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Exact Change.cpp
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/*
Solution by Rahul Surana
***********************************************************
You are given an array a1,a2,…,an consisting of integers from 0 to 9. A subarray al,al+1,al+2,…,ar−1,ar is good if the sum of elements of this subarray is equal to the length of this subarray (∑i=lrai=r−l+1).
For example, if a=[1,2,0], then there are 3 good subarrays: a1…1=[1],a2…3=[2,0] and a1…3=[1,2,0].
Calculate the number of good subarrays of the array a.
Input
The first line contains one integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains one integer n (1≤n≤105) — the length of the array a.
The second line of each test case contains a string consisting of n decimal digits, where the i-th digit is equal to the value of ai.
It is guaranteed that the sum of n over all test cases does not exceed 105.
Output
For each test case print one integer — the number of good subarrays of the array a.
***********************************************************
*/
#include <bits/stdc++.h>
#define ll long long
#define vl vector<ll>
#define vi vector<int>
#define pi pair<int,int>
#define pl pair<ll,ll>
#define all(a) a.begin(),a.end()
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define FOR(i,a) for(int i = 0; i < a; i++)
#define trace(x) cerr<<#x<<" : "<<x<<endl;
#define trace2(x,y) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<endl;
#define trace3(x,y,z) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<" | "<<#z<<" : "<<z<<endl;
#define fast_io std::ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
using namespace std;
vector<int> ar;
bool p(int v, int i, int j, int k){
FOR(s,i+1){
FOR(t,j+1){
if(s+2*t > v) continue;
if((v-s-2*t)%3 != 0) continue;
if((v-s-2*t)/3 <= k) return true;
}
}
return false;
}
bool pp(int i, int j, int k){
for(auto v: ar){
if(!p(v,i,j,k)) return false;
}
return true;
}
int main()
{
fast_io;
int t=1;
cin >> t;
while(t--) {
int n;
cin >> n;
int m = 0;
int ans = 1e9;
ar.resize(n);
FOR(i,n) { cin >> ar[i]; m = max(ar[i],m); }
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
int k = max(0,(int)ceil((m-i-2*j)/3));
if(pp(i,j,k)){
ans = min(ans,i+k+j);
}
}
}
cout << ans <<"\n";
}
}