Program to find number of operations needed to decrease n to 0 in C++

Suppose we have a number n. Now consider an operation where we can either 1. Decrement n by one 2. If n is even number, then decrement it by n / 2 3. If n is divisible by 3, then decrement by 2 * (n / 3) Finally find the minimum number of operations required to decrement n to zero.

So, if the input is like n = 16, then the output will be 5, as n = 16 even then decreases it by n/2 4 times, it will be 1. Then reduce it by 1 to get 0. So total 5 operations.

To solve this, we will follow these steps −

Define one map dp
Define a function dfs(), this will take x,
ret := x
if x is in dp, then −
- return dp[x]
if x <= 0, then −
- return x
if x is same as 1, then −
- return 1
md2 := x mod 2
md3 := x mod 3
ret := minimum of {ret, md2 + 1 + dfs((x - md2) / 2), md3 + 1 + dfs((x - md3) / 3)}
return dp[x] = ret
From the main method call and return dfs(n)

Example

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
unordered_map <int, int> dp;
int dfs(int x){
   int ret = x;
   if(dp.count(x))
      return dp[x];
   if(x <= 0)
      return x;
   if(x == 1)
      return 1;
   int md2 = x % 2;
   int md3 = x % 3;
   ret = min({ret, md2 + 1 + dfs((x - md2) / 2), md3 + 1 + dfs((x - md3) / 3)});
   return dp[x] = ret;
}
int solve(int n) {
   return dfs(n);
}
int main(){
   int n = 16;
   cout << solve(n);
}