Welcome to the realm of algorithmic design paradigms! In this chapter, we'll explore powerful strategies for tackling complex problems by breaking them down into manageable pieces. Our first stop is the 'Greedy Algorithm' paradigm. Imagine you're presented with a series of choices, and at each step, you can make the choice that seems best right now, without looking too far ahead. This is the essence of a greedy algorithm: making a locally optimal choice with the hope that this will lead to a globally optimal solution.
Greedy algorithms are often simple to design and implement. They work by iteratively making the best possible choice at the current step without considering the future consequences of that choice. If the problem exhibits optimal substructure and a greedy choice property, a greedy algorithm can indeed yield the best overall solution. However, it's crucial to understand that greedy algorithms don't always work. For some problems, a locally optimal choice might trap you into a suboptimal global solution. The art of using greedy algorithms lies in identifying problems where this strategy is guaranteed to succeed.
Let's consider a classic example: the Coin Change Problem. Given a set of coin denominations and a target amount, what is the minimum number of coins needed to make that amount? A greedy approach would be to always pick the largest denomination coin that is less than or equal to the remaining amount, and repeat this process until the target amount is reached.
function greedyCoinChange(denominations, amount) {
denominations.sort((a, b) => b - a); // Sort denominations in descending order
let coins = [];
let remainingAmount = amount;
for (let denom of denominations) {
while (remainingAmount >= denom) {
coins.push(denom);
remainingAmount -= denom;
}
}
if (remainingAmount === 0) {
return coins;
} else {
return "Cannot make exact change";
}
}While the greedy approach works for standard US currency denominations (e.g., 1, 5, 10, 25), it fails for other sets. For instance, if denominations were {1, 3, 4} and the target amount was 6, the greedy algorithm would pick 4, then 1, then 1 (total 3 coins). However, the optimal solution is two 3-unit coins (total 2 coins).
graph TD;
Start(Start)
A{Select largest coin <= remaining amount}
B[Add coin to solution]
C{Is remaining amount 0?}
D[Return solution]
E[No more coins to consider]
Start --> A
A --> B
B --> C
C -- Yes --> D
C -- No --> A
A --> E
E --> C