Down to Zero II


Problem Statement :


You are given Q queries. Each query consists of a single number N. You can perform any of the  2 operations N on  in each move:

1: If we take 2 integers  a and b  where , N = a * b , then we can change  N = max( a, b )

2: Decrease the value of N by 1.

Determine the minimum number of moves required to reduce the value of  N to 0.


Input Format

The first line contains the integer Q.
The next Q lines each contain an integer, N.


Output Format

Output Q lines. Each line containing the minimum number of moves required to reduce the value of N to 0 .



Solution :



title-img


                            Solution in C :

In C ++ :




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


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int test;
    cin >> test;
    while (test--){
        int n ;
        cin >> n ;
        int steps = 0;
        if (n==0){
            cout << 0 << endl;
            continue;
        }
        if (n==1){
            cout << 1 << endl;
            continue;
        }
        vector<int> dist(n+1,0);
        queue<int> q;
        q.push(n) ;
        dist[n] = 1 ;
        while (1){
            int element = q.front(); 
            q.pop();
            if(element == 2){
                cout << dist[2] + 1 << endl;
                break ;
            }
            if (dist[element-1] == 0 ){
                dist [element-1] = dist[element]+1;
                q.push(element-1);
            }
            for (int i=2; i*i<=element; i++){
                if (element%i == 0){
                    int maxfrac = element/i; 
                    if (dist[maxfrac] == 0) dist [maxfrac] = dist[element] + 1, q.push(maxfrac);
                }
            }
        }
    }
    return 0;
}







In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    private static int MAXSZ = (int)(1E6+1);
    private static int[] dp;
    
    private static int minimumMove(int n) {
        if(n == 0) return 0;
        if(dp[n] != -1) return dp[n];
        
        int minMove = Integer.MAX_VALUE;
        int sq = (int) Math.sqrt(n);
        
        for(int i = 2; i <= sq; i++) {
            if(n % i == 0) {
                minMove = Math.min(minMove, 1+minimumMove(n/i));
            }
        }
        minMove = Math.min(minMove, 1+ minimumMove(n-1));
        return (dp[n] = minMove);
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        dp = new int[MAXSZ];
        Arrays.fill(dp, -1);
        dp[0] = 0; dp[1] = 1; dp[2] = 2;
        
        int Q = sc.nextInt();
        for(int q = 1; q <= Q; q++) {
            int n = sc.nextInt();
            System.out.println(minimumMove(n));
        }
    }
}






In C :




#include <stdio.h>
#include <stdio.h>
#include <string.h>
unsigned long long min(unsigned long long x, unsigned long long y);
unsigned long long ans[1000001];

int main(){
  int Q,N,i,j;
  memset(ans,-1,sizeof(ans));
  ans[0]=0;
  for(i=0;i<1000000;i++){
    ans[i+1]=min(ans[i+1],ans[i]+1);
    for(j=2;j<=i && i*(unsigned long long)j<1000001;j++)
      ans[i*j]=min(ans[i*j],ans[i]+1);
  }
  scanf("%d",&Q);
  while(Q--){
    scanf("%d",&N);
    printf("%llu\n",ans[N]);
  }
  return 0;
}
unsigned long long min(unsigned long long x, unsigned long long y){
  return (x>y)?y:x;
}








In Python3 :





import math

import time
start = time.time()

def Sol( N ):
    if N == 0: return 0
    Q = [ (N,0) ]
    setQ = [ 0 ] * N
    while Q:
        N, steps = Q.pop(0)
        if N == 1: return steps+1
        div = int(math.sqrt( N ))
        while div > 1:
            if N % div == 0 and not setQ[N // div]:
                Q.append( (N // div, steps+1) )
                setQ[ N // div ] = 1
            div -= 1
        if not setQ[N-1]:
            Q.append( (N-1, steps+1) )
            setQ[ N-1 ] = 1
            
    
Q = int(input())
for _ in range(Q):
    N = int(input())
    print(Sol(N))
                        








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