LeetCode 2830: Maximize the Profit as the Salesman Solution
Master LeetCode problem 2830 (Maximize the Profit as the Salesman), 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.
2830. Maximize the Profit as the Salesman
Problem Explanation
Explanation
To maximize the profit as the salesman, we can use dynamic programming to solve this problem efficiently. We can iterate over the offers and for each offer, we can update the maximum profit that can be achieved by selling houses up to the end of that offer. We can maintain a dp array where dp[i] represents the maximum profit that can be achieved by selling houses up to house i.
At each offer, we have two options:
- Sell the houses covered by the current offer and update the maximum profit.
- Skip the current offer and retain the previous maximum profit.
We choose the option that gives us the maximum profit. After iterating through all the offers, the final value in the dp array will represent the maximum profit that can be achieved.
Time complexity: O(n) where n is the number of houses. Space complexity: O(n) for the dp array.
Solution Code
class Solution {
public int maximizeProfit(int n, int[][] offers) {
int[] dp = new int[n];
for (int i = 0; i < offers.length; i++) {
int start = offers[i][0];
int end = offers[i][1];
int gold = offers[i][2];
for (int j = start; j <= end; j++) {
dp[j] = Math.max(dp[j], j == 0 ? gold : dp[j - 1] + gold);
}
}
return dp[n - 1];
}
}Try It Yourself
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