数组和能否通过指定变换等于k？求优于指数时间的解法

Great question! Let's break this down and work through a far more efficient solution than the exponential-time approach you're currently using.

First, let's analyze what each element can be transformed into using the allowed operations:

• For the i-th element (with index i), the three operations let us:
  1. Flip its sign once (since flipping twice cancels out)
  2. Add or subtract i any number of times (since adding i multiple times gives A[i] + t*i for integer t, same for subtraction)

Combining these, each element has two possible base values (before adjusting with i multiples):

• A[i] (no sign flip) → can become A[i] + t*i for any integer t
• -A[i] (sign flipped once) → can become -A[i] + s*i for any integer s

Now, let's look at the total sum requirement. Let T be the sum of choosing either A[i] or -A[i] for each element. To reach total sum k, we need:

sum(c_i * i) = k - T

where c_i are integers (representing net add/subtract counts of i for each element). By Bezout's identity, the left-hand side can only be a multiple of the greatest common divisor (GCD) of all indices i (let's call this GCD G).

This simplifies our problem to:

Is there a way to choose ±A[i] for each element such that T ≡ k mod G?

If this is true, we can always adjust the sum to k using add/subtract i operations (since k-T is a multiple of G, and we can form any multiple of G via combinations of the indices).

If all indices are 0 (e.g., a single-element array with index 0):

• Adding/subtracting 0 does nothing, so the element can only be A[0] or -A[0]. We just check if k equals either value.

For arrays with non-zero indices:

1. Calculate G, the GCD of all indices.
2. Use dynamic programming to track all possible remainders modulo G that we can get by summing ±A[i] values.
3. Check if k mod G is in the set of achievable remainders.

Implementation Steps #

• Initialize a set dp with {0} (starting sum remainder is 0).
• For each element a in A:
  • For each remainder r in the current dp, compute the new remainders from adding a or -a (mod G).
  • Update dp to be the set of these new remainders.
• After processing all elements, check if k % G is in dp. If yes, return True; else, False.

Time Complexity #

This runs in O(n*G) time, where n is the array length. Since G is often small (e.g., G=1 for consecutive indices), this is drastically faster than exponential-time approaches.

Example Code (Python) #

import math
from functools import reduce

def compute_gcd(numbers):
    return reduce(math.gcd, numbers)

def can_reach_target_sum(A, k):
    n = len(A)
    indices = list(range(n))
    
    # Calculate GCD of all indices
    gcd_indices = compute_gcd(indices)
    
    if gcd_indices == 0:
        # All indices are 0; only A[0] or -A[0] are possible
        return k == A[0] or k == -A[0]
    
    target_remainder = k % gcd_indices
    achievable_remainders = {0}
    
    for num in A:
        next_remainders = set()
        num_mod = num % gcd_indices
        neg_num_mod = (-num) % gcd_indices
        
        for rem in achievable_remainders:
            next_remainders.add((rem + num_mod) % gcd_indices)
            next_remainders.add((rem + neg_num_mod) % gcd_indices)
        
        achievable_remainders = next_remainders
    
    return target_remainder in achievable_remainders

# Test cases
print(can_reach_target_sum([3], 5))       # True (index 1, GCD=1; adjust with +1 twice)
print(can_reach_target_sum([4], 7))       # False (index 2, GCD=2; 7 is odd, can't reach)
print(can_reach_target_sum([2, 3], 10))   # True (indices 1&2, GCD=1; any sum is possible)
print(can_reach_target_sum([5], 3))       # False (index 0; only 5/-5 are allowed)

内容的提问来源于stack exchange，提问作者Nilesh Kathait