How do you solve Inverse Coin Change on LeetCode?

Inverse Coin Change (LeetCode #3592) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The smallest denomination must have exactly one way to form it.

What is the time complexity of Inverse Coin Change?

The optimal solution for Inverse Coin Change runs in O(n) time and uses O(n) space. This complexity is efficient as we only traverse the numWays array once to find valid denominations.

What is the brute force solution for Inverse Coin Change?

The brute force approach for Inverse Coin Change is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check all possible coin denominations and see if they can generate the numWays array. This involves checking combinations of denominations and their contributions to each amount. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Inverse Coin Change?

Inverse Coin Change uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Inverse Coin Change?

Clarify the problem requirements before diving into coding. Think about edge cases, such as when numWays has only zeros. Explain your thought process clearly as you work through the solution.

What are common mistakes when solving Inverse Coin Change?

Assuming all indices with non-zero values are valid denominations. Overlooking the requirement for unique denominations.

#3592

Inverse Coin Change

Medium

ArrayDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can identify the smallest denomination by finding the first index where numWays[i] == 1. This denomination must be included, and we can derive others from it.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty list for denominations.
2Step 2: Find the smallest index c where numWays[c] == 1 and add c to denominations.
3Step 3: For each subsequent amount, check if numWays[i] can be formed using previously found denominations.

solution.py6 lines

1def recoverDenominations(numWays):
2    denominations = []
3    for i in range(1, len(numWays)):
4        if numWays[i] == 1:
5            denominations.append(i)
6    return sorted(denominations)

ℹ

Complexity note: This complexity is efficient as we only traverse the numWays array once to find valid denominations.

1The smallest denomination must have exactly one way to form it.
2Denominations can be derived from the structure of the numWays array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.