Minimum steps to reduce N to 0 by given operations
Last Updated :
15 Jul, 2025
Give an integer N, the task is to find the minimum number of moves to reduce N to 0 by one of the following operations:
- Reduce N by 1.
- Reduce N to (N/2), if N is divisible by 2.
- Reduce N to (N/3), if N is divisible by 3.
Examples:
Input: N = 10
Output: 4
Explanation:
Here N = 10
Step 1: Reducing N by 1 i.e., 10 - 1 = 9.
Step 2: Since 9 is divisible by 3, reduce it to N/3 = 9/3 = 3
Step 3: Since again 3 is divisible by 3 again repeating step 2, i.e., 3/3 = 1.
Step 4: 1 can be reduced by the step 1, i.e., 1-1 = 0
Hence, 4 steps are needed to reduce N to 0.
Input: N = 6
Output: 3
Explanation:
Here N = 6
Step 1: Since 6 is divisible by 2, then 6/2 =3
Step 2: since 3 is divisible by 3, then 3/3 = 1.
Step 3: Reduce N to N-1 by 1, 1-1 = 0.
Hence, 3 steps are needed to reduce N to 0.
Naive Approach: The idea is to use recursion for all the possible moves. Below are the steps:
- Observe that base case for the problem, if N < 2 then for all the cases the answer will be N itself.
- At every value of N, choose between 2 possible cases:
- Reduce n till n % 2 == 0 and then update n /= 2 with count = 1 + n%2 + f(n/2)
- Reduce n till n % 3 == 0 and then update n /= 3 with count = 1 + n%3 + f(n/3)
- Both the computation results in the recurrence relation as:
count = 1 + min(n%2 + f(n/2), n%3 + f(n/3))
where, f(n) is the minimum of moves to reduce N to 0.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum
// number to steps to reduce N to 0
int minDays(int n)
{
// Base case
if (n < 1)
return n;
// Recursive call to count the
// minimum steps needed
int cnt = 1 + min(n % 2 + minDays(n / 2),
n % 3 + minDays(n / 3));
// Return the answer
return cnt;
}
// Driver Code
int main()
{
// Given number N
int N = 6;
// Function call
cout << minDays(N);
return 0;
}
// This code is contributed by 29AjayKumar
// Java program for the above approach
class GFG{
// Function to find the minimum
// number to steps to reduce N to 0
static int minDays(int n)
{
// Base case
if (n < 1)
return n;
// Recursive Call to count the
// minimum steps needed
int cnt = 1 + Math.min(n % 2 + minDays(n / 2),
n % 3 + minDays(n / 3));
// Return the answer
return cnt;
}
// Driver Code
public static void main(String[] args)
{
// Given Number N
int N = 6;
// Function Call
System.out.print(minDays(N));
}
}
// This code is contributed by PrinciRaj1992
# Python3 program for the above approach
# Function to find the minimum
# number to steps to reduce N to 0
def minDays(n):
# Base case
if n < 1:
return n
# Recursive Call to count the
# minimum steps needed
cnt = 1 + min(n % 2 + minDays(n // 2),
n % 3 + minDays(n // 3))
# Return the answer
return cnt
# Driver Code
# Given Number N
N = 6
# Function Call
print(str(minDays(N)))
// C# program for the above approach
using System;
class GFG{
// Function to find the minimum
// number to steps to reduce N to 0
static int minDays(int n)
{
// Base case
if (n < 1)
return n;
// Recursive call to count the
// minimum steps needed
int cnt = 1 + Math.Min(n % 2 + minDays(n / 2),
n % 3 + minDays(n / 3));
// Return the answer
return cnt;
}
// Driver Code
public static void Main(String[] args)
{
// Given number N
int N = 6;
// Function call
Console.Write(minDays(N));
}
}
// This code is contributed by Rajput-Ji
<script>
// JavaScript program for the above approach
// Function to find the minimum
// number to steps to reduce N to 0
function minDays(n)
{
// Base case
if (n < 1)
return n;
// Recursive call to count the
// minimum steps needed
var cnt = 1 + Math.min(n % 2 + minDays(parseInt(n / 2)),
n % 3 + minDays(parseInt(n / 3)));
// Return the answer
return cnt;
}
// Driver Code
// Given number N
var N = 6;
// Function call
document.write( minDays(N));
</script>
Time Complexity: O(2N)
Auxiliary Space: O(1)
Efficient Approach: The idea is to use Dynamic Programming. The above recursive approach results in TLE because of the number of repeated subproblems. To optimize the above method using a dictionary to keep track of values whose recursive call is already performed to reduce the further computation such that value can be accessed faster.
Below is the implementation of the above approach:
C++
// C++ program for
// the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the minimum
// number to steps to reduce N to 0
int count(int n)
{
// Dictionary for storing
// the precomputed sum
map<int, int> dp;
// Bases Cases
dp[0] = 0;
dp[1] = 1;
// Check if n is not in dp then
// only call the function so as
// to reduce no of recursive calls
if ((dp.find(n) == dp.end()))
dp[n] = 1 + min(n % 2 +
count(n / 2), n % 3 +
count(n / 3));
// Return the answer
return dp[n];
}
// Driver Code
int main()
{
// Given number N
int N = 6;
// Function call
cout << count(N);
}
// This code is contributed by gauravrajput1
// Java program for the above approach
import java.util.HashMap;
class GFG{
// Function to find the minimum
// number to steps to reduce N to 0
static int count(int n)
{
// Dictionary for storing
// the precomputed sum
HashMap<Integer,
Integer> dp = new HashMap<Integer,
Integer>();
// Bases Cases
dp.put(0, 0);
dp.put(1, 1);
// Check if n is not in dp then
// only call the function so as
// to reduce no of recursive calls
if (!dp.containsKey(n))
dp.put(n, 1 + Math.min(n % 2 +
count(n / 2), n % 3 +
count(n / 3)));
// Return the answer
return dp.get(n);
}
// Driver Code
public static void main(String[] args)
{
// Given number N
int N = 6;
// Function call
System.out.println(String.valueOf((count(N))));
}
}
// This code is contributed by Amit Katiyar
# Python3 program for the above approach
# Function to find the minimum
# number to steps to reduce N to 0
def count(n):
# Dictionary for storing
# the precomputed sum
dp = dict()
# Bases Cases
dp[0] = 0
dp[1] = 1
# Check if n is not in dp then
# only call the function so as
# to reduce no of recursive calls
if n not in dp:
dp[n] = 1 + min(n % 2 + count(n//2), n % 3 + count(n//3))
# Return the answer
return dp[n]
# Driver Code
# Given Number N
N = 6
# Function Call
print(str(count(N)))
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Function to find the minimum
// number to steps to reduce N to 0
static int count(int n)
{
// Dictionary for storing
// the precomputed sum
Dictionary<int,
int> dp = new Dictionary<int,
int>();
// Bases Cases
dp.Add(0, 0);
dp.Add(1, 1);
// Check if n is not in dp then
// only call the function so as
// to reduce no of recursive calls
if (!dp.ContainsKey(n))
dp.Add(n, 1 + Math.Min(n % 2 + count(n / 2),
n % 3 + count(n / 3)));
// Return the answer
return dp[n];
}
// Driver Code
public static void Main(String[] args)
{
// Given number N
int N = 6;
// Function call
Console.WriteLine(String.Join("",
(count(N))));
}
}
// This code is contributed by Rajput-Ji
<script>
// JavaScript program for
// the above approach
// Function to find the minimum
// number to steps to reduce N to 0
function count(n)
{
// Dictionary for storing
// the precomputed sum
var dp = new Map();
// Bases Cases
dp.set(0, 0);
dp.set(1, 1);
// Check if n is not in dp then
// only call the function so as
// to reduce no of recursive calls
if (!dp.has(n))
dp.set(n, 1 + Math.min(n % 2 +
count(parseInt(n / 2)), n % 3 +
count(parseInt(n / 3))));
// Return the answer
return dp.get(n);
}
// Driver Code
// Given number N
var N = 6;
// Function call
document.write( count(N));
</script>
Time Complexity: O(Nlog N), where N represents the given integer.
Auxiliary Space: O(N), where N represents the given integer.
Efficient approach: Using DP Tabulation method
This method Provided further improves on this by avoiding the use of a map and instead using an array to store the previously computed results. This is more efficient as array access is faster than map access.
Implementation steps :
- Create a vector dp of size n+1 and initialize dp[0] to 0 and dp[1] to 1.
- Loop from i=2 to i=n and for each i, compute the minimum number of steps to reduce i to 0 using the recurrence relation dp[i] = 1 + min(dp[i-1], dp[i/2] (if i is even), dp[i/3] (if i is divisible by 3)).
- Return dp[n] as the result.
Implementation:
C++
// C++ program for above approach
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6+5;
int dp[MAXN];
int count(int n)
{
// Base Cases
if(n == 0)
return 0;
if(n == 1)
return 1;
// Check if n is already computed
if(dp[n] != 0)
return dp[n];
// Compute the answer recursively
int ans = 1 + min(n % 2 + count(n / 2), n % 3 + count(n / 3));
// Store the computed answer
dp[n] = ans;
// Return the answer
return ans;
}
int main()
{
// Given number N
int N = 6;
// Initialize the dp array to 0
memset(dp, 0, sizeof(dp));
// Function call
cout << count(N);
}
// this code is contributed by bhardwajji
// Java program for above approach
import java.util.*;
class Main {
static int[] dp = new int[(int)1e6 + 5];
public static int count(int n)
{
// Base Cases
if (n == 0)
return 0;
if (n == 1)
return 1;
// Check if n is already computed
if (dp[n] != 0)
return dp[n];
// Compute the answer recursively
int ans = 1
+ Math.min(n % 2 + count(n / 2),
n % 3 + count(n / 3));
// Store the computed answer
dp[n] = ans;
// Return the answer
return ans;
}
public static void main(String[] args)
{
// Given number N
int N = 6;
// Initialize the dp array to 0
Arrays.fill(dp, 0);
// Function call
System.out.println(count(N));
}
}
// This code is contributed by sarojmcy2e
MAXN = 1000005
dp = [0] * MAXN
def count(n):
# Base Cases
if n == 0:
return 0
if n == 1:
return 1
# Check if n is already computed
if dp[n] != 0:
return dp[n]
# Compute the answer recursively
ans = 1 + min(n % 2 + count(n // 2), n % 3 + count(n // 3))
# Store the computed answer
dp[n] = ans
# Return the answer
return ans
# Given number N
N = 6
# Initialize the dp array to 0
dp = [0] * MAXN
# Function call
print(count(N))
using System;
class MainClass {
const int MAXN = 1000005;
static int[] dp = new int[MAXN];
static int count(int n) {
// Base Cases
if (n == 0)
return 0;
if (n == 1)
return 1;
// Check if n is already computed
if (dp[n] != 0)
return dp[n];
// Compute the answer recursively
int ans = 1 + Math.Min(n % 2 + count(n / 2), n % 3 + count(n / 3));
// Store the computed answer
dp[n] = ans;
// Return the answer
return ans;
}
public static void Main() {
// Given number N
int N = 6;
// Initialize the dp array to 0
Array.Fill(dp, 0);
// Function call
Console.WriteLine(count(N));
}
}
// Javascript program for above approach
let dp = new Array(1000005).fill(0);
function count(n) {
// Base Cases
if (n == 0)
return 0;
if (n == 1)
return 1;
// Check if n is already computed
if (dp[n] != 0)
return dp[n];
// Compute the answer recursively
let ans = 1 + Math.min(n % 2 + count(Math.floor(n / 2)),
n % 3 + count(Math.floor(n / 3)));
// Store the computed answer
dp[n] = ans;
// Return the answer
return ans;
}
// Given number N
let N = 6;
// Function call
console.log(count(N));
Time Complexity: O(N)
Auxiliary Space: O(N),
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