Minimum number of given operations required to reduce a number to 2

Minimum number of given operations required to reduce a number to 2

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; }

Time Complexity: O(n)
Auxiliary Space: O(n)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

Adblock test (Why?)

Original page link

Best Cool Tech Gadgets

Top favorite technology gadgets