Minimum Operations To Reduce An Integer To 0 - Leetcode Solution

Problem Statement:

Given an integer n, we need to reduce it to 0 using the following operations:

1. If n is even, divide it by 2.
2. If n is odd, either add 1 or subtract 1 from it.

Return the minimum number of operations needed to reduce n to 0.

Solution:

The problem statement is asking us to perform some simple operations on a given number n and reduce it to 0. The operations provided to us are:

1. If n is even, divide it by 2.
2. If n is odd, either add 1 or subtract 1 from it.

We need to perform the minimum number of operations in order to reduce the number to 0.

One approach to solve this problem can be to use recursion. We can break down the problem into sub-problems and keep reducing the number based on the operations given in the problem statement.

If the given number n is already 0, we don't need to perform any further operations, so we can return 0. If the given number n is even, we can divide it by 2 and recursively call the minimumOperations function with the quotient. If the given number n is odd, we can either add 1 or subtract 1 from it and then recursively call the minimumOperations function.

Let's see the code implementation for this approach:

def minimumOperations(n: int) -> int: if n == 0: return 0 if n % 2 == 0: return 1 + minimumOperations(n // 2) else: return 1 + min(minimumOperations(n-1), minimumOperations(n+1))

The time complexity of this algorithm is O(log n) as we are dividing the given number by 2 at each step. The space complexity of this algorithm is O(log n) as well, as the recursion stack can hold up to log2 n function calls.

We can optimize this algorithm further by using dynamic programming and memoization. We can store the minimum number of operations required to reduce a number in a table and use it during the recursion to avoid redundant calculations. This can reduce the time complexity to O(n) and space complexity to O(n).

Let's see the code implementation for this optimized approach:

def minimumOperations(n: int) -> int: table = [None] * (n+1) def dp(num): if num == 0: return 0 if table[num] is not None: return table[num] if num % 2 == 0: res = 1 + dp(num // 2) else: res = 1 + min(dp(num-1), dp(num+1)) table[num] = res return res return dp(n)

In this approach, we are maintaining a table to store the minimum number of operations required to reduce a number. We are checking if the value for the given number already exists in the table, if yes, then we are returning it. Otherwise, we are calculating the minimum number of operations required and storing it in the table before returning the value.

This algorithm is much more efficient than the earlier one, especially for larger values of n.

Conclusion:

We have seen two different approaches to solve the Minimum Operations To Reduce An Integer To 0 problem on Leetcode. The first approach used recursion and the second approach used dynamic programming with memoization. Both approaches have different performance characteristics and can be used based on the constraints of the problem.

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