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main.cpp
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/// Source : https://leetcode.com/problems/split-array-with-same-average/description/
/// Author : liuyubobobo
/// Time : 2018-03-24
#include <iostream>
#include <vector>
#include <numeric>
#include <unordered_set>
using namespace std;
/// Dynamic Programming
///
/// Suppose the same average is ave, then ave = sum(A) / len(A)
/// Try how many items B contains from 1 to len(A)/2
/// Suppose B contains i items, then sum(B) = i * ave
/// This sub-problem can be solved by dynamic programming
///
/// Time Complexity: O(len(A) * len(A) * sum(A))
/// Space Complexity: O(len(A) * sum(A))
class Solution {
public:
bool splitArraySameAverage(vector<int>& A) {
int m = A.size();
int sum = accumulate(A.begin(), A.end(), 0);
for(int i = 1 ; i <= m/2 ; i ++)
if((sum * i) % m == 0 && ok(A, (sum * i) / m, i))
return true;
return false;
}
private:
bool ok(const vector<int>& A, int c, int num){
vector<unordered_set<int>> dp(c + 1, unordered_set<int>());
dp[0].insert(0);
for(int a: A)
for(int i = c ; i >= a ; i --)
if(dp[i - a].size() != 0)
if(a == 0) { // dp[i-a] is dp[i]
vector<int> tmp(dp[i].begin(), dp[i].end());
for(int i = 0 ; i < tmp.size() ; i ++)
tmp[i] ++;
for(int e: tmp)
dp[i].insert(e);
}
else{
for(int x: dp[i - a])
dp[i].insert(x + 1);
}
return dp[c].find(num) != dp[c].end();
}
};
void print_bool(bool res){
cout << (res ? "True" : "False") << endl;
}
int main() {
vector<int> A1 = {1,2,3,4,5,6,7,8};
print_bool(Solution().splitArraySameAverage(A1));
// true
vector<int> A2 = {3, 1};
print_bool(Solution().splitArraySameAverage(A2));
// false
vector<int> A3 = {18, 10, 5, 3};
print_bool(Solution().splitArraySameAverage(A3));
// false
vector<int> A4 = {2,0,5,6,16,12,15,12,4};
print_bool(Solution().splitArraySameAverage(A4));
// true
return 0;
}