Problem Description

You are provided with an array nums where each element is a non-negative power of 2. A target integer is also given. Your task is to make a subsequence of nums that adds up to target. A subsequence can be formed by removing some elements from the array without changing the order of the remaining elements.

Operations that you are allowed to perform involve choosing an element nums[i] greater than 1, removing it from the array, and adding two occurrences of nums[i] / 2 to the end of the array. The goal is to find the minimum number of such operations required to reach the target sum using a subsequence of the modified array. If it's not possible to reach the target, you should return -1.

To clarify:

• The sequence is 0-indexed.
• Operations can only be performed on elements greater than 1.
• You are adding two halves of the number you remove back to the array.
• We are interested in returning the smallest possible number of operations.
• We are only creating subsequences, not permutations. Order matters.

Intuition

The solution approach requires decomposing the target into powers of 2, similar to binary representation. Since every element in nums is a power of 2, the sum that we need to hit (the target) can be seen as a sum of certain powers of 2. What we want is to find which powers of 2 are needed and in what quantity to reach the target.

The intuition behind the solution comes from the binary representation of the target. Since every number in nums is a power of 2, the target's binary representation indicates which powers of 2 are required. What we do is build up smaller powers of 2 to form the larger ones needed. This is akin to converting coins you possess into coins of lower denominations to make exact change.

Here’s a breakdown of the intuition:

• Check if the sum of all elements in nums is less than the target. If yes, we can't possibly hit the target, so return -1.
• Count how many times each power of 2 appears in nums. This is prepared for knowing whether we have enough "coins" of each power to reach each bit in target's binary representation.
• Iterate over each bit of target from LSB to MSB.
• For each bit set in target, we need to make sure we have a power of 2 available to match this bit.
  • If not enough of that power is available, we build it up from smaller powers by doing our defined operation, which "breaks" a number in half and adds two of the halves.
  • Each build-up is counted as a conversion operation.
• Add the number of operations at each step, and continue until the entire binary representation of target is processed.
• If we process the whole representation successfully, the total count of operations is the answer we return.

Learn more about Greedy patterns.

Solution Approach

The algorithm takes advantage of bit manipulation and greedy approach to minimize the number of operations required to achieve a subsequence that sums to target. It initializes a frequency array cnt to represent the count of each power of 2 present in nums. Since any num in the array is a non-negative power of 2, up to 2^31 could be included; hence the size of cnt is 32, for the 32 bits.

Here's a detailed walk-through of the implementation:

1. Sum Check: Before we even start, we ensure that the sum of all elements in nums is at least target. If not, we immediately return -1.

2. Count Powers of 2: Next, we count occurrences of each power of 2, which is done through bitwise operations. For each x in nums, we use a bitwise shift and bitwise AND to find which power of 2 it represents and increment the corresponding count in cnt.

3. Iterate Over Target Bits:

   • We then iterate through the bits of target, starting with the least significant bit (LSB).
   • If the current bit in the target is not set (0), there is nothing to match, so we move on to the next more significant bit.
   • When we encounter a bit set to 1, it indicates we need a power of 2 of that magnitude to contribute to the sum.

4. Build Required Powers:

   • If the necessary power of 2 is not available, we need to build it up from the available smaller powers by performing operations which effectively "combine" two elements of a smaller power of 2 into one of the next higher power.
   • We ensure that if we are looking for 2^i, all lower powers are combined as much as possible (cnt[j] is halved and cnt[j + 1] incremented accordingly).

5. Optimize and Perform Operations:

   • If there's no count available (cnt[j] is 0) for the required power 2^j, we convert all available powers. This is where the greedy strategy comes in. We keep running our conversion operation until we've built the required power of 2.
   • Each action of converting smaller powers to bigger ones is a single operation. The difference j - i contributes to the total operations performed, as it reflects converting 2^i to 2^j.
   • When building up a required power, we subtract one from the higher power and reset j to i since we just used up one power of 2^j to match a set bit in the target.

6. Result: The iteration continues until all bits in target are processed (did we match all bits?). If we achieved our goal, the total number of operations (ans) is returned. If it was impossible to meet a bit requirement, the first part of the code would have already returned -1.

The pattern used in this solution is a mix of bit manipulation (to read individual bits in the target and convert nums into count of powers of 2) and a greedy approach that always opts to convert the available smallest powers of 2 to the necessary ones in the minimum number of operations.

Using the Python bitwise shift >> and bitwise AND &, each number's contribution to the target sum is analyzed bit by bit, and the array cnt is updated to reflect how many operations are needed to "break down" larger powers of 2 into smaller ones to match the target's bit pattern.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose we have the following nums array and target:

• nums = [4, 8, 1, 2]
• target = 11 (binary representation: 1011)

Initial Preparation

• Calculate the sum of the nums: 4 + 8 + 1 + 2 = 15. Since 15 is greater than 11, we can proceed.
• Construct the frequency array cnt: [1, 1, 1, 0, 1]. We have one 1 (2^0), one 2 (2^1), one 4 (2^2), zero 8 (2^3), and one 16 (2^4).

Iterate Over Target Bits (from LSB to MSB)

• Bit 0 (2^0): Target is 1. We have one 1 in cnt, so we use it. No operation needed.
• Bit 1 (2^1): Target is 1. We have one 2 in cnt, so we use it. No operation needed.
• Bit 2 (2^2): Target is 0. Move to the next bit.
• Bit 3 (2^3): Target is 1, but cnt for 2^3 is 0. We need to build this power up.
  • Since we need 2^3 and we have one 2^2, we do not have enough of 2^2 to double and make one 2^3. However, we have 2^4 available which can be broken down.

Building Required Powers

• We 'break' our 2^4 into two 2^3s. One operation is performed. Now cnt is: [1, 1, 1, 2, 0].
• We use one 2^3 for our target. Now cnt is: [1, 1, 1, 1, 0].

Final Steps

• We have used all bits set in target, and the number of operations performed is 1.
• We return the total number of operations: 1.

The process shows how we break down powers of 2 only when necessary to reach our target and keep the operations to a minimum. In this case, converting a 2^4 to two 2^3s ensured that we could create the target sum of 11 from the subsequence of the modified array.

Solution Implementation

Time and Space Complexity

The given code intends to count the minimum number of operations required to reduce a list of numbers such that their sum equals a given target. An operation consists of dividing a non-zero element of the array by 2 and taking the floor.

Time Complexity

The time complexity of the code is determined by several nested loops and operations on the cnt list which has fixed size of 32 (since we're considering the binary representation of integers within 32 bits).

• The first for loop iterates over all the elements in the given nums list. Its complexity is O(n), where n is the number of elements in nums.

• Inside this loop, there is a nested for loop iterating at most 32 times (for each bit in the integer's binary representation). This makes the complexity of code inside the first loop O(32n) = O(n), because the 32 is a constant factor.

• After initializing i and j, we have a while loop that continues until i equals 32. In the worst case, this loop could run up to 32 times.

• Within the said while loop, we have several other while loops that increase i and j. Since i and j index through the bit positions at most 32 times, these while loops contribute at most O(32), which can be considered a fixed size and thus a constant factor.

• Each operation within the loops has a constant time complexity.

Combining these, the overall time complexity of this code snippet is O(n), where n is the length of nums because we can treat all the other operations as constants.

Space Complexity

The space complexity is determined by the additional space used by the algorithm excluding the input size.

• We have a constant-sized list cnt of size 32 to keep track of the count of bits set at different positions.

• Variables i, j, and ans also take up space, but they are constant-sized and do not depend on the input size.

Given this, the space complexity of the algorithm is O(1), as the space used does not scale with the input size; it remains constant.

Learn more about how to find time and space complexity quickly using problem constraints.