LeetCode-in-Java
3165. Maximum Sum of Subsequence With Non-adjacent Elements
Hard
You are given an array nums consisting of integers. You are also given a 2D array queries, where queries[i] = [posi, xi].
For query i, we first set nums[posi] equal to xi, then we calculate the answer to query i which is the maximum sum of a subsequence of nums where no two adjacent elements are selected.
Return the sum of the answers to all queries.
Since the final answer may be very large, return it modulo 109 + 7.
A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: nums = [3,5,9], queries = [[1,-2],[0,-3]]
Output: 21
Explanation:
After the 1st query, nums = [3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 3 + 9 = 12.
After the 2nd query, nums = [-3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 9.
Example 2:
Input: nums = [0,-1], queries = [[0,-5]]
Output: 0
Explanation:
After the 1st query, nums = [-5,-1] and the maximum sum of a subsequence with non-adjacent elements is 0 (choosing an empty subsequence).
Constraints:
1 <= nums.length <= 5 * 104-105 <= nums[i] <= 1051 <= queries.length <= 5 * 104queries[i] == [posi, xi]0 <= posi <= nums.length - 1-105 <= xi <= 105
Solution
public class Solution {
private static final int YY = 0;
private static final int YN = 1;
private static final int NY = 2;
private static final int NN = 3;
private static final int MOD = 1_000_000_007;
public int maximumSumSubsequence(int[] nums, int[][] queries) {
long[][] tree = build(nums);
long result = 0;
for (int i = 0; i < queries.length; ++i) {
result += set(tree, queries[i][0], queries[i][1]);
result %= MOD;
}
return (int) result;
}
private static long[][] build(int[] nums) {
final int len = nums.length;
int size = 1;
while (size < len) {
size <<= 1;
}
long[][] tree = new long[size * 2][4];
for (int i = 0; i < len; ++i) {
tree[size + i][YY] = nums[i];
}
for (int i = size - 1; i > 0; --i) {
tree[i][YY] =
Math.max(
tree[2 * i][YY] + tree[2 * i + 1][NY],
tree[2 * i][YN] + Math.max(tree[2 * i + 1][YY], tree[2 * i + 1][NY]));
tree[i][YN] =
Math.max(
tree[2 * i][YY] + tree[2 * i + 1][NN],
tree[2 * i][YN] + Math.max(tree[2 * i + 1][YN], tree[2 * i + 1][NN]));
tree[i][NY] =
Math.max(
tree[2 * i][NY] + tree[2 * i + 1][NY],
tree[2 * i][NN] + Math.max(tree[2 * i + 1][YY], tree[2 * i + 1][NY]));
tree[i][NN] =
Math.max(
tree[2 * i][NY] + tree[2 * i + 1][NN],
tree[2 * i][NN] + Math.max(tree[2 * i + 1][YN], tree[2 * i + 1][NN]));
}
return tree;
}
private static long set(long[][] tree, int idx, int val) {
int size = tree.length / 2;
tree[size + idx][YY] = val;
for (int i = (size + idx) / 2; i > 0; i /= 2) {
tree[i][YY] =
Math.max(
tree[2 * i][YY] + tree[2 * i + 1][NY],
tree[2 * i][YN] + Math.max(tree[2 * i + 1][YY], tree[2 * i + 1][NY]));
tree[i][YN] =
Math.max(
tree[2 * i][YY] + tree[2 * i + 1][NN],
tree[2 * i][YN] + Math.max(tree[2 * i + 1][YN], tree[2 * i + 1][NN]));
tree[i][NY] =
Math.max(
tree[2 * i][NY] + tree[2 * i + 1][NY],
tree[2 * i][NN] + Math.max(tree[2 * i + 1][YY], tree[2 * i + 1][NY]));
tree[i][NN] =
Math.max(
tree[2 * i][NY] + tree[2 * i + 1][NN],
tree[2 * i][NN] + Math.max(tree[2 * i + 1][YN], tree[2 * i + 1][NN]));
}
return Math.max(tree[1][YY], Math.max(tree[1][YN], Math.max(tree[1][NY], tree[1][NN])));
}
}