Description
Find X Value of Array II
HardYou are given an array of positive integers nums and a positive integer k. You are also given a 2D array queries, where queries[i] = [indexi, valuei, starti, xi].
You are allowed to perform an operation once on nums, where you can remove any suffix from nums such that nums remains non-empty.
The x-value of nums for a given x is defined as the number of ways to perform this operation so that the product of the remaining elements leaves a remainder of x modulo k.
For each query in queries you need to determine the x-value of nums for xi after performing the following actions:
- Update
nums[indexi]tovaluei. Only this step persists for the rest of the queries. - Remove the prefix
nums[0..(starti - 1)](wherenums[0..(-1)]will be used to represent the empty prefix).
Return an array result of size queries.length where result[i] is the answer for the ith query.
A prefix of an array is a subarray that starts from the beginning of the array and extends to any point within it.
A suffix of an array is a subarray that starts at any point within the array and extends to the end of the array.
Note that the prefix and suffix to be chosen for the operation can be empty.
Note that x-value has a different definition in this version.
Example 1:
Input: nums = [1,2,3,4,5], k = 3, queries = [[2,2,0,2],[3,3,3,0],[0,1,0,1]]
Output: [2,2,2]
Explanation:
- For query 0,
numsbecomes[1, 2, 2, 4, 5], and the empty prefix must be removed. The possible operations are:- Remove the suffix
[2, 4, 5].numsbecomes[1, 2]. - Remove the empty suffix.
numsbecomes[1, 2, 2, 4, 5]with a product 80, which gives remainder 2 when divided by 3.
- Remove the suffix
- For query 1,
numsbecomes[1, 2, 2, 3, 5], and the prefix[1, 2, 2]must be removed. The possible operations are:- Remove the empty suffix.
numsbecomes[3, 5]. - Remove the suffix
[5].numsbecomes[3].
- Remove the empty suffix.
- For query 2,
numsbecomes[1, 2, 2, 3, 5], and the empty prefix must be removed. The possible operations are:- Remove the suffix
[2, 2, 3, 5].numsbecomes[1]. - Remove the suffix
[3, 5].numsbecomes[1, 2, 2].
- Remove the suffix
Example 2:
Input: nums = [1,2,4,8,16,32], k = 4, queries = [[0,2,0,2],[0,2,0,1]]
Output: [1,0]
Explanation:
- For query 0,
numsbecomes[2, 2, 4, 8, 16, 32]. The only possible operation is:- Remove the suffix
[2, 4, 8, 16, 32].
- Remove the suffix
- For query 1,
numsbecomes[2, 2, 4, 8, 16, 32]. There is no possible way to perform the operation.
Example 3:
Input: nums = [1,1,2,1,1], k = 2, queries = [[2,1,0,1]]
Output: [5]
Constraints:
1 <= nums[i] <= 1091 <= nums.length <= 1051 <= k <= 51 <= queries.length <= 2 * 104queries[i] == [indexi, valuei, starti, xi]0 <= indexi <= nums.length - 11 <= valuei <= 1090 <= starti <= nums.length - 10 <= xi <= k - 1
Solutions
I thought yesterday was too much, until I saw today.
/** By Leetcode */
class SegmentTree {
constructor(nums, k) {
this.k = k;
const n = nums.length;
const size = 2 << n.toString(2).length;
this.tree = Array.from({ length: size }, () =>
new Array(k + 1).fill(0),
);
this.build(nums, 1, 0, n - 1);
}
makeLeaf(o, value) {
const info = new Array(this.k + 1).fill(0);
const r = value % this.k;
info[r] = 1;
info[this.k] = r;
this.tree[o] = info;
}
mergePre(left, right) {
const pre = new Array(this.k + 1).fill(0);
const mulL = left[this.k];
const mulR = right[this.k];
pre[this.k] = (mulL * mulR) % this.k;
for (let x = 0; x < this.k; x++) pre[x] = left[x];
for (let x = 0; x < this.k; x++) {
pre[(mulL * x) % this.k] += right[x];
}
return pre;
}
maintain(o) {
this.tree[o] = this.mergePre(this.tree[o * 2], this.tree[o * 2 + 1]);
}
build(nums, o, l, r) {
if (l === r) {
this.makeLeaf(o, nums[l]);
return;
}
const m = Math.floor((l + r) / 2);
this.build(nums, o * 2, l, m);
this.build(nums, o * 2 + 1, m + 1, r);
this.maintain(o);
}
update(o, l, r, index, value) {
if (l === r) {
this.makeLeaf(o, value);
return;
}
const m = Math.floor((l + r) / 2);
if (index <= m) this.update(o * 2, l, m, index, value);
else this.update(o * 2 + 1, m + 1, r, index, value);
this.maintain(o);
}
query(o, l, r, L, R) {
if (L <= l && r <= R) return this.tree[o];
const m = Math.floor((l + r) / 2);
if (R <= m) return this.query(o * 2, l, m, L, R);
if (L > m) return this.query(o * 2 + 1, m + 1, r, L, R);
const left = this.query(o * 2, l, m, L, R);
const right = this.query(o * 2 + 1, m + 1, r, L, R);
return this.mergePre(left, right);
}
}
var resultArray = function (nums, k, queries) {
const n = nums.length;
const seg = new SegmentTree(nums, k);
const ans = [];
for (const [index, value, start, x] of queries) {
seg.update(1, 0, n - 1, index, value);
const pre = seg.query(1, 0, n - 1, start, n - 1);
ans.push(pre[x]);
}
return ans;
};