What is the brute force approach and complexity for Perfect Squares?

For an amount left, try every square that fits, then solve the smaller amount with time complexity O(√N · 2^N) and space complexity O(N).

What is the optimal approach and complexity for Perfect Squares?

dp[i] = 1 + the best dp[i − k·k] over every square ≤ i; fill from 0 up to n with optimal time complexity O(N√N) and space complexity O(N).

How do you identify the pattern for Perfect Squares?

Look for key signals in the prompt: "the fewest perfect squares" indicates A minimum count, so dp holds numbers and combines them with min. Word Break combined yes-or-no answers with OR.; "add up to exactly `n`" indicates The pieces must sum to n. Track the amount left, not a position in a string.; "a square can be used more than once" indicates Reuse is allowed, so dp[i − k·k] is read as it is, with no "used" flags..

Perfect Squares

Medium·LeetCode #279 Pro OnlyPattern Transfer

LeetCode

Problem Statement

You are given a whole number n. A perfect square is a number like 1, 4, 9, 16 and so on, which is a whole number multiplied by itself.

Find the fewest perfect squares that add up to exactly n. A square can be used more than once.

Return that smallest count.

Input

A whole number n.

Output

The fewest perfect squares whose sum is n.

Example 1

Input: n = 12

Output: 3

12 = 4 + 4 + 4. No two squares make 12 (for example 9 + 3 and 4 + 8 do not work), so 3 is the least.

Example 2

Input: n = 13

Output: 2

13 = 4 + 9. 13 is not a square itself, so one square is not enough.

Constraints

• 1 ≤ n ≤ 10^4

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“the fewest perfect squares”

→ A minimum count, so dp holds numbers and combines them with min. Word Break combined yes-or-no answers with OR.

“add up to exactly `n`”

→ The pieces must sum to n. Track the amount left, not a position in a string.

“a square can be used more than once”

→ Reuse is allowed, so dp[i − k·k] is read as it is, with no "used" flags.

Approach Complexity Ladder

Brute Force — try every square, then solve the restO(√N · 2^N)O(N)

For an amount left, try every square that fits, then solve the smaller amount

Optimal — bottom-up table over 0 … nO(N√N)O(N)

dp[i] = 1 + the best dp[i − k·k] over every square ≤ i; fill from 0 up to n

Brute Force Intuition

For an amount left, ask how few squares make it. Try every square k·k that is at most left. Take it, and ask the same question for left − k·k. Add one for the square taken, and keep the smallest total. When left is 0, no squares are needed.

Recognition rule for this sub-pattern:

dp[i] = can the first i characters be split validly? Try every last piece.

Remember this · DP — 1D Linear

Trigger

"Number of ways", "maximum/minimum up to i", choices at each step that depend on the previous one or two.

Think

If I knew the answer for every smaller prefix, how would I get the answer for i?

Key question

What does dp[i] mean in one sentence?

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