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New Problem Solution -"1711. Count Good Meals"
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// Source : https://leetcode.com/problems/count-good-meals/ | ||
// Author : Hao Chen | ||
// Date : 2021-03-30 | ||
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/***************************************************************************************************** | ||
* | ||
* 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 | ||
******************************************************************************************************/ | ||
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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] << ", "; | ||
} | ||
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unordered_map<int, int> stat; | ||
int big = 0; | ||
for(auto& d: deliciousness){ | ||
stat[d]++; | ||
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} | ||
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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]; | ||
} | ||
} | ||
} | ||
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// remove the duplication - m/2, | ||
// because both [1,3] and [3,1] are counted. | ||
return (m/2) % 1000000007; | ||
} | ||
}; |