3413. Maximum Coins From K Consecutive Bags - Detailed Explanation
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
 
- coins = 
- Output: 15
- Explanation:
 The optimal choice is to select the bags with coins[4, 5, 6]. Their sum is4 + 5 + 6 = 15.
Example 2:
- Input:
- coins = [3, 2, 1, 0, 4]
- k = 2
 
- coins = 
- 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
 
- coins = 
- Output: 40
- Explanation:
 Sincekis equal to the total number of bags, you take all of them. The sum is10 + 9 + 8 + 7 + 6 = 40.
Constraints:
- 1 ≤ coins.length ≤ 10⁵
- 1 ≤ k ≤ coins.length
- Each element in coinsis 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 lengthk. 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 firstkelements 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 lengthk, 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 sizek:- Compute the sum of the first kelements.
- 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.
- Track and update the maximum sum encountered.
 
- Compute the sum of the first 
- Benefits:
 This approach runs in O(n) time with O(1) extra space, making it efficient even for large inputs.
Code Implementations
Python Code
Java Code
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 kelements. This is your initial window.
- Initialize a variable max_sumwith this initial sum.
 
- Compute the sum of the first 
- 
Sliding the Window: - For each index istarting fromkup 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_sumif the new window sum is greater.
 
- Subtract the element at 
 
- For each index 
- 
Result: - After processing the entire array, the max_sumholds the maximum sum of any contiguous subarray of lengthk.
 
- After processing the entire array, the 
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 add4(the 4th element)
 New sum = 6 - 1 + 4 = 9
 
- Remove 
- 
Next window: - Remove 2and add5
 New sum = 9 - 2 + 5 = 12
 
- Remove 
- 
Next window: - Remove 3and add6
 New sum = 12 - 3 + 6 = 15
 
- Remove 
- 
Final window: - Remove 4and add1
 New sum = 15 - 4 + 1 = 12
 
- Remove 
- 
Maximum Sum: 
 The maximum among these sums is 15.
Common Mistakes
- Not Handling Edge Cases:
 Forgetting to check whenkis 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 ofkelements 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 andkis 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 lengthkwill have the same sum.
Alternative Variations and Related Problems
- 
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.
Related Problems for Further Practice
- Maximum Sum Subarray
- Sliding Window Maximum
- Subarray Sum Equals k
- Best Time to Buy and Sell Stock
GET YOUR FREE
Coding Questions Catalog
$197

$78
$78