Given two integers A and B, convert A to B by performing one of the following operations any number of times:
- A = A + K
- A = A – K, where K belongs to [1, 10]
The task is to find the minimum number of operations required to convert A to B using the above operations.
Examples:
Input: A = 13, B = 42
Output: 3
Explanation:
The following sequence of moves can be performed: 13 ? 23 ? 32 ? 42(add 10, add 9, add 10).Input: A = 18, B = 4
Output: 2
Explanation:
The following sequence of moves can be performed: 18 ? 10 ? 4 (subtract 8, subtract 6).
Approach: The idea is to simply calculate the required number of moves by dividing the absolute difference of A and B by all the numbers in the range [1…10] and adding it to the resultant variable. Follow the steps below to solve the problem:
- Initialize a variable required_moves to store the minimum count of moves required.
- Find the absolute difference of A and B.
- Iterate over the range [1, 10] and perform the following operations:
- Divide the number by i and add it to the resultant variable.
- Calculate modulo of absolute difference by i.
- Finally, print the value of required_moves.
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 minimum number// of moves to obtained B from Avoid convertBfromA(int a, int b){ // Stores the minimum // number of moves int moves = 0; // Absolute difference int x = abs(a - b); // K is in range [0, 10] for (int i = 10; i > 0; i--) { moves += x / i; x = x % i; } // Print the required moves cout << moves << " ";}// Driver Codeint main(){ int A = 188, B = 4; convertBfromA(A, B); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{// Function to find minimum number// of moves to obtained B from Astatic void convertBfromA(int a, int b){ // Stores the minimum // number of moves int moves = 0; // Absolute difference int x = Math.abs(a - b); // K is in range [0, 10] for(int i = 10; i > 0; i--) { moves += x / i; x = x % i; } // Print the required moves System.out.print(moves + " ");}// Driver Codepublic static void main (String[] args){ int A = 188, B = 4; convertBfromA(A, B);}}// This code is contributed by code_hunt |
Python3
# Python3 program for the above approach# Function to find minimum number# of moves to obtained B from Adef convertBfromA(a, b): # Stores the minimum # number of moves moves = 0 # Absolute difference x = abs(a - b) # K is in range [0, 10] for i in range(10, 0, -1): moves += x // i x = x % i # Print the required moves print(moves, end = " ")# Driver CodeA = 188B = 4convertBfromA(A, B)# This code is contributed by code_hunt |
C#
// C# program for the above approach using System;class GFG{// Function to find minimum number// of moves to obtained B from Astatic void convertBfromA(int a, int b){ // Stores the minimum // number of moves int moves = 0; // Absolute difference int x = Math.Abs(a - b); // K is in range [0, 10] for(int i = 10; i > 0; i--) { moves += x / i; x = x % i; } // Print the required moves Console.Write(moves + " ");}// Driver Codepublic static void Main (){ int A = 188, B = 4; convertBfromA(A, B);}}// This code is contributed by code_hunt |
Javascript
<script>// Javascript program to implement// the above approach// Function to find minimum number// of moves to obtained B from Afunction convertBfromA(a, b){ // Stores the minimum // number of moves let moves = 0; // Absolute difference let x = Math.abs(a - b); // K is in range [0, 10] for(let i = 10; i > 0; i--) { moves += Math.floor(x / i); x = x % i; } // Print the required moves document.write(moves + " ");} // Driver Code let A = 188, B = 4; convertBfromA(A, B); </script> |
19
Time Complexity: O(K), where K is in the range [0, 10]
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
