How can I memoize this digit dp problem's rec solution?

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Hello codeforces community!

Problem:

Given a lowerlimit and upperlimit (Values on coupon).

A coupon containing the sumofdigits= S will win the lottery. Find S in such a way that max no of people win the lottery.

In short return the count of sum (S) that is repeating max no of times. And the ways to choose that count.

Ex: LL:1 UP:5

array:[1,2,3,4,5]

Sumof digits: [1,2,3,4,5]

Count:1

No of ways to choose that count :5 (max count =1 for 5 )

Ex: LL:1 UP: 10

Array [1,2,3,4,5,6,7,8,9,10]

Sumofdigits: [1,2,3,4,5,6,7,8,9,1]

Count: 2 (sum=1 is repeating 2 times)

No of ways to choose :1

Constraints: 1<=lowerlimit<upperlimit<=10^18

My solution[C++]

#include <bits/stdc++.h>
using namespace std;

int n;
string s;
unordered_map<int, int> mp;

void helper(int idx, int sum, int tight)
{
	if(idx==n)
	{
		mp[sum]++;
		return;
	}
	if(tight)
	{
		for(int i=0;i<=s[idx]-'0';i++)
		{
			helper(idx+1, sum+i, (i==s[idx]-'0'));
		}
	}
	else
	{
		for(int i=0;i<10;i++)
		{
			helper(idx+1, sum+i, 0);
		}
	}
}

void solve()
{
	cin>>s;

	n=s.size();

	helper(0, 0, 1);

	int mx=0, cnt=0;

	for(auto x: mp)
	{
		if(x.second>mx)
		{
			mx=x.second;
			cnt=1;
		}
		else if(x.second==mx)
		{
			cnt++;
		}
	}

	cout<<mx<<" "<<cnt<<endl;
}
 
signed main() {

	solve();
 
return 0;
}

Now I want to memoize this rec solution but can't find a way to do it(as currently the func's return type is void). Can someone pls help :) Thanks!

Comments (3)

Write comment?

hijibijee

3 years ago, # |

Didn't get what's going on in your code. Here is my solution for the problem: https://pastebin.com/Uh5CXjc5 May be it will help. :D

→ Reply

dv.jakhar_

3 years ago, # ^ |

Wow, we can go sumwise too...Got this...Thanks a ton, bro! Actually in my solution I was only taking the upper limit for simplicity.

vedant-z

4 months ago, # |

My Code

#include <bits/stdc++.h>
using namespace std;
 
const int M=1e9+7;

#define rep(i,a,b) for(int i = a; i < b; i++)
#define fr(a,b) for(int i = a; i < b; i++)
#define all(x) (x).begin(),(x).end()
#define ll long long int

long long int dp[20][165][2][2];

long long int solve(int i, int sum, int tight1, int tight2, string& l, string& r) {
    if(i>=l.length()) return sum==0;
    if(dp[i][sum][tight1][tight2]!=-1) return dp[i][sum][tight1][tight2];
    long long int ans=0;
    for(char dig='0';dig<='9';dig++){
        if(sum<dig-'0') break;
        if((tight1 && dig<l[i]) || (tight2 && dig>r[i])) continue;
        ans+=solve(i+1,sum-(dig-'0'),tight1 && dig==l[i],tight2 && dig==r[i],l,r);
    }
    return dp[i][sum][tight1][tight2]=ans;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  long long int lowLimit,highLimit;cin>>lowLimit>>highLimit;
  string temp=to_string(lowLimit),l,r=to_string(highLimit);
  int dif=r.length()-temp.length();
  while(dif--) l+='0';
  l+=temp;
  unordered_map<int,long long int> mp;
  memset(dp,-1,sizeof(dp));
  for(int sum=1;sum<163;sum++){
    long long int res=solve(0,sum,1,1,l,r);
    if(res) mp[sum]+=res;
  }
  long long int mx=0,ct=0;
  for(auto i:mp) mx=max(mx,i.second);
  for(auto i:mp){
    if(i.second==mx) ++ct;
  }
  cout<<ct<<" "<<mx;

  return 0;
}

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