Mastering LeetCode's Coin Change Problem: A Comprehensive Guide

Efficient Strategies and Solutions to Minimize the Number of Coins Needed

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6 min read

Mastering LeetCode's Coin Change Problem: A Comprehensive Guide

Introduction

The "Coin Change" problem is a classic algorithmic challenge that often appears in coding interviews and competitive programming. Solving this problem efficiently is crucial for aspiring software engineers as it tests one's understanding of dynamic programming, breadth-first search, and recursive memoization.

In this guide, we will explore different strategies to tackle the Coin Change problem, analyze their complexities, and provide Python implementations for each approach.

Problem Statement

The problem is defined as follows:

You are given an integer array coins representing coins of different denominations and an integer amountrepresenting a total amount of money.

Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Approach Overview

There are several approaches to solving the Coin Change problem:

  1. Dynamic Programming (DP)

  2. Breadth-First Search (BFS)

  3. Recursive Memoization

We will start with the Dynamic Programming approach, which is often the most intuitive and commonly used method for this type of problem.

Dynamic Programming Solution

Explanation

The dynamic programming approach involves creating a list dp where dp[i] represents the minimum number of coins needed to make up the amount i. We initialize dp[0] to 0 since zero coins are needed to make up the amount 0. For all other amounts, we initialize dp[i] to infinity (float('inf')) to signify that initially, the amount cannot be made up with the given coins.

We then iterate over each coin and for each coin, update the dp list for all amounts that can be reached with the current coin. The state transition is defined as dp[x] = min(dp[x], dp[x - coin] + 1).

Code Implementation

from typing import List

def coinChange(coins: List[int], amount: int) -> int:
    # Initialize a list for storing the minimum coins needed for each amount
    dp = [float('inf')] * (amount + 1)
    # Base case: 0 coins are needed to make the amount 0
    dp[0] = 0

    # Iterate over each coin in the coins list
    for coin in coins:
        # Update the dp list for each amount that can be reached with the current coin
        for x in range(coin, amount + 1):
            dp[x] = min(dp[x], dp[x - coin] + 1)

    # If dp[amount] is still inf, it means the amount cannot be made up by any combination of coins
    return dp[amount] if dp[amount] != float('inf') else -1

# Example usage:
# coins = [1, 2, 5]
# amount = 11
# print(coinChange(coins, amount))  # Output: 3 (11 = 5 + 5 + 1)

Edge Cases and Considerations

  • Amount is 0: The function should return 0 because no coins are needed to make up an amount of 0.

  • No possible solution: If it is impossible to form the amount with the given coins, the function should return -1.

  • Single coin types: The function should handle cases where only one type of coin is provided.

  • Large amount values: The function should be efficient enough to handle large values for the amount without excessive computation time.

Optimizations

Early Termination and Sorted Coins

Sorting the coins in descending order can sometimes help find the solution faster, especially if we encounter larger denominations first. Additionally, early termination can be implemented to stop the iterations once we find the minimum number of coins needed.

Code Implementation

from typing import List

def coinChange(coins: List[int], amount: int) -> int:
    coins.sort(reverse=True)
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0

    for coin in coins:
        for x in range(coin, amount + 1):
            dp[x] = min(dp[x], dp[x - coin] + 1)
            if dp[x] == dp[coin] + (x // coin):  # Early termination
                break

    return dp[amount] if dp[amount] != float('inf') else -1

# Example usage:
# coins = [1, 2, 5]
# amount = 11
# print(coinChange(coins, amount))  # Output: 3 (11 = 5 + 5 + 1)

Alternative Approaches

Breadth-First Search (BFS) Approach

The BFS approach is suitable for large search spaces and varied coin denominations. This method explores all possible combinations level by level, ensuring that the shortest path (i.e., the fewest number of coins) is found.

Code Implementation

from typing import List
from collections import deque

def coinChange(coins: List[int], amount: int) -> int:
    if amount == 0:
        return 0
    coins.sort(reverse=True)
    queue = deque([(0, 0)])  # (current amount, number of coins)
    visited = set()

    while queue:
        current_amount, num_coins = queue.popleft()

        for coin in coins:
            next_amount = current_amount + coin
            if next_amount == amount:
                return num_coins + 1
            if next_amount > amount or next_amount in visited:
                continue
            visited.add(next_amount)
            queue.append((next_amount, num_coins + 1))

    return -1

# Example usage:
# coins = [1, 2, 5]
# amount = 11
# print(coinChange(coins, amount))  # Output: 3 (11 = 5 + 5 + 1)

Recursive Memoization Approach

The top-down recursive approach with memoization leverages recursive function calls and stores previously computed results to avoid redundant calculations. This approach is often easier to understand and implement but can have different performance characteristics.

Code Implementation

from typing import List
from functools import lru_cache

def coinChange(coins: List[int], amount: int) -> int:
    @lru_cache(None)
    def dp(n):
        if n == 0:
            return 0
        if n < 0:
            return float('inf')
        min_coins = float('inf')
        for coin in coins:
            res = dp(n - coin)
            if res != float('inf'):
                min_coins = min(min_coins, res + 1)
        return min_coins

    result = dp(amount)
    return result if result != float('inf') else -1

# Example usage:
# coins = [1, 2, 5]
# amount = 11
# print(coinChange(coins, amount))  # Output: 3 (11 = 5 + 5 + 1)

Comparative Analysis

  • Dynamic Programming (DP): Time complexity O(n * m), space complexity O(n). Efficient for most cases but can be optimized further with early termination and sorting.

  • Breadth-First Search (BFS): Explores all possible combinations level by level, ensuring the shortest path. Can be more memory-intensive but useful for large and varied inputs.

  • Recursive Memoization: Uses recursive calls and memoization to avoid redundant calculations. Can be easier to understand and implement but may have higher overhead due to recursion.

Practical Applications

The Coin Change problem and its variations appear frequently in real-world scenarios, such as:

  • Financial applications: Calculating the minimum number of coins or bills needed for a given amount.

  • Resource allocation: Optimizing the allocation of resources with different denominations or units.

  • Inventory management: Minimizing the number of items needed to fulfill orders.

Conclusion

Solving the Coin Change problem efficiently requires a deep understanding of different algorithmic approaches. While dynamic programming is often the go-to method, exploring alternative approaches like BFS and recursive memoization can provide additional insights and optimization opportunities.

Practice solving similar problems on platforms like LeetCode to enhance your problem-solving skills and prepare for coding interviews.

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