LeetCode 1799: Maximize Score After N Operations
LeetCode 1799 Solution Explanation
Explanation:
- Algorithmic Idea:
- We can use dynamic programming to solve this problem efficiently. The key idea is to consider the state of the problem as the elements left in the array after each operation and the current operation number.
- We can iterate through all possible pairs of elements from the remaining array, calculate the score for each pair, and recursively call the function for the next operation with the updated array. We keep track of the maximum score obtained so far.
- Step-by-Step Iterations:
- Initialize a 2D DP array to store the maximum score for each state (remaining array and current operation).
- Start with the initial state (original array and operation 1).
- Iterate through all possible pairs of elements, calculate the score, and update the DP array.
- Recursively call the function for the next operation with the updated array and operation number.
- Return the maximum score obtained after n operations.
- Time Complexity: O(n^2 * 2^n) where n is the number of elements in the original array.
- Space Complexity: O(n * 2^n) for the DP array.
:
LeetCode 1799 Solutions in Java, C++, Python
class Solution {
public int maxScore(int[] nums) {
int n = nums.length / 2;
int[][] dp = new int[1 << (2 * n)][n + 1];
return dfs(nums, dp, 0, 1);
}
private int dfs(int[] nums, int[][] dp, int mask, int operation) {
if (operation > nums.length / 2) {
return 0;
}
if (dp[mask][operation] != 0) {
return dp[mask][operation];
}
int maxScore = 0;
for (int i = 0; i < nums.length; i++) {
if ((mask & (1 << i)) == 0) {
for (int j = i + 1; j < nums.length; j++) {
if ((mask & (1 << j)) == 0) {
int newMask = mask | (1 << i) | (1 << j);
int score = operation * gcd(nums[i], nums[j]);
maxScore = Math.max(maxScore, score + dfs(nums, dp, newMask, operation + 1));
}
}
}
}
dp[mask][operation] = maxScore;
return maxScore;
}
private int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
}Interactive Code Editor for LeetCode 1799
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