Minimum steps to minimize n as per given condition
Given a number n, count minimum steps to minimize it to 1 according to the following criteria:
- If n is divisible by 2 then we may reduce n to n/2.
- If n is divisible by 3 then you may reduce n to n/3.
- Decrement n by 1.
Examples:
Input : n = 10 Output : 3 Input : 6 Output : 2
Greedy Approach (Doesn’t work always) :
As per greedy approach we may choose the step that makes n as low as possible and continue the same, till it reaches 1.
while ( n > 1)
{
if (n % 3 == 0)
n /= 3;
else if (n % 2 == 0)
n /= 2;
else
n--;
steps++;
}
If we observe carefully, the greedy strategy doesn’t work here.
Eg: Given n = 10 , Greedy –> 10 /2 = 5 -1 = 4 /2 = 2 /2 = 1 ( 4 steps ).
But the optimal way is –> 10 -1 = 9 /3 = 3 /3 = 1 ( 3 steps ).
So, we must think of dynamic approach for optimal solution.
Dynamic Approach:
For finding minimum steps we have three possibilities for n and they are:
f(n) = 1 + f(n-1) f(n) = 1 + f(n/2) // if n is divisible by 2 f(n) = 1 + f(n/3) // if n is divisible by 3
Below is memoization based implementation of above recursive formula.
C++
// CPP program to minimize n to 1 by given // rule in minimum steps #include <bits/stdc++.h> using namespace std; // function to calculate min steps int getMinSteps(int n, int *memo) { // base case if (n == 1) return 0; if (memo[n] != -1) return memo[n]; // store temp value for n as min( f(n-1), // f(n/2), f(n/3)) +1 int res = getMinSteps(n-1, memo); if (n%2 == 0) res = min(res, getMinSteps(n/2, memo)); if (n%3 == 0) res = min(res, getMinSteps(n/3, memo)); // store memo[n] and return memo[n] = 1 + res; return memo[n]; } // This function mainly initializes memo[] and // calls getMinSteps(n, memo) int getMinSteps(int n) { int memo[n+1]; // initialize memoized array for (int i=0; i<=n; i++) memo[i] = -1; return getMinSteps(n, memo); } // driver program int main() { int n = 10; cout << getMinSteps(n); return 0; } |
Java
// Java program to minimize n to 1 // by given rule in minimum steps import java.io.*; class GFG { // function to calculate min steps static int getMinSteps(int n, int memo[]) { // base case if (n == 1) return 0; if (memo[n] != -1) return memo[n]; // store temp value for // n as min( f(n-1), // f(n/2), f(n/3)) +1 int res = getMinSteps(n - 1, memo); if (n % 2 == 0) res = Math.min(res, getMinSteps(n / 2, memo)); if (n % 3 == 0) res = Math.min(res, getMinSteps(n / 3, memo)); // store memo[n] and return memo[n] = 1 + res; return memo[n]; } // This function mainly // initializes memo[] and // calls getMinSteps(n, memo) static int getMinSteps(int n) { int memo[] = new int[n + 1]; // initialize memoized array for (int i = 0; i <= n; i++) memo[i] = -1; return getMinSteps(n, memo); } // Driver Code public static void main (String[] args) { int n = 10; System.out.println(getMinSteps(n)); } } // This code is contributed by anuj_67. |
Python3
# Python program to minimize # n to 1 by given # rule in minimum steps # function to calculate min steps def getMinSteps(n, memo): # base case if (n == 1): return 0 if (memo[n] != -1): return memo[n] # store temp value for n as min(f(n-1), # f(n//2), f(n//3)) + 1 res = getMinSteps(n-1, memo) if (n%2 == 0): res = min(res, getMinSteps(n//2, memo)) if (n%3 == 0): res = min(res, getMinSteps(n//3, memo)) # store memo[n] and return memo[n] = 1 + res return memo[n] # This function mainly # initializes memo[] and # calls getMinSteps(n, memo) def getsMinSteps(n): memo = [0 for i in range(n+1)] # initialize memoized array for i in range(n+1): memo[i] = -1 return getMinSteps(n, memo) # driver program n = 10print(getsMinSteps(n)) # This code is contributed by Soumen Ghosh. |
C#
// C# program to minimize n to 1 // by given rule in minimum steps using System; class GFG { // function to calculate min steps static int getMinSteps(int n, int []memo) { // base case if (n == 1) return 0; if (memo[n] != -1) return memo[n]; // store temp value for // n as min( f(n-1), // f(n/2), f(n/3)) +1 int res = getMinSteps(n - 1, memo); if (n % 2 == 0) res = Math.Min(res, getMinSteps(n / 2, memo)); if (n % 3 == 0) res = Math.Min(res, getMinSteps(n / 3, memo)); // store memo[n] and return memo[n] = 1 + res; return memo[n]; } // This function mainly // initializes memo[] and // calls getMinSteps(n, memo) static int getMinSteps(int n) { int []memo = new int[n + 1]; // initialize memoized array for (int i = 0; i <= n; i++) memo[i] = -1; return getMinSteps(n, memo); } // Driver Code public static void Main () { int n = 10; Console.WriteLine(getMinSteps(n)); } } // This code is contributed by anuj_67. |
PHP
<?php // PHP program to minimize n to 1 by // given rule in minimum steps // function to calculate min steps function getMinSteps( $n, $memo) { // base case if ($n == 1) return 0; if ($memo[$n] != -1) return $memo[$n]; // store temp value for n // as min( f(n-1), // f(n/2), f(n/3)) +1 $res = getMinSteps($n - 1, $memo); if ($n % 2 == 0) $res = min($res, getMinSteps($n / 2, $memo)); if ($n % 3 == 0) $res = min($res, getMinSteps($n / 3, $memo)); // store memo[n] and return $memo[$n] = 1 + $res; return $memo[$n]; } // This function mainly initializes // memo[] and calls getMinSteps(n, memo) function g_etMinSteps( $n) { $memo= array(); // initialize memoized array for($i = 0; $i <= $n; $i++) $memo[$i] = -1; return getMinSteps($n, $memo); } // Driver Code $n = 10; echo g_etMinSteps($n); // This code is contributed by anuj_67. ?> |
Output:
3
Below is a tabulation based solution :
C++
// A tabulation based solution in C++ #include <bits/stdc++.h> using namespace std; int getMinSteps(int n) { int table[n+1]; for (int i=0; i<=n; i++) table[i] = n-i; for (int i=n; i>=1; i--) { if (!(i%2)) table[i/2] = min(table[i]+1, table[i/2]); if (!(i%3)) table[i/3] = min(table[i]+1, table[i/3]); } return table[1]; } // driver program int main() { int n = 10; cout << getMinSteps(n); return 0; } // This code is contributed by Jaydeep Rathore |
Java
// A tabulation based // solution in Java import java.io.*; class GFG { static int getMinSteps(int n) { int table[] = new int[n + 1]; for (int i = 0; i <= n; i++) table[i] = n - i; for (int i = n; i >= 1; i--) { if (!(i % 2 > 0)) table[i / 2] = Math.min(table[i] + 1, table[i / 2]); if (!(i % 3 > 0)) table[i / 3] = Math.min(table[i] + 1, table[i / 3]); } return table[1]; } // Driver Code public static void main (String[] args) { int n = 10; System.out.print(getMinSteps(n)); } } // This code is contributed // by anuj_67. |
Python3
# A tabulation based solution in Python3 def getMinSteps(n) : table = [0] * (n + 1) for i in range(n + 1) : table[i] = n-i for i in range(n, 0, -1) : if (not(i%2)) : table[i//2] = min(table[i]+1, table[i//2]) if (not(i%3)) : table[i//3] = min(table[i]+1, table[i//3]) return table[1] # driver program if __name__ == "__main__" : n = 10 print(getMinSteps(n)) # This code is contributed by Ryuga |
C#
// A tabulation based // solution in C# using System; class GFG { static int getMinSteps(int n) { int []table = new int[n + 1]; for (int i = 0; i <= n; i++) table[i] = n - i; for (int i = n; i >= 1; i--) { if (!(i % 2 > 0)) table[i / 2] = Math.Min(table[i] + 1, table[i / 2]); if (!(i % 3 > 0)) table[i / 3] = Math.Min(table[i] + 1, table[i / 3]); } return table[1]; } // Driver Code public static void Main () { int n = 10; Console.WriteLine(getMinSteps(n)); } } // This code is contributed // by anuj_67. |
PHP
<?php // A tabulation based solution in PHP function getMinSteps( $n) { $table = array(); for ($i = 0; $i <= $n; $i++) $table[$i] = $n - $i; for ($i = $n; $i >= 1; $i--) { if (!($i % 2)) $table[$i / 2] = min($table[$i] + 1, $table[$i / 2]); if (!($i % 3)) $table[$i / 3] = min($table[$i] + 1, $table[$i / 3]); } return $table[1]; } // Driver Code $n = 10; echo getMinSteps($n); // This code is contributed by anuj_67. ?> |
Output:
3
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