Repository navigation
Expand file tree
/
Copy pathCountGoodMeals.cpp
More file actions
69 lines (63 loc) · 2.17 KB
/
Copy pathCountGoodMeals.cpp
File metadata and controls
69 lines (63 loc) · 2.17 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
// Source : https://leetcode.com/problems/count-good-meals/
// Author : Hao Chen
// Date : 2021-03-30
/*****************************************************************************************************
*
* A good meal is a meal that contains exactly two different food items with a sum of deliciousness
* equal to a power of two.
*
* You can pick any two different foods to make a good meal.
*
* Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the i^
* th item of food, return the number of different good meals you can make from this list
* modulo 10^9 + 7.
*
* Note that items with different indices are considered different even if they have the same
* deliciousness value.
*
* Example 1:
*
* Input: deliciousness = [1,3,5,7,9]
* Output: 4
* Explanation: The good meals are (1,3), (1,7), (3,5) and, (7,9).
* Their respective sums are 4, 8, 8, and 16, all of which are powers of 2.
*
* Example 2:
*
* Input: deliciousness = [1,1,1,3,3,3,7]
* Output: 15
* Explanation: The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.
*
* Constraints:
*
* 1 <= deliciousness.length <= 10^5
* 0 <= deliciousness[i] <= 2^20
******************************************************************************************************/
class Solution {
public:
int countPairs(vector<int>& deliciousness) {
const int MAX_EXP = 22;
int pow2[MAX_EXP];
for (int i=0; i<MAX_EXP; i++){
pow2[i] = 1 << i;
//cout << pow2[i] << ", ";
}
unordered_map<int, int> stat;
int big = 0;
for(auto& d: deliciousness){
stat[d]++;
}
long m = 0;
for(auto& d: deliciousness){
for(int i=MAX_EXP-1; i>=0 && pow2[i] >= d; i--){
int x = pow2[i] - d;
if ( stat.find(x) != stat.end() ){
m += (x==d) ? stat[x]-1 : stat[x];
}
}
}
// remove the duplication - m/2,
// because both [1,3] and [3,1] are counted.
return (m/2) % 1000000007;
}
};