LeetCode 2835 - Minimum Operations to Form Subsequence With Target Sum

The problem presents a 0-indexed array nums of non-negative powers of 2 and an integer target. Each element in nums is guaranteed to be a power of two, which is important because it allows us to reason about sums and splits in terms of binary representation.

leetcodehardarraygreedybit-manipulation

LeetCode Problem 2835

Difficulty: 🔴 Hard
Topics: Array, Greedy, Bit Manipulation

Solution

Problem Understanding

The problem presents a 0-indexed array nums of non-negative powers of 2 and an integer target. Each element in nums is guaranteed to be a power of two, which is important because it allows us to reason about sums and splits in terms of binary representation. You are allowed to perform a specific operation any number of times: choose an element greater than 1, remove it, and replace it with two copies of half its value, effectively splitting a power of two into smaller powers of two.

The goal is to determine the minimum number of such operations required so that some subsequence of nums sums exactly to target. A subsequence is any selection of elements that maintains relative order, meaning you can skip elements but cannot reorder them. If it is impossible to reach the target sum, the function should return -1.

Constraints clarify the input size and values: nums has up to 1000 elements, elements can be as large as 2^30, and the target can be as large as 2^31 - 1. These limits rule out exponential brute-force searches over all subsequences, so an efficient approach is required. Important edge cases include:

  • The sum of all elements in nums is less than target, making it impossible.
  • target requires elements not initially present in nums unless we perform splits.
  • Elements equal to 1 cannot be split further, limiting flexibility in forming the target sum.

Approaches

Brute Force

A naive approach would consider generating all possible sequences of operations and then checking every subsequence sum to see if it equals target. This approach guarantees correctness because it exhaustively explores all possibilities. However, it is infeasible due to the combinatorial explosion: each element can be split multiple times, leading to potentially billions of sequences for nums of length up to 1000.

Optimal Approach

The key insight comes from binary representation. Since all elements are powers of two, any sum can be represented as a combination of powers of two. We can maintain a frequency count of each power of two in the array and attempt to match the binary representation of the target. Starting from the smallest power of two (1) upwards:

  1. Count how many of each power we have.
  2. Attempt to cover the target's set bits from smallest to largest.
  3. If we need a higher power but only have a lower power, we can perform splits, counting each split as an operation.

This greedy approach works because splitting larger powers into smaller powers directly increases flexibility for forming the target sum. By working from the lowest power upwards, we can satisfy each bit in the target using minimal operations.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Explore all possible splits and subsequences. Exponential and infeasible.
Optimal O(31 + n) = O(n) O(31) = O(1) Count powers of two and satisfy target bits greedily. Linear in array size, constant extra space for power counts.

Algorithm Walkthrough

  1. Initialize an array count of length 31 (powers from 2^0 to 2^30) to zero. Iterate through nums and increment count[log2(num)] for each num. This gives the frequency of each power of two in the array.
  2. Initialize a variable operations to zero.
  3. Iterate through powers of two from 0 to 30. For each power i:

a. Check if the i-th bit of target is set. If it is, we need one element of value 2^i.

b. If count[i] > 0, use one element to cover this bit. Decrement count[i].

c. If count[i] == 0, find the next larger power j > i that has count[j] > 0. For each power we must split down from 2^j to 2^i, increment operations for every split (each split doubles the element count but halves the value). After splitting, update count[i] to 1.

d. After covering the i-th bit, carry over any remaining elements to the next power by count[i+1] += count[i] // 2. This ensures we can use surplus elements in higher powers if needed. 4. After iterating all bits, if we have successfully covered all set bits of target, return operations. If at any step we cannot find a larger power to split, return -1 because forming the target is impossible.

Why it works: The algorithm guarantees minimal operations because it only splits larger powers when smaller powers are required by the target. Surplus elements are carried upward, ensuring no unnecessary splits. Since every element is a power of two, each bit in the target can be independently satisfied, making this greedy approach correct.

Python Solution

from typing import List
import math

class Solution:
    def minOperations(self, nums: List[int], target: int) -> int:
        count = [0] * 31
        for num in nums:
            power = int(math.log2(num))
            count[power] += 1
        
        operations = 0
        for i in range(31):
            if (target >> i) & 1:  # if the i-th bit is set in target
                if count[i] > 0:
                    count[i] -= 1
                else:
                    j = i + 1
                    while j < 31 and count[j] == 0:
                        j += 1
                    if j == 31:
                        return -1  # no larger power to split
                    while j > i:
                        count[j] -= 1
                        count[j - 1] += 2
                        operations += 1
                        j -= 1
                    count[i] -= 1  # now we have the needed power
            # carry over surplus to next power
            if i < 30:
                count[i + 1] += count[i] // 2
        
        return operations

Implementation Explanation: We first count occurrences of each power of two. Then, we iterate through each bit of the target. If the bit is set, we attempt to use the corresponding power from nums. If it does not exist, we split the next higher power until we get the required element. Any leftover elements are carried to the next higher power for future bits. The operations counter tracks every split performed.

Go Solution

import (
    "math/bits"
)

func minOperations(nums []int, target int) int {
    count := make([]int, 31)
    for _, num := range nums {
        power := bits.TrailingZeros(uint(num))
        count[power]++
    }

    operations := 0
    for i := 0; i < 31; i++ {
        if (target>>i)&1 == 1 {
            if count[i] > 0 {
                count[i]--
            } else {
                j := i + 1
                for j < 31 && count[j] == 0 {
                    j++
                }
                if j == 31 {
                    return -1
                }
                for j > i {
                    count[j]--
                    count[j-1] += 2
                    operations++
                    j--
                }
                count[i]--
            }
        }
        if i < 30 {
            count[i+1] += count[i] / 2
        }
    }

    return operations
}

Go Implementation Notes: bits.TrailingZeros is used to compute log2 for powers of two. Slices are used to store counts. The logic mirrors Python closely. Care is taken with integer division and indexing to avoid out-of-range errors.

Worked Examples

Example 1: nums = [1,2,8], target = 7

Step Count Array (powers 0-3) Target Bit Operations Action
Initial [1,1,0,1] bit0=1 0 use 1 → count[0]=0
bit1 [0,1,0,1] 1 0 use 2 → count[1]=0
bit2 [0,0,0,1] 1 1 split 8→4,4 → count[2]=1, operations=1
Done Subsequence [1,2,4] sum=7 - 1 return 1

Example 2: nums = [1,32,1,2], target = 12

Step Count Array Target Bit Operations Action
Initial [2,1,0,0,0,0,1] bit0=0 0 skip
bit1 [2,1,0,0,0,0,1] 1 0 use 2 → count[1]=0
bit2 [2,0,0,0,0,0,1] 1 2 split 32→16→8→4→2? Actually, 32 split twice → 16,16 then