Greedy trains you to think: “What is the most progress I can make right now without ruining the future?” — a skill that transfers well to DP state transitions, graph traversal heuristics, and resource allocation problems.

(Easy) 860. Lemonade Change, NeetCode 250

https://leetcode.com/problems/lemonade-change/

Last review: Aug 29, 2025 Next review: You got a little hint to solve it, maybe, practice again.

Greedy

• Time Complexity O(n)
• Space Complexity O(1)

func lemonadeChange(bills []int) bool {
	// give the change before earning the profit

	var note5 int
	var note10 int

	for _, bill := range bills {
		switch bill {
		case 5:
			note5++
		case 10:
			if note5 < 1 {
				return false
			}
			note5--
			note10++
		case 20:
			if note5 < 3 && (note5 < 1 || note10 < 1) {
				return false
			}
			if note5 >= 1 && note10 >= 1 {
				note5--
				note10--
			} else {
				note5 -= 3
			}
		}
	}

	return true
}

⭐️ (Easy) 605. Can Place Flowers

https://leetcode.com/problems/can-place-flowers/description/

Last review: Feb 13, 2026 Next review: .

DFS + DP

• Time Complexity O(n!)
  • At each position before it caches, it will try all further possibilities.
• Space Complexity O(n)

func canPlaceFlowers(flowerbed []int, n int) bool {
	if n == 0 {
		return true
	}

	length := len(flowerbed)
	noPathCache := make(map[int]bool)

	var dfs func(i, planted int) bool
	dfs = func(i, planted int) bool {
		for j := i; j < length; j++ {
			if flowerbed[j] == 1 ||
				(j > 0 && flowerbed[j-1] == 1) ||
				(j < length-1 && flowerbed[j+1] == 1) ||
				noPathCache[j] {
				continue
			}

			planted++
			flowerbed[j] = 1
			if planted == n {
				return true
			}
			if dfs(j+1, planted) {
				return true
			}
			planted--
			flowerbed[j] = 0
			noPathCache[j] = true
		}
		return false
	}

	return dfs(0, 0)
}

Greedy

3 contiguous empty pots can plant a flower.

When we see three consecutive empty plots (0 0 0), plant immediately in the middle.

[1,0,1]     -- X
[1,0,0,1]   -- X
[1,0,0,0,1] -- O

About edge cases: