Given three numbers A, B, and K where K is 0 initially. The task is to find the minimum operations to convert K into B using the below operations:
- Add 1 to K i.e K = K + 1
- Add A * 10c in K i.e K = K + A * 10c, where c is any integer (c >= 0).
Examples:
Input: A = 2, B = 7
Output: 4
Explanation: Initially K = 0, following operations are done on K:
- K = K + A * 100 => K = 0 + 2 * 1 => K= 2
- K = K + A * 100 => K = 2 + 2 * 1 => K= 4
- K = K + A * 100 => K = 4 + 2 * 1 => K= 6
- Add 1 to K => K = 7
Therefore, minimum operations needed are 4
Input: A = 25, B = 1337
Output: 20
Approach: A number can be represented as B = X * A + Y, where A is the maximum number that is multiplied with X and their product is nearest to B and Y is that number that is less than A. Follow the below steps to solve the problem:
- Initialize a variable K and assign X to it.
- Multiply K with 10 until it is greater than B.
- Initialize the variable ans with 0.
- Store Y part in ans using modulus operator ans = B % A.
- And subtract ans from B say B = B – ans.
- Now module B by K until it is K is greater or equal to A.
- And store division of B/K in ans.
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 operationsint minOperations(int A, int B){ // Initialize K and assign A into K int K = A; // Calculate the nearest // smaller number with B while (K < B) { K = K * 10; // If K is larger than B divide // by 10 and break the loop // We can get the nearest number if (K > B) { K = K / 10; break; } } // Now according to approach // Y part is B % A // which is smaller than X int ans = B % A; // Subtract Y part from B B = B - ans; // Iterate until K is // Greater than or equal to A while (K >= A) { // store ans which is division number ans = ans + B / K; // Modulus B by K B = B % K; // Divide K by 10 K = K / 10; } return ans;}// Driver Codeint main(){ int A = 25, B = 1337; int ans = minOperations(A, B); cout << ans; return 0;} |
Java
// Java code for the above approachimport java.util.*;class GFG{ // Function to find the minimum operations static int minOperations(int A, int B) { // Initialize K and assign A into K int K = A; // Calculate the nearest // smaller number with B while (K < B) { K = K * 10; // If K is larger than B divide // by 10 and break the loop // We can get the nearest number if (K > B) { K = K / 10; break; } } // Now according to approach // Y part is B % A // which is smaller than X int ans = B % A; // Subtract Y part from B B = B - ans; // Iterate until K is // Greater than or equal to A while (K >= A) { // store ans which is division number ans = ans + B / K; // Modulus B by K B = B % K; // Divide K by 10 K = K / 10; } return ans; } // Driver Code public static void main(String[] args) { int A = 25, B = 1337; int ans = minOperations(A, B); System.out.print(ans); }}// This code is contributed by sanjoy_62. |
Python
# Python program for the above approach# Function to find the minimum operationsdef minOperations(A, B): # Initialize K and assign A into K K = A # Calculate the nearest # smaller number with B while (K < B): K = K * 10 # If K is larger than B divide # by 10 and break the loop # We can get the nearest number if (K > B): K = K // 10 break # Now according to approach # Y part is B % A # which is smaller than X ans = B % A # Subtract Y part from B B = B - ans # Iterate until K is # Greater than or equal to A while (K >= A): # store ans which is division number ans = ans + B // K # Modulus B by K B = B % K # Divide K by 10 K = K // 10 return ans# Driver CodeA = 25B = 1337ans = minOperations(A, B)print(ans) # This code is contributed by Samim Hossain Mondal. |
C#
// C# code for the above approachusing System;class GFG{ // Function to find the minimum operations static int minOperations(int A, int B) { // Initialize K and assign A into K int K = A; // Calculate the nearest // smaller number with B while (K < B) { K = K * 10; // If K is larger than B divide // by 10 and break the loop // We can get the nearest number if (K > B) { K = K / 10; break; } } // Now according to approach // Y part is B % A // which is smaller than X int ans = B % A; // Subtract Y part from B B = B - ans; // Iterate until K is // Greater than or equal to A while (K >= A) { // store ans which is division number ans = ans + B / K; // Modulus B by K B = B % K; // Divide K by 10 K = K / 10; } return ans; } // Driver Code public static void Main() { int A = 25, B = 1337; int ans = minOperations(A, B); Console.Write(ans); }}// This code is contributed by Saurabh Jaiswal |
Javascript
<script> // JavaScript program for the above approach // Function to find the minimum operations const minOperations = (A, B) => { // Initialize K and assign A into K let K = A; // Calculate the nearest // smaller number with B while (K < B) { K = K * 10; // If K is larger than B divide // by 10 and break the loop // We can get the nearest number if (K > B) { K = parseInt(K / 10); break; } } // Now according to approach // Y part is B % A // which is smaller than X let ans = B % A; // Subtract Y part from B B = B - ans; // Iterate until K is // Greater than or equal to A while (K >= A) { // store ans which is division number ans = ans + parseInt(B / K); // Modulus B by K B = B % K; // Divide K by 10 K = parseInt(K / 10); } return ans; } // Driver Code let A = 25, B = 1337; let ans = minOperations(A, B); document.write(ans);// This code is contributed by rakeshsahni</script> |
Output
20
Time Complexity: O(1)
Auxiliary Space: O(1)
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