array (number of bags) and the value of k (number of consecutive bags)?
Can the coin values in the bags be negative, zero, or only positive?
If k is greater than the number of bags, what should I return?
If there are multiple sets of k consecutive bags that yield the same maximum number of coins, is any one of them acceptable, or is there a specific set I should prioritize returning?
Can I modify the input array, or should I assume it's immutable?

Brute Force Solution

Approach

The brute force method to maximize coins selects `k` consecutive bags, trying every single possible starting point. It calculates the total coins for each consecutive selection, and remembers the selection that yields the most coins.

Here's how the algorithm would work step-by-step:

First, consider the first `k` bags.
Calculate the total number of coins in these `k` bags.
Now, move to the next set of `k` consecutive bags, starting from the second bag.
Calculate the total number of coins in this new set of `k` bags.
Repeat this process, shifting the starting point by one bag each time, until you reach the end where there are not enough bags to form a group of `k`.
Keep track of the total number of coins obtained for each set of `k` consecutive bags.
Finally, compare the total coins from all the sets, and identify the set that gave you the maximum number of coins. That's your answer.

Code Implementation

def max_coins_from_k_bags_brute_force(coin_bags, k_consecutive):
    max_coins = 0
    # Iterate through all possible starting positions.
    for starting_position in range(len(coin_bags) - k_consecutive + 1):

        current_coins = 0
        # Calculate the total coins for the current consecutive k bags.
        for i in range(k_consecutive):
            current_coins += coin_bags[starting_position + i]

        # Update max_coins if the current total is greater.
        if current_coins > max_coins:

            max_coins = current_coins
    return max_coins

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array of n bags, selecting k consecutive bags at each step. For each starting position, it calculates the sum of k bags. The outer loop iterates approximately n-k+1 times. The inner loop summing k elements runs a fixed k iterations. Therefore, the overall time complexity is proportional to (n-k+1)*k. Since k is a constant, the time complexity simplifies to O(n).

Space Complexity

O(1) – The brute force method calculates the sum of K consecutive bags at a time. It only requires a variable to store the current sum of coins for the K bags and another variable to store the maximum sum found so far. The space used for these variables is constant, regardless of the number of bags (N). Therefore, the space complexity is O(1).

Optimal Solution

Approach

The best way to solve this is using dynamic programming. Imagine we're building up the solution step by step, remembering the best choices we've made so far. Instead of recalculating things, we reuse previous results to make faster decisions.

Here's how the algorithm would work step-by-step:

Think about the problem backward. Instead of picking from the start, consider what happens if we *don't* pick the last bag.
If we don't pick the last bag, our problem becomes smaller: find the maximum coins from 'k' consecutive bags in the range excluding the last bag.
If we *do* pick the last bag, then we have to consider the bags before it. Since we need 'k' consecutive bags, the amount of coins we can obtain depends on what happened before the last bag.
Keep track of the best amount of coins you can have up to each bag position. This means: 'What is the maximum amount of coins I can get if I stop at this bag, considering all possible consecutive picks?'
To find the best result for a given bag, decide: Do I include it as part of a consecutive set, or do I ignore it and use the best result from before?
Compare all the possible maximum amounts and pick the overall highest result. This will be the largest sum of coins you can achieve.

Code Implementation

def max_coins_from_bags(coin_bags, consecutive_bags):
    number_of_bags = len(coin_bags)

    # Store the maximum coins obtainable up to each bag.
    max_coins_upto = [0] * (number_of_bags + 1)

    for i in range(1, number_of_bags + 1):
        # Option 1: Exclude the current bag.
        max_coins_upto[i] = max_coins_upto[i - 1]

        # Option 2: Include the current bag.
        if i >= consecutive_bags:
            current_sum = sum(coin_bags[i - consecutive_bags:i])

            #Determine max coins if we include the current bag
            max_coins_upto[i] = max(max_coins_upto[i],
                                   max_coins_upto[i - consecutive_bags] + current_sum)
        elif i < consecutive_bags:
            current_sum = sum(coin_bags[0:i])

            #Base case if the number of bags is less than consecutive
            max_coins_upto[i] = max(max_coins_upto[i], current_sum)

    # Result is the maximum coins up to the last bag.
    return max_coins_upto[number_of_bags]

Big(O) Analysis

Time Complexity

O(n*k) – The algorithm iterates through the bags array of size n. For each bag, it considers whether to include it in a consecutive sequence of k bags. In the worst case, for each of the n bags, we may need to look back k steps to find the optimal starting point for a consecutive k-bag sequence ending at that bag. This nested operation results in approximately n*k operations. Thus, the time complexity is O(n*k).

Space Complexity

O(N) – The described dynamic programming approach requires keeping track of the best amount of coins up to each bag position. This implies storing intermediate results in an array or similar data structure, where the size of this data structure is directly proportional to the number of bags, N. Therefore, the auxiliary space used is proportional to N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no coins can be collected.
k is 0	Return 0 immediately, as no bags can be selected.
k is greater than the array length	Return the sum of all elements in the array as we can pick all bags.
Array contains only negative numbers	The sliding window approach will still correctly identify the k consecutive elements that give the maximum (least negative) sum.
Array contains very large numbers, potential integer overflow	Use a data type that can accommodate larger sums, such as long, to avoid integer overflow.
k equals the array length	Return the sum of all the elements in the input array.
Array contains zeros	The algorithm should correctly handle zeros, as they contribute 0 to the sum, possibly impacting the maximum sum found within the k-sized windows.
All elements in the array are identical	The sliding window will find the sum of any k consecutive elements, all having the same value, and return that as the result.

Null or empty input array

How to Handle:

Return 0 immediately as no coins can be collected.

k is 0

How to Handle:

Return 0 immediately, as no bags can be selected.

k is greater than the array length

How to Handle:

Return the sum of all elements in the array as we can pick all bags.

Array contains only negative numbers

How to Handle:

The sliding window approach will still correctly identify the k consecutive elements that give the maximum (least negative) sum.

Array contains very large numbers, potential integer overflow

How to Handle:

Use a data type that can accommodate larger sums, such as long, to avoid integer overflow.

k equals the array length

How to Handle:

Return the sum of all the elements in the input array.

Array contains zeros

How to Handle:

The algorithm should correctly handle zeros, as they contribute 0 to the sum, possibly impacting the maximum sum found within the k-sized windows.

All elements in the array are identical

How to Handle:

The sliding window will find the sum of any k consecutive elements, all having the same value, and return that as the result.