Leetcode 2944: Minimum Number of Coins for Fruits
You are given a list of integers ‘prices’, where each element ‘prices[i]’ represents the cost to purchase the ‘i’-th fruit. If you buy the ‘i’-th fruit at prices[i] coins, you get all subsequent fruits for free. Your task is to determine the minimum number of coins required to purchase all the fruits.
Problem
Approach
Steps
Complexity
Input: You are given an array of integers, 'prices', where each element represents the price to purchase that fruit.
Example: prices = [2, 5, 3]
Constraints:
• 1 <= prices.length <= 1000
• 1 <= prices[i] <= 10^5
Output: Return the minimum number of coins required to purchase all the fruits.
Example: For prices = [2, 5, 3], the output is 5.
Constraints:
Goal: The goal is to minimize the number of coins required to buy all the fruits while maximizing the rewards from purchasing fruits.
Steps:
• Use dynamic programming (dp) to track the minimum coins needed to buy all the fruits.
• At each fruit, determine whether to buy it to get all subsequent fruits for free or skip it based on the cost of purchasing it.
Goal: The length of the prices array is between 1 and 1000, and each price is between 1 and 100000.
Steps:
• prices.length <= 1000
• prices[i] <= 100000
Assumptions:
• The prices array always has at least one element.
• Each price in the array is distinct and positive.
• Input: prices = [2, 5, 3]
• Explanation: The optimal strategy is to buy the first fruit for 2 coins and the second fruit for 5 coins, getting the third fruit for free. The total cost is 2 + 5 = 7.
• Input: prices = [1, 8, 2, 3]
• Explanation: By buying the first fruit for 1 coin and the third fruit for 2 coins, you get the second and fourth fruits for free, resulting in a total cost of 3.
Approach: The problem can be solved using dynamic programming, where at each fruit you decide whether to buy it based on the rewards it provides for subsequent fruits.
Observations:
• The dynamic programming approach tracks the cost of buying all fruits considering whether you buy a fruit or use a reward from a previous purchase.
• This problem is optimized by considering the cost at each fruit and how buying it minimizes the overall cost, including rewards from purchasing subsequent fruits.
Steps:
• Initialize a dynamic programming array to track the minimum coins required for each fruit.
• Iterate over the fruits, and at each fruit, decide whether to buy it or use a previous reward to minimize the total cost.
Empty Inputs:
• The input prices array cannot be empty since the problem guarantees at least one fruit.
Large Inputs:
• Even though the prices array can be as large as 1000 elements, the problem is still manageable with dynamic programming.
Special Values:
• If all prices are equal, the optimal strategy may involve purchasing fruits based on their position in the array.
Constraints:
• The length of prices is manageable, but the size of individual prices can be quite large, so efficient algorithms are necessary.
int minimumCoins(vector<int>& A) {
int n = A.size();
vector<int> dp(n + 1, 0);
deque<int> q;
for (int i = 0; i < n; i++) {
while (!q.empty() && (q.front() + 1) * 2 < i + 1)
q.pop_front();
while (!q.empty() && dp[q.back()] + A[q.back()] >= dp[i] + A[i])
q.pop_back();
q.push_back(i);
dp[i + 1] = dp[q.front()] + A[q.front()];
}
return dp[n];
}
1 : Function Definition
int minimumCoins(vector<int>& A) {
Defines the function 'minimumCoins' that takes a vector 'A' of coin values and returns the minimum number of coins needed to achieve the desired total.
2 : Initialize Variables
int n = A.size();
Initializes the variable 'n' to store the size of the input vector 'A', representing the number of available coins.
3 : Vector Initialization
vector<int> dp(n + 1, 0);
Creates a dynamic programming array 'dp' with 'n + 1' elements, initialized to zero. 'dp[i]' will store the minimum number of coins needed for the amount 'i'.
4 : Deque Initialization
deque<int> q;
Initializes a deque 'q' that will be used to efficiently manage indices of the 'dp' array during the dynamic programming process.
5 : Loop Initialization
for (int i = 0; i < n; i++) {
Starts a loop that iterates over each coin in the vector 'A'.
6 : Pop Front of Deque
while (!q.empty() && (q.front() + 1) * 2 < i + 1)
Checks if the front of the deque represents an index that is no longer relevant for computing the minimum number of coins. If so, it removes that index from the front.
7 : Pop Back of Deque
q.pop_front();
Removes the front element from the deque, ensuring only relevant indices remain for the computation.
8 : Pop Back of Deque (Condition)
while (!q.empty() && dp[q.back()] + A[q.back()] >= dp[i] + A[i])
Checks if the current coin value, combined with the previous best solution (stored in 'dp'), is better or worse than the existing solution. If the current solution is better, it pops elements from the back of the deque.
9 : Pop Back of Deque
q.pop_back();
Removes the back element from the deque, as it represents a suboptimal solution compared to the current coin.
10 : Push Current Index to Deque
q.push_back(i);
Adds the current index 'i' to the back of the deque to be considered for future iterations.
11 : Update DP Array
dp[i + 1] = dp[q.front()] + A[q.front()];
Updates the dynamic programming array 'dp' by setting 'dp[i + 1]' to the minimum number of coins needed by taking the value from the front of the deque.
12 : Return Result
return dp[n];
Returns the final result, which is the minimum number of coins needed to make the value 'n', stored in 'dp[n]'.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear since we only iterate over the fruits once while updating the dynamic programming array.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the dp array that tracks the minimum coins for each fruit.
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