What is the Greedy pattern?

The Greedy pattern is a systematic approach to solving a class of coding interview problems. By recognizing the Greedy structure, you can solve problems in optimal time complexity using proven techniques.

When should I use the Greedy pattern?

Use the Greedy pattern when the problem structure matches its recognition signals — typically when brute force would be O(n²) or worse, and the pattern reduces complexity to O(n log n) typically.

What is the time complexity of Greedy?

The Greedy pattern typically runs in O(n log n) typically time and uses O(1) space. The exact complexity depends on the specific sub-variant being applied.

Is Greedy important for coding interviews?

Yes — the Greedy pattern is frequently tested at FAANG companies and competitive engineering interviews. Mastering this pattern gives you a reliable framework to solve many problem variations quickly.

How do I practice the Greedy pattern?

Start by studying the pattern's mental model and recognition signals on DSAPatterns, then practice the curated LeetCode problems below in order from Easy to Hard. Aim for pattern recognition before memorizing solutions.

← back to patterns

pattern / MEDIUM

Greedy.

Make locally optimal choices for global optimum

complexity

time

O(n log n) typically

space

O(1)

practice questions

browse and solve all 7 pattern questions →

02' — visual

interactive

Sliding window — k = 3

Maximum sum subarray

Window

[0, 2]

Sum

Step

0 / 3

03'

mental
model.

Sort by some criteria, then make the locally optimal choice at each step.

Greedy algorithms build solution piece by piece, always choosing the option that looks best at the moment.

when to use ✓

01Activity selection
02Fractional knapsack
03Huffman coding
04Job sequencing
05Minimum coins (canonical systems)

when not ✗

01When local choice doesn't lead to global optimum
02DP is needed

04'

recognition signals

Optimization problem

Locally optimal = globally optimal

Sorting + iteration

Activity/interval selection

Interval schedulingSorting + greedyHeap-based greedyTwo-pointer greedy

05'

templates

python

1# Beginner: Fixed-size sliding window
2def max_sum_k(nums, k):
3   n = len(nums)
4   if n < k:
5      return 0
6   window_sum = sum(nums[:k])
7   max_sum = window_sum
8   for i in range(k, n):
9      window_sum += nums[i] - nums[i - k]
10      max_sum = max(max_sum, window_sum)
11   return max_sum

common mistakes

01Applying when not applicable
02Wrong sorting criteria
03Missing edge cases

interview tips

01Prove correctness with exchange argument
02Sort by the right key
03Try examples to verify

06'

practice this pattern

7 curated problems from LeetCode

See all

Jump Game II# · Greedy

MEDIUM

Minimum Add to Make Parentheses Valid# · Greedy

MEDIUM

Gas Station# · Greedy

MEDIUM

Task Scheduler# · Greedy

MEDIUM

Minimum Cost to Hire K Workers# · Greedy

HARD

Candy# · Greedy

HARD

Minimum Number of Refueling Stops# · Greedy

HARD

07'

faq

← back to patterns

pattern / MEDIUM

Greedy.

Make locally optimal choices for global optimum

complexity

time

O(n log n) typically

space

O(1)

practice questions

browse and solve all 7 pattern questions →

02' — visual

interactive

Sliding window — k = 3

Maximum sum subarray

Window

[0, 2]

Sum

Step

0 / 3

03'

mental
model.

Sort by some criteria, then make the locally optimal choice at each step.

Greedy algorithms build solution piece by piece, always choosing the option that looks best at the moment.

when to use ✓

01Activity selection
02Fractional knapsack
03Huffman coding
04Job sequencing
05Minimum coins (canonical systems)

when not ✗

01When local choice doesn't lead to global optimum
02DP is needed

04'

recognition signals

Optimization problem

Locally optimal = globally optimal

Sorting + iteration

Activity/interval selection

Interval schedulingSorting + greedyHeap-based greedyTwo-pointer greedy

05'

templates

python

1# Beginner: Fixed-size sliding window
2def max_sum_k(nums, k):
3   n = len(nums)
4   if n < k:
5      return 0
6   window_sum = sum(nums[:k])
7   max_sum = window_sum
8   for i in range(k, n):
9      window_sum += nums[i] - nums[i - k]
10      max_sum = max(max_sum, window_sum)
11   return max_sum