Given a initial number x and two operations which are given below:
- Multiply number by 2.
- Subtract 1 from the number.
The task is to find out minimum number of operation required to convert number x into y using only above two operations. We can apply these operations any number of times.
Constraints:
1 <= x, y <= 1000
Example:
Input : x = 4, y = 7 Output : 2 We can transform x into y using following two operations. 1. 4*2 = 8 2. 8-1 = 7 Input : x = 2, y = 5 Output : 4 We can transform x into y using following four operations. 1. 2*2 = 4 2. 4-1 = 3 3. 3*2 = 6 4. 6-1 = 5 Answer = 4 Note that other sequences of two operations would take more operations.
The idea is to use BFS for this. We run a BFS and create nodes by multiplying with 2 and subtracting by 1, thus we can obtain all possible numbers reachable from starting number.
Important Points :
- When we subtract 1 from a number and if it becomes < 0 i.e. Negative then there is no reason to create next node from it (As per input constraints, numbers x and y are positive).
- Also, if we have already created a number then there is no reason to create it again. i.e. we maintain a visited array.
Implementation:
C++
// C++ program to find minimum number of steps needed// to convert a number x into y with two operations// allowed : (1) multiplication with 2 (2) subtraction// with 1.#include <bits/stdc++.h>using namespace std;// A node of BFS traversalstruct node { int val; int level;};// Returns minimum number of operations// needed to convert x into y using BFSint minOperations(int x, int y){ // To keep track of visited numbers // in BFS. set<int> visit; // Create a queue and enqueue x into it. queue<node> q; node n = { x, 0 }; q.push(n); // Do BFS starting from x while (!q.empty()) { // Remove an item from queue node t = q.front(); q.pop(); // If the removed item is target // number y, return its level if (t.val == y) return t.level; // Mark dequeued number as visited visit.insert(t.val); // If we can reach y in one more step if (t.val * 2 == y || t.val - 1 == y) return t.level + 1; // Insert children of t if not visited // already if (visit.find(t.val * 2) == visit.end()) { n.val = t.val * 2; n.level = t.level + 1; q.push(n); } if (t.val - 1 >= 0 && visit.find(t.val - 1) == visit.end()) { n.val = t.val - 1; n.level = t.level + 1; q.push(n); } }}// Driver codeint main(){ int x = 4, y = 7; cout << minOperations(x, y); return 0;} |
Java
// Java program to find minimum// number of steps needed to// convert a number x into y// with two operations allowed :// (1) multiplication with 2// (2) subtraction with 1.import java.util.HashSet;import java.util.LinkedList;import java.util.Set;class GFG { int val; int steps; public GFG(int val, int steps) { this.val = val; this.steps = steps; }}public class GeeksForGeeks { private static int minOperations(int src, int target) { Set<Integer> visited = new HashSet<>(1000); LinkedList<GFG> queue = new LinkedList<GFG>(); GFG node = new GFG(src, 0); queue.offer(node); while (!queue.isEmpty()) { GFG temp = queue.poll(); if(visited.contains(temp.val)) { continue; } visited.add(temp.val); if (temp.val == target) { return temp.steps; } int mul = temp.val * 2; int sub = temp.val - 1; // given constraints if (mul > 0 && mul < 1000) { GFG nodeMul = new GFG(mul, temp.steps + 1); queue.offer(nodeMul); } if (sub > 0 && sub < 1000) { GFG nodeSub = new GFG(sub, temp.steps + 1); queue.offer(nodeSub); } } return -1; } // Driver code public static void main(String[] args) { // int x = 2, y = 5; int x = 4, y = 7; GFG src = new GFG(x, y); System.out.println(minOperations(x, y)); }}// This code is contributed by Rahul |
Python3
# Python3 program to find minimum number of# steps needed to convert a number x into y# with two operations allowed :# (1) multiplication with 2# (2) subtraction with 1.import queue# A node of BFS traversalclass node: def __init__(self, val, level): self.val = val self.level = level# Returns minimum number of operations# needed to convert x into y using BFSdef minOperations(x, y): # To keep track of visited numbers # in BFS. visit = set() # Create a queue and enqueue x into it. q = queue.Queue() n = node(x, 0) q.put(n) # Do BFS starting from x while (not q.empty()): # Remove an item from queue t = q.get() # If the removed item is target # number y, return its level if (t.val == y): return t.level # Mark dequeued number as visited visit.add(t.val) # If we can reach y in one more step if (t.val * 2 == y or t.val - 1 == y): return t.level+1 # Insert children of t if not visited # already if (t.val * 2 not in visit): n.val = t.val * 2 n.level = t.level + 1 q.put(n) if (t.val - 1 >= 0 and t.val - 1 not in visit): n.val = t.val - 1 n.level = t.level + 1 q.put(n)# Driver codeif __name__ == '__main__': x = 4 y = 7 print(minOperations(x, y))# This code is contributed by PranchalK |
C#
// C# program to find minimum// number of steps needed to// convert a number x into y// with two operations allowed :// (1) multiplication with 2// (2) subtraction with 1.using System;using System.Collections.Generic;public class GFG { public int val; public int steps; public GFG(int val, int steps) { this.val = val; this.steps = steps; }}public class GeeksForGeeks { private static int minOperations(int src, int target) { HashSet<GFG> visited = new HashSet<GFG>(1000); List<GFG> queue = new List<GFG>(); GFG node = new GFG(src, 0); queue.Add(node); visited.Add(node); while (queue.Count != 0) { GFG temp = queue[0]; queue.RemoveAt(0); visited.Add(temp); if (temp.val == target) { return temp.steps; } int mul = temp.val * 2; int sub = temp.val - 1; // given constraints if (mul > 0 && mul < 1000) { GFG nodeMul = new GFG(mul, temp.steps + 1); queue.Add(nodeMul); } if (sub > 0 && sub < 1000) { GFG nodeSub = new GFG(sub, temp.steps + 1); queue.Add(nodeSub); } } return -1; } // Driver code public static void Main(String[] args) { // int x = 2, y = 5; int x = 4, y = 7; GFG src = new GFG(x, y); Console.WriteLine(minOperations(x, y)); }}// This code is contributed by aashish1995 |
Javascript
// JavaScript program to find minimum number of// steps needed to convert a number x into y// with two operations allowed :// (1) multiplication with 2// (2) subtraction with 1.// A node of BFS traversalclass node {constructor(val, level) {this.val = val;this.level = level;}}// Returns minimum number of operations// needed to convert x into y using BFSfunction minOperations(x, y) {// To keep track of visited numbers// in BFS.const visit = new Set();// Create a queue and enqueue x into it.const q = [];const n = new node(x, 0);q.push(n);// Do BFS starting from xwhile (q.length > 0) {// Remove an item from queueconst t = q.shift();// If the removed item is target// number y, return its levelif (t.val == y) { return t.level;}// Mark dequeued number as visitedvisit.add(t.val);// If we can reach y in one more stepif (t.val * 2 == y || t.val - 1 == y) { return t.level + 1;}// Insert children of t if not visited// alreadyif (!visit.has(t.val * 2)) { n.val = t.val * 2; n.level = t.level + 1; q.push(Object.assign({}, n));}if (t.val - 1 >= 0 && !visit.has(t.val - 1)) { n.val = t.val - 1; n.level = t.level + 1; q.push(Object.assign({}, n));}}}// Driver codeconst x = 4;const y = 7;console.log(minOperations(x, y));// This code is contributed by lokeshpotta20. |
Output
2
This article is contributed by Vipin Khushu. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Optimized solution:
In the second approach, we will check the least most bit of the number and take a decision according to the value of that bit.
Instead of converting x into y, we will convert y into x and will reverse the operations which will take the same number of operations as converting x into y.
So, reversed operations for y will be:
- Divide number by 2
- Increment number by 1
Implementation:
C++14
#include <iostream>using namespace std;int min_operations(int x, int y) { // If both are equal then return 0 if (x == y) return 0; // Check if conversion is possible or not if (x <= 0 && y > 0) return -1; // If x > y then we can just increase y by 1 // Therefore return the number of increments required if (x > y) return x - y; // If last bit is odd // then increment y so that we can make it even if (y & 1) return 1 + min_operations(x, y + 1); // If y is even then divide it by 2 to make it closer to // x else return 1 + min_operations(x, y / 2);}// Driver codesigned main() { cout << min_operations(4, 7) << endl; return 0;} |
C
#include <stdio.h>int min_operations(int x, int y){ // If both are equal then return 0 if (x == y) return 0; // Check if conversion is possible or not if (x <= 0 && y > 0) return -1; // If x > y then we can just increase y by 1 // Therefore return the number of increments required if (x > y) return x - y; // If last bit is odd // then increment y so that we can make it even if (y & 1) return 1 + min_operations(x, y + 1); // If y is even then divide it by 2 to make it closer to // x else return 1 + min_operations(x, y / 2);}// Driver codesigned main(){ printf("%d", min_operations(4, 7)); return 0;}// This code is contributed by Rohit Pradhan |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG{ static int minOperations(int x, int y) { // If both are equal then return 0 if (x == y) return 0; // Check if conversion is possible or not if (x <= 0 && y > 0) return -1; // If x > y then we can just increase y by 1 // Therefore return the number of increments // required if (x > y) return x - y; // If last bit is odd // then increment y so that we can make it even if (y % 2 != 0) return 1 + minOperations(x, y + 1); // If y is even then divide it by 2 to make it // closer to x else return 1 + minOperations(x, y / 2); } public static void main(String[] args) { System.out.println(minOperations(4, 7)); }}// This code is contributed by Shobhit Yadav |
Python3
def min_operations(x, y): # If both are equal then return 0 if x == y: return 0 # Check if conversion is possible or not if x <= 0 and y > 0: return -1 # If x > y then we can just increase y by 1 # Therefore return the number of increments required if x > y: return a-b # If last bit is odd # then increment y so that we can make it even if y & 1 == 1: return 1+min_operations(x, y+1) # If y is even then divide it by 2 to make it closer to x else: return 1+min_operations(x, y//2)# Driver codeprint(min_operations(4, 7)) |
C#
using System;class GFG { static int min_operations(int x, int y) { // If both are equal then return 0 if (x == y) return 0; // Check if conversion is possible or not if (x <= 0 && y > 0) return -1; // If x > y then we can just increase y by 1 // Therefore return the number of increments // required if (x > y) return x - y; // If last bit is odd // then increment y so that we can make it even if (y % 2 == 1) return 1 + min_operations(x, y + 1); // If y is even then divide it by 2 to make it // closer to // x else return 1 + min_operations(x, y / 2); } // Driver code public static int Main() { Console.WriteLine(min_operations(4, 7)); return 0; }}// This code is contributed by Taranpreet |
Javascript
<script>function min_operations(x,y){ // If both are equal then return 0 if (x == y) return 0; // Check if conversion is possible or not if (x <= 0 && y > 0) return -1; // If x > y then we can just increase y by 1 // Therefore return the number of increments required if (x > y) return x - y; // If last bit is odd // then increment y so that we can make it even if (y & 1) return 1 + min_operations(x, y + 1); // If y is even then divide it by 2 to make it closer to // x else return 1 + min_operations(x, y / 2);}// Driver codedocument.write(min_operations(4, 7));// This code is contributed by Taranpreet</script> |
Output
2
Time complexity:O(Y-X), where X, Y is the given number in the problem.
Space complexity: O(1), since no extra space used.
The optimized solution is contributed by BurningTiles. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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