LeetCode 3181: Maximum Total Reward Using Operations II Solution
Master LeetCode problem 3181 (Maximum Total Reward Using Operations II), a hard 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.
3181. Maximum Total Reward Using Operations II
Problem Explanation
Explanation:
To solve this problem, we can iterate through the reward values while keeping track of the maximum reward we can achieve at each index. We can maintain a dynamic programming array where dp[i] represents the maximum reward that can be achieved if we stop at index i.
At each index i, we have two choices:
- Either we skip the current reward value and keep the previous maximum reward achieved.
- Or we take the current reward value and add it to the maximum reward achieved at the previous index where the reward value was less than the current reward value.
We update dp[i] accordingly and continue this process until we reach the end of the array. The final answer will be the maximum reward achieved at the last index.
- Time complexity: O(n) where n is the length of the input array.
- Space complexity: O(n) for the dynamic programming array.
:
Solution Code
class Solution {
public int maxReward(int[] rewardValues) {
int n = rewardValues.length;
int[] dp = new int[n];
dp[0] = rewardValues[0];
int maxReward = dp[0];
for (int i = 1; i < n; i++) {
dp[i] = Math.max(rewardValues[i], dp[i - 1] + rewardValues[i]);
maxReward = Math.max(maxReward, dp[i]);
}
return maxReward;
}
}Try It Yourself
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