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923. 3Sum With Multiplicity
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[ LeetCode 시즌 3 ] 2022년 4월 6일 문제 입니다.
[Medium] 923. 3Sum With Multiplicity
[Medium] 923. 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
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class Solution {
public:
int threeSumMulti(vector<int>& arr, int target) {
int vectorSize=arr.size();
int mod=1e9+7;
unsigned long int counter=0;
std::map<int,unsigned long int> m;
std::map<int,unsigned long int> sum;
for(int i=0;i<vectorSize;i++){
m[arr[i]]++;
};
for(std::map<int,unsigned long int>::iterator itr1=m.begin();itr1!=m.end();itr1++){
for(std::map<int,unsigned long int>::iterator itr2=itr1;itr2!=m.end();itr2++){
for(std::map<int,unsigned long int>::iterator itr3=itr2;itr3!=m.end();itr3++){
if((itr1->first+itr2->first+itr3->first)==target){
if(itr1->first != itr2->first){
if(itr2->first != itr3->first){
//std::cout<<"case1"<<std::endl;
counter=counter+(itr1->second * itr2->second * itr3->second);
}else{
if(itr2->second >= 2){
//std::cout<<"case2"<<std::endl;
counter=counter+(itr1->second * (itr2->second * (itr2->second-1))/2);
};
};
}else{
if(itr2->first != itr3->first){
if(itr1->second >= 2){
//std::cout<<"case3"<<std::endl;
counter=counter+((itr1->second * (itr1->second-1))/2 * itr3->second);
};
}else{
if(itr1->second >= 3){
//std::cout<<"case4"<<std::endl;
counter=counter+(((itr1->second)*(itr1->second-1) * (itr1->second-2) / 6 ))%mod;
};
};
};
};
};
};
};
return counter%mod;
}
};