3592. Inverse Coin Change

Problem Link

Description

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You are given a 1-indexed integer array numWays, where numWays[i] represents the number of ways to select a total amount i using an infinite supply of some fixed coin denominations. Each denomination is a positive integer with value at most numWays.length.

However, the exact coin denominations have been lost. Your task is to recover the set of denominations that could have resulted in the given numWays array.

Return a sorted array containing unique integers which represents this set of denominations.

If no such set exists, return an empty array.

 

Example 1:

Input: numWays = [0,1,0,2,0,3,0,4,0,5]

Output: [2,4,6]

Explanation:

Amount	Number of ways	Explanation
1	0	There is no way to select coins with total value 1.
2	1	The only way is [2].
3	0	There is no way to select coins with total value 3.
4	2	The ways are [2, 2] and [4].
5	0	There is no way to select coins with total value 5.
6	3	The ways are [2, 2, 2], [2, 4], and [6].
7	0	There is no way to select coins with total value 7.
8	4	The ways are [2, 2, 2, 2], [2, 2, 4], [2, 6], and [4, 4].
9	0	There is no way to select coins with total value 9.
10	5	The ways are [2, 2, 2, 2, 2], [2, 2, 2, 4], [2, 4, 4], [2, 2, 6], and [4, 6].

Example 2:

Input: numWays = [1,2,2,3,4]

Output: [1,2,5]

Explanation:

Amount	Number of ways	Explanation
1	1	The only way is [1].
2	2	The ways are [1, 1] and [2].
3	2	The ways are [1, 1, 1] and [1, 2].
4	3	The ways are [1, 1, 1, 1], [1, 1, 2], and [2, 2].
5	4	The ways are [1, 1, 1, 1, 1], [1, 1, 1, 2], [1, 2, 2], and [5].

Example 3:

Input: numWays = [1,2,3,4,15]

Output: []

Explanation:

No set of denomination satisfies this array.

 

Constraints:

• 1 <= numWays.length <= 100
• 0 <= numWays[i] <= 2 * 108

Solution

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Python3

class Solution:
    def findCoins(self, numWays: List[int]) -> List[int]:
        N = len(numWays)
        coins = []
        dp = [0] * (N + 1)
        dp[0] = 1
 
        for i in range(1, N + 1):
            target = numWays[i - 1]
 
            if target - dp[i] == 1:
                coins.append(i)
                for j in range(i, N + 1):
                    dp[j] += dp[j - i]
 
            if target != dp[i]:
                return []
 
        return coins