Leetcode #2952: Minimum Number of Coins to be Added
In this guide, we solve Leetcode #2952 Minimum Number of Coins to be Added in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed integer array coins, representing the values of the coins available, and an integer target. An integer x is obtainable if there exists a subsequence of coins that sums to x.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: coins = [1,4,10], target = 19
Output: 2
Explanation: We need to add coins 2 and 8. The resulting array will be [1,2,4,8,10].
It can be shown that all integers from 1 to 19 are obtainable from the resulting array, and that 2 is the minimum number of coins that need to be added to the array.
Python Solution
class Solution:
def minimumAddedCoins(self, coins: List[int], target: int) -> int:
coins.sort()
s = 1
ans = i = 0
while s <= target:
if i < len(coins) and coins[i] <= s:
s += coins[i]
i += 1
else:
s <<= 1
ans += 1
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.