2022-04-06 Daily-Challenge
Today I have done leetcode's April LeetCoding Challenge with cpp
.
April LeetCoding Challenge 6
Description
3Sum With Multiplicity
Given an integer array arr
, and an integer target
, return the number of tuples i, j, k
such that i < j < k
and arr[i] + arr[j] + arr[k] == target
.
As the answer can be very large, return it modulo 109 + 7
.
Example 1:
Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation:
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.
Example 2:
Input: arr = [1,1,2,2,2,2], target = 5
Output: 12
Explanation:
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.
Constraints:
3 <= arr.length <= 3000
0 <= arr[i] <= 100
0 <= target <= 300
Solution
const int MOD = 1e9 + 7;
constexpr int pick(int total, int need) {
if(total < need) return 0;
long long answer = 1;
for(int i = 1; i <= need; ++i) {
answer *= (total + 1 - i);
answer /= i;
answer %= MOD;
}
return answer;
}
class Solution {
public:
int threeSumMulti(vector<int>& arr, int target) {
map<int, int> count;
for(auto num : arr) count[num] += 1;
long long answer = 0;
for(auto it = count.begin(); it != count.end(); ++it) {
for(auto jt = it; jt != count.end(); ++jt) {
int rest = target - it->first - jt->first;
if(rest < jt->first || !count.count(rest)) continue;
if(it->first == jt->first && it->first == rest) {
answer += pick(it->second, 3);
} else if(it->first == jt->first) {
answer += pick(it->second, 2) * count[rest];
} else if(jt->first == rest) {
answer += pick(jt->second, 2) * it->second;
} else {
answer += it->second * jt->second * count[rest];
}
answer %= MOD;
}
}
return answer;
}
};
// Accepted
// 71/71 cases passed (12 ms)
// Your runtime beats 92.78 % of cpp submissions
// Your memory usage beats 62.54 % of cpp submissions (10.6 MB)