Minimum number of operation required to convert number x into y
Last Updated :
22 Mar, 2023
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++
#include <bits/stdc++.h>
using namespace std;
struct node {
int val;
int level;
};
int minOperations( int x, int y)
{
set< int > visit;
queue<node> q;
node n = { x, 0 };
q.push(n);
while (!q.empty()) {
node t = q.front();
q.pop();
if (t.val == y)
return t.level;
visit.insert(t.val);
if (t.val * 2 == y || t.val - 1 == y)
return t.level + 1;
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);
}
}
}
int main()
{
int x = 4, y = 7;
cout << minOperations(x, y);
return 0;
}
|
Java
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 ;
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 ;
}
public static void main(String[] args)
{
int x = 4 , y = 7 ;
GFG src = new GFG(x, y);
System.out.println(minOperations(x, y));
}
}
|
Python3
import queue
class node:
def __init__( self , val, level):
self .val = val
self .level = level
def minOperations(x, y):
visit = set ()
q = queue.Queue()
n = node(x, 0 )
q.put(n)
while ( not q.empty()):
t = q.get()
if (t.val = = y):
return t.level
visit.add(t.val)
if (t.val * 2 = = y or t.val - 1 = = y):
return t.level + 1
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)
if __name__ = = '__main__' :
x = 4
y = 7
print (minOperations(x, y))
|
C#
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;
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;
}
public static void Main(String[] args)
{
int x = 4, y = 7;
GFG src = new GFG(x, y);
Console.WriteLine(minOperations(x, y));
}
}
|
Javascript
class node {
constructor(val, level) {
this .val = val;
this .level = level;
}
}
function minOperations(x, y) {
const visit = new Set();
const q = [];
const n = new node(x, 0);
q.push(n);
while (q.length > 0) {
const t = q.shift();
if (t.val == y) {
return t.level;
}
visit.add(t.val);
if (t.val * 2 == y || t.val - 1 == y) {
return t.level + 1;
}
if (!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));
}
}
}
const x = 4;
const y = 7;
console.log(minOperations(x, y));
|
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 (x == y)
return 0;
if (x <= 0 && y > 0)
return -1;
if (x > y)
return x - y;
if (y & 1)
return 1 + min_operations(x, y + 1);
else
return 1 + min_operations(x, y / 2);
}
signed main() {
cout << min_operations(4, 7) << endl;
return 0;
}
|
C
#include <stdio.h>
int min_operations( int x, int y)
{
if (x == y)
return 0;
if (x <= 0 && y > 0)
return -1;
if (x > y)
return x - y;
if (y & 1)
return 1 + min_operations(x, y + 1);
else
return 1 + min_operations(x, y / 2);
}
signed main()
{
printf ( "%d" , min_operations(4, 7));
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int minOperations( int x, int y)
{
if (x == y)
return 0 ;
if (x <= 0 && y > 0 )
return - 1 ;
if (x > y)
return x - y;
if (y % 2 != 0 )
return 1 + minOperations(x, y + 1 );
else
return 1 + minOperations(x, y / 2 );
}
public static void main(String[] args)
{
System.out.println(minOperations( 4 , 7 ));
}
}
|
Python3
def min_operations(x, y):
if x = = y:
return 0
if x < = 0 and y > 0 :
return - 1
if x > y:
return a - b
if y & 1 = = 1 :
return 1 + min_operations(x, y + 1 )
else :
return 1 + min_operations(x, y / / 2 )
print (min_operations( 4 , 7 ))
|
C#
using System;
class GFG {
static int min_operations( int x, int y)
{
if (x == y)
return 0;
if (x <= 0 && y > 0)
return -1;
if (x > y)
return x - y;
if (y % 2 == 1)
return 1 + min_operations(x, y + 1);
else
return 1 + min_operations(x, y / 2);
}
public static int Main()
{
Console.WriteLine(min_operations(4, 7));
return 0;
}
}
|
Javascript
<script>
function min_operations(x,y)
{
if (x == y)
return 0;
if (x <= 0 && y > 0)
return -1;
if (x > y)
return x - y;
if (y & 1)
return 1 + min_operations(x, y + 1);
else
return 1 + min_operations(x, y / 2);
}
document.write(min_operations(4, 7));
</script>
|
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.
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