70 lines
2.2 KiB
C++
70 lines
2.2 KiB
C++
// Source : https://leetcode.com/problems/count-good-meals/
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// Author : Hao Chen
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// Date : 2021-03-30
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/*****************************************************************************************************
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*
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* A good meal is a meal that contains exactly two different food items with a sum of deliciousness
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* equal to a power of two.
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*
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* You can pick any two different foods to make a good meal.
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*
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* Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the i^
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* th item of food, return the number of different good meals you can make from this list
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* modulo 10^9 + 7.
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*
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* Note that items with different indices are considered different even if they have the same
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* deliciousness value.
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*
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* Example 1:
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*
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* Input: deliciousness = [1,3,5,7,9]
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* Output: 4
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* Explanation: The good meals are (1,3), (1,7), (3,5) and, (7,9).
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* Their respective sums are 4, 8, 8, and 16, all of which are powers of 2.
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*
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* Example 2:
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*
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* Input: deliciousness = [1,1,1,3,3,3,7]
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* Output: 15
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* Explanation: The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.
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*
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* Constraints:
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*
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* 1 <= deliciousness.length <= 10^5
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* 0 <= deliciousness[i] <= 2^20
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******************************************************************************************************/
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class Solution {
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public:
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int countPairs(vector<int>& deliciousness) {
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const int MAX_EXP = 22;
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int pow2[MAX_EXP];
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for (int i=0; i<MAX_EXP; i++){
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pow2[i] = 1 << i;
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//cout << pow2[i] << ", ";
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}
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unordered_map<int, int> stat;
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int big = 0;
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for(auto& d: deliciousness){
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stat[d]++;
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}
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long m = 0;
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for(auto& d: deliciousness){
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for(int i=MAX_EXP-1; i>=0 && pow2[i] >= d; i--){
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int x = pow2[i] - d;
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if ( stat.find(x) != stat.end() ){
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m += (x==d) ? stat[x]-1 : stat[x];
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}
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}
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}
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// remove the duplication - m/2,
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// because both [1,3] and [3,1] are counted.
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return (m/2) % 1000000007;
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}
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};
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