Keep Multiplying Found Values By Two - Leetcode Solution

Problem Statement:

Given an integer array nums, you start with an element nums[0]. You can perform a series of operations on nums such that:

• multiply each element by 2.
• add 1 to each element.

Return the minimum number of operations needed to return the array to its initial value.

Solution:

The goal of the problem is to find the minimum number of operations required to restore the array to its initial value. We can solve this problem using dynamic programming.

Let's define dp[i] as the minimum number of operations required to restore the array to its initial value if we start with nums[i].

Now, let's take an example to understand how we can compute dp[i]. Suppose we have an array nums=[5,3,1]. If we start with nums[0], we can either multiply it by 2 or add 1 to it. If we multiply 5 by 2, we get 10, and if we add 1 to it, we get 6. Therefore, we have two options, and we need to choose the one that leads to the minimum number of operations.

If we multiply 5 by 2, we need to find out the minimum number of operations required to restore the array if we start with nums[1]=3. Similarly, if we add 1 to 5, we need to find out the minimum number of operations required to restore the array if we start with nums[1]=6.

We can compute dp[i] as follows:

1. Initialize dp[n-1]=0, where n is the length of the array.
2. Loop through the array in reverse order from n-2 to 0.
3. For each element, compute the minimum number of operations required if we multiply it by 2 or add 1 to it.
4. Store the minimum value in dp[i].
5. Return dp[0], which is the minimum number of operations required to restore the array to its initial value if we start with nums[0].

The time complexity of this solution is O(n*log(max)), where max is the maximum value in the array. This is because we need to perform log(max) operations to multiply a number by 2.

Code:

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        n=len(nums)
        dp=[0]*n
        dp[n-1]=0
        for i in range(n-2,-1,-1):
            dp[i]=dp[i+1]+1
            if nums[i]*2<=nums[i+1]:
                dp[i]=min(dp[i],dp[i+1]-1)
            temp=nums[i]
            while temp%2==0:
                temp//=2
                j=bisect.bisect_right(nums,temp)-1
                if j>i:
                    dp[i]=min(dp[i],dp[j]+j-i)
        return dp[0]

Explanation:

1. First, we initialize dp[n-1]=0, where n is the length of the array.
2. Loop through the array in reverse order from n-2 to 0.
3. For each element in the array, we compute the minimum number of operations required if we multiply it by 2 or add 1 to it. We store the minimum value in dp[i].
4. In the for loop, we first assume that we will add 1 to nums[i]. Therefore, dp[i] will be dp[i+1]+1.
5. Then, we check if multiplying nums[i] by 2 will result in a value that is less than or equal to nums[i+1]. If it is, then we can choose to multiply nums[i] by 2 instead of adding 1. In this case, dp[i]=dp[i+1]-1, as we are only performing one operation instead of two.
6. Finally, we check if there is any element in the array that is a factor of nums[i] after dividing it by 2. We store the index of this element in j. If j is greater than i, then we can choose to multiply nums[i] by 2 and then perform operations till j to reach nums[j]. In this case, dp[i]=dp[j]+j-i.
7. We return dp[0], which is the minimum number of operations required to restore the array to its initial value if we start with nums[0].

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