LeetCode 2944: Minimum Number of Coins for Fruits Solution
Master LeetCode problem 2944 (Minimum Number of Coins for Fruits), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2944. Minimum Number of Coins for Fruits
Problem Explanation
Explanation
To solve this problem, we can use a dynamic programming approach. We will iterate through the prices array and for each fruit, we will consider two scenarios - either we buy the fruit or we take it for free as a reward. We will maintain an array dp to store the minimum number of coins needed to acquire fruits up to the current index. For each fruit, we update the dp array with the optimal cost.
Algorithm
- Initialize a dp array of size n+1 where n is the length of the prices array.
- Initialize dp[0] to 0 as we don't need any coins to acquire the first fruit.
- Iterate from 1 to n:
- For each fruit at index i, consider two scenarios:
- Buy the fruit: cost[i] + dp[i-1]
- Take the fruit for free as a reward: dp[j] where j is the index of the fruit we got for free
- Update dp[i] with the minimum of the above two scenarios.
- For each fruit at index i, consider two scenarios:
- Return dp[n] which represents the minimum number of coins needed to acquire all the fruits.
Time Complexity
The time complexity of this algorithm is O(n^2) where n is the length of the prices array.
Space Complexity
The space complexity of this algorithm is O(n) where n is the length of the prices array.
Solution Code
class Solution {
public int minCoinsForFruits(int[] prices) {
int n = prices.length;
int[] dp = new int[n+1];
dp[0] = 0;
for (int i = 1; i <= n; i++) {
dp[i] = prices[i-1] + dp[i-1];
for (int j = 0; j < i-1; j++) {
dp[i] = Math.min(dp[i], dp[j]);
}
}
return dp[n];
}
}Try It Yourself
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