Given an integer X, the task is to convert 1 into X by using the below-given operations:
- Multiply the number by 2.
- Multiply the number by 3.
- Add 1 to the number.
The task is to print the minimum number of operations needed to convert 1 into X using these three operations and also print the sequence of operations performed.
Examples:
Input: X = 5
Output:
3
1 3 4 5
Explanation:
Before any operation, X = 1
1st Operation: Multiply the number 1 by 3, hence X = 3.
2nd Operation: Add 1 to the number 3, hence X = 4.
3rd Operation: Add 1 to the number 4, hence X = 5.
Therefore, minimum 3 operations are required and the sequence is 1 3 4 5.Input: X = 20
Output:
4
1 3 9 10 20
Explanation:
Before any operation, X = 1
1st Operation: Multiply the number 1 by 3, hence X = 3.
2nd Operation: Multiply the number 3 by 3, hence X = 9.
3rd Operation: Add 1 to the number 9, hence X = 10.
4th Operation: Multiply the number 10 by 2, hence X = 20.
Therefore, minimum 4 operations are required and the sequence is 1 3 9 10 20 .
Naive Approach: The simplest approach is to find all possible sequences of operations to convert 1 into X and return the one with the minimum number of operations.
Time Complexity: O(3N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above naive approach the idea is to use Dynamic Programming to find the minimum number of operations and then backtrack to find the required sequence of operations. Below are the steps:
- Initialize the dp[] to store the minimum number of operations for every index i to reach 1 to i.
- Now dp[X] store the minimum number of operations required to make X from 1. We will fill the dp[] array in the Bottom-Up Approach. For the any value X we can reduce in one of the below ways:
- If X is divisible by 2, then divide it by 2 and count this operation.
- If X is divisible by 3, then divide it by 3 and count this operation.
- Else subtract 1 from X.
- From the above steps, the recurrence relation formed is given by:
dp[X] = min(dp[X/2] + 1, // If X is divisible by 2
dp[X/3] + 1, // If X is divisible by 3
dp[X-1] + 1)
Base Condition:
dp[1] = 0 // As no operation required when X = 1
- After the dp[] array is created, backtrack using the values stored in the array and store the sequence in the list.
- Print the minimum operations and the sequence stored in the list.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include<bits/stdc++.h>using namespace std;// Function to print the Minimum number // of operations required to convert 1 // into X by using three given operationsvoid printMinOperations(int N){ // Initialize a DP array to store //min operations for sub-problems int dp[N + 1]; for(int i = 0; i < N + 1; i++) dp[i] = N; // Base Condition: // No operation required // when N is 1 dp[1] = 0; for(int i = 2; i < N + 1; i++) { // Multiply the number by 2 if (i % 2 == 0 && dp[i] > dp[i / 2] + 1) dp[i] = dp[i / 2] + 1; // Multiply the number by 3 if (i % 3 == 0 && dp[i] > dp[i / 3] + 1) dp[i] = dp[i / 3] + 1; // Add 1 to the number. if (dp[i] > dp[i - 1] + 1) dp[i] = dp[i - 1] + 1; } // Print the minimum operations cout << dp[N] << endl; // Initialize a list to store // the sequence vector<int>seq; // Backtrack to find the sequence while (N > 1) { seq.push_back(N); // If add by 1 if(dp[N - 1] == dp[N] - 1) N = N - 1; // If multiply by 2 else if (N % 2 == 0 && dp[N / 2] == dp[N] - 1) N = N / 2; // If multiply by 3 else N = N / 3; } seq.push_back(1); // Print the sequence of operation for(int i = seq.size() - 1; i >= 0; i--) cout << seq[i] << " "; cout << endl;}// Driver codeint main(){ // Given number X int X = 96234; // Function call printMinOperations(X);}// This code is contributed by Stream_Cipher |
Java
// Java program for the above approach import java.util.*; class GFG{// Function to print the Minimum number // of operations required to convert 1 // into X by using three given operationsstatic void printMinOperations(int N){ // Initialize a DP array to store //min operations for sub-problems int dp[] = new int[N + 1]; for(int i = 0; i < N + 1; i++) dp[i] = N; // Base Condition: // No operation required // when N is 1 dp[1] = 0; for(int i = 2; i < N + 1; i++) { // Multiply the number by 2 if (i % 2 == 0 && dp[i] > dp[i / 2] + 1) dp[i] = dp[i / 2] + 1; // Multiply the number by 3 if (i % 3 == 0 && dp[i] > dp[i / 3] + 1) dp[i] = dp[i / 3] + 1; // Add 1 to the number. if (dp[i] > dp[i - 1] + 1) dp[i] = dp[i - 1] + 1; } // Print the minimum operations System.out.println(dp[N]); // Initialize a list to store // the sequence Vector<Integer> seq = new Vector<Integer>(); // Backtrack to find the sequence while (N > 1) { seq.add(N); // If add by 1 if(dp[N - 1] == dp[N] - 1) N = N - 1; // If multiply by 2 else if (N % 2 == 0 && dp[N / 2] == dp[N] - 1) N = N / 2; // If multiply by 3 else N = N / 3; } seq.add(1); // Print the sequence of operation for(int i = seq.size() - 1; i >= 0; i--) System.out.print(seq.get(i) + " ");}// Driver codepublic static void main(String args[]) { // Given number X int X = 96234; // Function call printMinOperations(X);}}// This code is contributed by Stream_Cipher |
Python3
# Python3 program for the above approach # Function to print the Minimum number # of operations required to convert 1 # into X by using three given operations def printMinOperations(N): # Initialize a DP array to store # min operations for sub-problems dp = [N]*(N + 1) # Base Condition: # No operation required # when N is 1 dp[1] = 0 for i in range(2, N + 1): # Multiply the number by 2 if (i % 2 == 0 and dp[i] > dp[i//2]+1): dp[i] = dp[i//2]+1 # Multiply the number by 3 if (i % 3 == 0 and dp[i] > dp[i//3]+1): dp[i] = dp[i//3]+1 # Add 1 to the number. if (dp[i] > dp[i-1]+1): dp[i] = dp[i-1] + 1 # Print the minimum operations print(dp[N], end ="\n") # Initialize a list to store # the sequence seq = [] # Backtrack to find the sequence while (N > 1): seq.append(N) # If add by 1 if dp[N-1] == dp[N] - 1: N = N-1 # If multiply by 2 elif N % 2 == 0 and dp[N//2] == dp[N] - 1: N = N//2 # If multiply by 3 else: N = N//3 seq.append(1) # Print the sequence of operation for i in reversed(seq): print(i, end =" ") # Driver Code # Given number X X = 96234# Function Call printMinOperations(X) |
C#
// C# program for the above approach using System; using System.Collections.Generic;class GFG{ // Function to print the Minimum number // of operations required to convert 1 // into X by using three given operationsstatic void printMinOperations(int N){ // Initialize a DP array to store //min operations for sub-problems int []dp = new int[N + 1]; for(int i = 0; i < N + 1; i++) dp[i] = N; // Base Condition: // No operation required // when N is 1 dp[1] = 0; for(int i = 2; i < N + 1; i++) { // Multiply the number by 2 if (i % 2 == 0 && dp[i] > dp[i / 2] + 1) dp[i] = dp[i / 2] + 1; // Multiply the number by 3 if (i % 3 == 0 && dp[i] > dp[i / 3] + 1) dp[i] = dp[i / 3] + 1; // Add 1 to the number. if (dp[i] > dp[i - 1] + 1) dp[i] = dp[i - 1] + 1; } // Print the minimum operations Console.WriteLine(dp[N]); // Initialize a list to store // the sequence List<int> seq = new List<int>(); // Backtrack to find the sequence while (N > 1) { seq.Add(N); // If add by 1 if(dp[N - 1] == dp[N] - 1) N = N - 1; // If multiply by 2 else if (N % 2 == 0 && dp[N / 2] == dp[N] - 1) N = N / 2; // If multiply by 3 else N = N / 3; } seq.Add(1); seq.Reverse(); // Print the sequence of operation foreach(var i in seq) { Console.Write(i + " "); }}// Driver codepublic static void Main() { // Given number X int X = 96234; // Function call printMinOperations(X);}}// This code is contributed by Stream_Cipher |
Javascript
<script>// JavaScript program for the above approach // Function to print the Minimum number // of operations required to convert 1 // into X by using three given operationsfunction printMinOperations(N){ // Initialize a DP array to store //min operations for sub-problems var dp = Array(N+1) for(var i = 0; i < N + 1; i++) dp[i] = N; // Base Condition: // No operation required // when N is 1 dp[1] = 0; for(var i = 2; i < N + 1; i++) { // Multiply the number by 2 if (i % 2 == 0 && dp[i] > dp[i / 2] + 1) dp[i] = dp[i / 2] + 1; // Multiply the number by 3 if (i % 3 == 0 && dp[i] > dp[i / 3] + 1) dp[i] = dp[i / 3] + 1; // Add 1 to the number. if (dp[i] > dp[i - 1] + 1) dp[i] = dp[i - 1] + 1; } // Print the minimum operations document.write( dp[N] + "<br>"); // Initialize a list to store // the sequence var seq = []; // Backtrack to find the sequence while (N > 1) { seq.push(N); // If add by 1 if(dp[N - 1] == dp[N] - 1) N = N - 1; // If multiply by 2 else if (N % 2 == 0 && dp[N / 2] == dp[N] - 1) N = N / 2; // If multiply by 3 else N = N / 3; } seq.push(1); // Print the sequence of operation for(var i = seq.length - 1; i >= 0; i--) document.write( seq[i] + " "); document.write("<br>");}// Driver code// Given number Xvar X = 96234; // Function callprintMinOperations(X);</script> |
Output:
14 1 3 9 10 11 33 99 297 891 2673 8019 16038 16039 48117 96234
Time Complexity: O(N)
Auxiliary Space: O(N)
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