3413. Maximum Coins From K Consecutive Bags - Detailed Explanation

Free Coding Questions Catalog

Boost your coding skills with our essential coding questions catalog. Take a step towards a better tech career now!

Problem Statement

Description:
You are given an array coins where each element represents the number of coins in a bag. You are also given an integer k. Your task is to choose exactly k consecutive bags such that the sum of coins in those bags is maximized. Return the maximum number of coins you can collect from these k consecutive bags.

Example 1:

Input:
- coins = [1, 2, 3, 4, 5, 6, 1]
- k = 3
Output: 15
Explanation:
The optimal choice is to select the bags with coins [4, 5, 6]. Their sum is 4 + 5 + 6 = 15.

Example 2:

Input:
- coins = [3, 2, 1, 0, 4]
- k = 2
Output: 5
Explanation:
The maximum sum can be obtained by selecting [3, 2] (sum = 5). Other pairs yield smaller sums.

Example 3:

Input:
- coins = [10, 9, 8, 7, 6]
- k = 5
Output: 40
Explanation:
Since k is equal to the total number of bags, you take all of them. The sum is 10 + 9 + 8 + 7 + 6 = 40.

Constraints:

1 ≤ coins.length ≤ 10⁵
1 ≤ k ≤ coins.length
Each element in coins is a non-negative integer.

Hints to Approach the Problem

Hint 1:
Notice that you are asked to find the maximum sum of a contiguous subarray (or window) of length k. This is a classic sliding window problem.
Hint 2:
Instead of calculating the sum for every possible subarray from scratch (which can be inefficient), consider using a sliding window technique where you update the sum by subtracting the element that goes out of the window and adding the new element that comes in.
Hint 3:
Think about initializing the window with the first k elements and then moving the window one position at a time while tracking the maximum sum encountered.

Approaches

Approach 1: Brute Force (Naive)

Idea:
Iterate over every possible starting index of a contiguous subarray of length k, calculate the sum, and keep track of the maximum.
Drawbacks:
This requires O(n * k) time, which is not efficient for large inputs (e.g., when n is 10⁵).

Approach 2: Optimal Sliding Window Technique

Idea:
Use a sliding window of fixed size k:
1. Compute the sum of the first k elements.
2. Slide the window by one element at a time. At each step, update the sum by subtracting the element that leaves the window and adding the element that enters.
3. Track and update the maximum sum encountered.
Benefits:
This approach runs in O(n) time with O(1) extra space, making it efficient even for large inputs.

Code Implementations

Python Code

Python3

. . . .

Java Code

Java

. . . .

Complexity Analysis

Time Complexity:
- Optimal Sliding Window Approach:
  The algorithm traverses the array once, performing O(1) work for each element.
  
  Time Complexity: O(n)
Space Complexity:
- Only a few variables are used for tracking sums and indices.
  
  Space Complexity: O(1)

Step-by-Step Walkthrough

Initialization:
- Compute the sum of the first k elements. This is your initial window.
- Initialize a variable max_sum with this initial sum.
Sliding the Window:
- For each index i starting from k up to the end of the array:
  - Subtract the element at coins[i - k] (the element that is no longer in the window).
  - Add the new element coins[i] that enters the window.
  - Update max_sum if the new window sum is greater.
Result:
- After processing the entire array, the max_sum holds the maximum sum of any contiguous subarray of length k.

Visual Example

Consider the array coins = [1, 2, 3, 4, 5, 6, 1] with k = 3.

Initial window (first 3 elements):
Sum = 1 + 2 + 3 = 6
Slide the window one element to the right:
- Remove 1 (the first element) and add 4 (the 4th element)
  New sum = 6 - 1 + 4 = 9
Next window:
- Remove 2 and add 5
  New sum = 9 - 2 + 5 = 12
Next window:
- Remove 3 and add 6
  New sum = 12 - 3 + 6 = 15
Final window:
- Remove 4 and add 1
  New sum = 15 - 4 + 1 = 12
Maximum Sum:
The maximum among these sums is 15.

Common Mistakes

Not Handling Edge Cases:
Forgetting to check when k is equal to the length of the array. In such a case, the answer is simply the sum of the entire array.
Inefficient Sum Calculation:
Recalculating the sum of k elements from scratch for every window leads to an O(n*k) solution, which is inefficient.

Edge Cases

Single Element Array:
If the array contains only one bag and k is 1, the answer is the value of that single element.
k Equals Array Length:
The optimal window is the entire array, so the answer is the sum of all elements.
Array with Identical Values:
Any contiguous subarray of length k will have the same sum.

Maximum Sum Subarray of Size at Most k:
A variation where the subarray size can vary up to k.
Sliding Window Maximum:
Instead of a sum, you might be asked to find the maximum value in each window.
Subarray Sum Equals k:
Counting the number of subarrays whose sum equals a given value.

Maximum Sum Subarray
Sliding Window Maximum
Subarray Sum Equals k
Best Time to Buy and Sell Stock