Minimum number of given operations required to reduce a number to 2
Last Updated :
06 Jul, 2021
Given a positive integer N, the task is to reduce N to 2 by performing the following operations minimum number of times:
- Operation 1: Divide N by 5, if N is exactly divisible by 5.
- Operation 2: Subtract 3 from N.
If it is not possible, print -1.
Examples:
Input: N = 28
Output: 3
Explanation: Operation 1: Subtract 3 from 28. Therefore, N becomes 28 – 3 = 25.
Operation 2: Divide 25 by 5. Therefore, N becomes 25 / 5 = 5.
Operation 3: Subtract 3 from 5. Therefore, N becomes 5 – 3 = 2.
Hence, the minimum number of operations required is 3.
Input: n=10
Output: 1
Explanation: Operation 1: Divide 10 by 5, so n becomes 10/5=2.
Hence, the minimum operations required is 1.
Naive Approach: The idea is to recursively compute the minimum number of steps required.
- If the number is not divisible by 5, then subtract 3 from n and recur for the modified value of n, adding 1 to the result.
- Else make two recursive calls, one by subtracting 3 from n and the other by diving n by 5 and return the one with the minimum number of operations, adding 1 to the result.
Time Complexity: O(2n)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use dynamic programming. Follow these steps to solve this problem.
- Create an array, say dp[n+1] to store minimum operations and initialize all the entries with INT_MAX, where dp[i] stores the minimum number of steps required to reach 2 from i.
- Handle the base case by initializing dp[2] as 0.
- Iterate in the range [2, n] using the variable i
- If the value of i*5 ? n, then update dp[i*5] to minimum of dp[i*5] and dp[i]+1.
- If the value of i+3 ? n, then update dp[i+3] to minimum of dp[i+3] and dp[i]+1.
- Print the value of dp[n] as the result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findMinOperations( int n)
{
int i, dp[n + 1];
for (i = 0; i < n + 1; i++) {
dp[i] = 999999;
}
dp[2] = 0;
for (i = 2; i < n + 1; i++) {
if (i * 5 <= n) {
dp[i * 5] = min(
dp[i * 5], dp[i] + 1);
}
if (i + 3 <= n) {
dp[i + 3] = min(
dp[i + 3], dp[i] + 1);
}
}
return dp[n];
}
int main()
{
int n = 28;
int m = findMinOperations(n);
if (m != 9999)
cout << m;
else
cout << -1;
return 0;
}
|
Java
public class GFG {
static int findMinOperations( int n)
{
int i = 0 ;
int dp[] = new int [n + 1 ];
for (i = 0 ; i < n + 1 ; i++) {
dp[i] = 999999 ;
}
dp[ 2 ] = 0 ;
for (i = 2 ; i < n + 1 ; i++) {
if (i * 5 <= n) {
dp[i * 5 ] = Math.min(dp[i * 5 ], dp[i] + 1 );
}
if (i + 3 <= n) {
dp[i + 3 ] = Math.min(dp[i + 3 ], dp[i] + 1 );
}
}
return dp[n];
}
public static void main(String[] args)
{
int n = 28 ;
int m = findMinOperations(n);
if (m != 9999 )
System.out.println(m);
else
System.out.println(- 1 );
}
}
|
Python3
def findMinOperations(n):
dp = [ 0 for i in range (n + 1 )]
for i in range (n + 1 ):
dp[i] = 999999
dp[ 2 ] = 0
for i in range ( 2 , n + 1 ):
if (i * 5 < = n):
dp[i * 5 ] = min (dp[i * 5 ],
dp[i] + 1 )
if (i + 3 < = n):
dp[i + 3 ] = min (dp[i + 3 ],
dp[i] + 1 )
return dp[n]
if __name__ = = '__main__' :
n = 28
m = findMinOperations(n)
if (m ! = 9999 ):
print (m)
else :
print ( - 1 )
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int findMinOperations( int n)
{
int i;
int []dp = new int [n + 1];
for (i = 0; i < n + 1; i++)
{
dp[i] = 999999;
}
dp[2] = 0;
for (i = 2; i < n + 1; i++)
{
if (i * 5 <= n)
{
dp[i * 5] = Math.Min(dp[i * 5],
dp[i] + 1);
}
if (i + 3 <= n)
{
dp[i + 3] = Math.Min(dp[i + 3],
dp[i] + 1);
}
}
return dp[n];
}
public static void Main()
{
int n = 28;
int m = findMinOperations(n);
if (m != 9999)
Console.Write(m);
else
Console.Write(-1);
}
}
|
Javascript
<script>
function findMinOperations(n)
{
let i = 0;
let dp = new Array(n + 1);
for (i = 0; i < n + 1; i++) {
dp[i] = 999999;
}
dp[2] = 0;
for (i = 2; i < n + 1; i++) {
if (i * 5 <= n) {
dp[i * 5] = Math.min(dp[i * 5], dp[i] + 1);
}
if (i + 3 <= n) {
dp[i + 3] = Math.min(dp[i + 3], dp[i] + 1);
}
}
return dp[n];
}
let n = 28;
let m = findMinOperations(n);
if (m != 9999)
document.write(m);
else
document.write(-1);
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(n)
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