Most Frequent IDs - Heap w/ Lazy Removal, Segment Tree [JS]
Solution 1: Segment Tree
Use a max segment tree to efficiently find and update the maximum frequency in the range (0, 10^5)
.
n = length of nums
, m = max(nums[i])
Time Complexity: O(m + n log(m))
Space Complexity: O(m)
(excluding output)
var mostFrequentIDs = function(nums, freq) {
let n = nums.length, m = Math.max(...nums), tree = new MaxSegmentTree(m);
let ans = Array(n);
for (let i = 0; i < n; i++) {
let id = nums[i] - 1;
tree.add(id, freq[i]);
ans[i] = tree.maxRange(0, m - 1);
}
return ans;
};
class MaxSegmentTree {
constructor(n) {
this.size = n;
this.segTree = Array(n * 2).fill(0);
}
add(index, value) {
let n = this.size, idx = index + n;
this.segTree[idx] += value;
idx = Math.floor(idx / 2);
while (idx > 0) {
this.segTree[idx] = Math.max(this.segTree[idx * 2], this.segTree[idx * 2 + 1]);
idx = Math.floor(idx / 2);
}
}
maxRange(left, right) {
if (left > right) return 0;
let n = this.size, max = 0;
let left_idx = left + n, right_idx = right + n;
while (left_idx <= right_idx) {
if (left_idx % 2 === 1) max = Math.max(max, this.segTree[left_idx++]);
if (right_idx % 2 === 0) max = Math.max(max, this.segTree[right_idx--]);
left_idx = Math.floor(left_idx / 2);
right_idx = Math.floor(right_idx / 2);
}
return max;
}
}
Solution 2: Max Heap w/ Lazy Removal
- Use an array
value
to keep track of the current value of each number. - Use a max heap with lazy removal to get the maximum value with constant updates.
When we update a value, don't remove from the heap immediately, but remove it from the top of the heap when we need it, and discard expired values (value that is different from current value).
n = length of nums
, m = max(nums[i])
Time Complexity: O(n log(n) + m)
Space Complexity: O(n + m)
var mostFrequentIDs = function(nums, freq) {
let n = nums.length, value = Array(Math.max(...nums) + 1).fill(0);
let heap = new Heap((a, b) => b[1] - a[1]); // [id, value]
let ans = Array(n);
for (let i = 0; i < n; i++) {
value[nums[i]] += freq[i];
heap.add([nums[i], value[nums[i]]]);
while (!heap.isEmpty() && heap.top()[1] !== value[heap.top()[0]]) {
heap.remove();
}
if (heap.isEmpty()) {
ans[i] = 0;
} else {
ans[i] = heap.top()[1];
}
}
return ans;
};
class Heap {
constructor(comparator = ((a, b) => a - b)) {
this.values = [];
this.comparator = comparator;
this.size = 0;
}
add(val) {
this.size++;
this.values.push(val);
let idx = this.size - 1, parentIdx = Math.floor((idx - 1) / 2);
while (parentIdx >= 0 && this.comparator(this.values[parentIdx], this.values[idx]) > 0) {
[this.values[parentIdx], this.values[idx]] = [this.values[idx], this.values[parentIdx]];
idx = parentIdx;
parentIdx = Math.floor((idx - 1) / 2);
}
}
remove() {
if (this.size === 0) return -1;
this.size--;
if (this.size === 0) return this.values.pop();
let removedVal = this.values[0];
this.values[0] = this.values.pop();
let idx = 0;
while (idx < this.size && idx < Math.floor(this.size / 2)) {
let leftIdx = idx * 2 + 1, rightIdx = idx * 2 + 2;
if (rightIdx === this.size) {
if (this.comparator(this.values[leftIdx], this.values[idx]) > 0) break;
[this.values[leftIdx], this.values[idx]] = [this.values[idx], this.values[leftIdx]];
idx = leftIdx;
} else if (this.comparator(this.values[leftIdx], this.values[idx]) < 0 || this.comparator(this.values[rightIdx], this.values[idx]) < 0) {
if (this.comparator(this.values[leftIdx], this.values[rightIdx]) <= 0) {
[this.values[leftIdx], this.values[idx]] = [this.values[idx], this.values[leftIdx]];
idx = leftIdx;
} else {
[this.values[rightIdx], this.values[idx]] = [this.values[idx], this.values[rightIdx]];
idx = rightIdx;
}
} else {
break;
}
}
return removedVal;
}
top() {
return this.values[0];
}
isEmpty() {
return this.size === 0;
}
}
Comments
Post a Comment