3943. Number of Pairs After Increment
You are given two integer arrays nums1 and nums2, and a 2D integer array queries.
Each queries[i] is one of the following types:
[1, x, y, val] – Add val to every element in nums2[x..y].[2, tot] – Compute the number of pairs (j, k) such that nums1[j] + nums2[k] == tot.
Return an integer array answer, where answer[j] is the number of pairs for the j^th query of type 2.
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struct Block {
long long base, l, r;
map<long long,long long> freq;
vector<long long> A;
Block(int l, int r) : base(0), l(l), r(r) {}
void append(int x) {
A.push_back(x);
freq[x]++;
}
int lookup(long long x) {
x -= base;
if(freq.count(x)) return freq[x];
return 0;
}
void udt(int le, int ri, int val) {
if(le <= l and r <= ri) base += val;
else {
int from = max(le * 1ll, l) - l;
int to = min(ri * 1ll, r) - l;
for(int i = from; i <= to and i < A.size(); i++) {
if(--freq[A[i]] == 0)
freq.erase(A[i]);
A[i] += val;
++freq[A[i]];
}
}
}
};
class Solution {
public:
vector<int> numberOfPairs(vector<int>& nums1, vector<int>& nums2, vector<vector<int>>& queries) {
vector<int> res;
int sq = sqrt(nums2.size());
vector<Block> b;
for(int i = 0; i < nums2.size(); i++) {
if(i % sq == 0) b.push_back(Block(i,i+sq-1));
b.back().append(nums2[i]);
}
for(auto& q : queries) {
if(q[0] == 1) {
int l = q[1], r = q[2], val = q[3];
for(int i = l / sq; i <= r / sq; i++) b[i].udt(l,r,val);
} else {
int now = 0, target = q[1];
for(auto& block : b) {
for(auto& n : nums1) {
now += block.lookup(target - n);
}
}
res.push_back(now);
}
}
return res;
}
};
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