Problem Description

Given an integer n, you need to find the minimum number of perfect square numbers that add up to exactly n.

A perfect square is a number that can be expressed as the product of an integer with itself. For example:

• 1 = 1 × 1 is a perfect square
• 4 = 2 × 2 is a perfect square
• 9 = 3 × 3 is a perfect square
• 16 = 4 × 4 is a perfect square
• Numbers like 3 and 11 are not perfect squares

The task is to determine the least number of these perfect squares needed to sum to the given number n.

For example:

• If n = 12, one possible way is 12 = 4 + 4 + 4 (three perfect squares)
• Another way is 12 = 4 + 8, but 8 is not a perfect square
• The optimal answer for n = 12 is 3

The solution uses dynamic programming with a 2D table f[i][j] where:

• i represents considering perfect squares from 1² to i²
• j represents the target sum we're trying to achieve
• f[i][j] stores the minimum number of perfect squares needed to sum to j using squares up to i²

The algorithm iterates through all possible perfect squares up to √n and for each target value, decides whether to include the current perfect square or not, choosing the option that gives the minimum count.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: We can model this problem as a graph where each number from 0 to n is a node, and there's an edge from number a to number b if b = a + (perfect square). We're looking for the shortest path from 0 to n.

Is it a tree?

• No: The graph is not a tree since multiple paths can lead to the same number (e.g., 12 can be reached as 0→4→8→12 or 0→9→12).

Is the problem related to directed acyclic graphs?

• No: While the graph is directed (we only add perfect squares, never subtract), the problem is fundamentally about finding the shortest path.

Is the problem related to shortest paths?

• Yes: We need to find the minimum number of perfect squares (minimum number of edges/steps) to reach n from 0.

Is the graph Weighted?

• No: Each edge has the same weight of 1 (adding one perfect square counts as one step).

BFS

• Yes: Since we have an unweighted graph and need the shortest path, BFS is the appropriate algorithm.

Conclusion: The flowchart suggests using BFS (Breadth-First Search) for finding the minimum number of perfect squares that sum to n. In the BFS approach, we would start from n and subtract all possible perfect squares, treating each valid subtraction as moving to a new node, until we reach 0. The level at which we reach 0 gives us the minimum number of perfect squares needed.

Intuition

The key insight is recognizing that this problem can be viewed as finding the shortest path in an implicit graph. Imagine starting at number n and trying to reach 0 by subtracting perfect squares at each step. Each subtraction represents moving to a new node in our graph.

For example, if n = 12:

• From 12, we can subtract 1 (1²), 4 (2²), or 9 (3²) to reach 11, 8, or 3
• From 8, we can subtract 1 or 4 to reach 7 or 4
• From 4, we can subtract 1 or 4 to reach 3 or 0
• When we reach 0, we've found a valid decomposition

Since each step (subtracting a perfect square) has equal weight, we want the path with the fewest steps. This naturally leads to BFS, which explores all possibilities level by level and guarantees finding the shortest path first.

However, the given solution uses dynamic programming instead of BFS. The DP approach builds up the answer systematically by considering: "If I can use perfect squares up to i², what's the minimum number needed to sum to j?"

The recurrence relation is:

• Either don't use the current perfect square i²: f[i][j] = f[i-1][j]
• Or use it: f[i][j] = f[i][j - i*i] + 1 (if j >= i*i)

We take the minimum of these two options. The beauty of this approach is that it considers all possible combinations systematically, similar to how the unbounded knapsack problem allows unlimited use of each item. Here, we can use each perfect square as many times as needed to reach our target sum.

Solution Approach

The solution implements a dynamic programming approach using a 2D table to solve this unbounded knapsack variant.

Data Structure Setup:

• Calculate m = int(sqrt(n)) to find the maximum perfect square we need to consider
• Create a 2D DP table f with dimensions (m+1) × (n+1) initialized with infinity
• f[i][j] represents the minimum number of perfect squares needed to sum to j using squares from 1² to i²
• Set f[0][0] = 0 as base case (0 squares needed to sum to 0)

Dynamic Programming Iteration: The algorithm iterates through each perfect square and each target sum:

for i in range(1, m + 1):       # Consider squares from 1² to i²
    for j in range(n + 1):       # For each target sum from 0 to n
        f[i][j] = f[i - 1][j]    # Option 1: Don't use i²
        if j >= i * i:           # If we can use i²
            f[i][j] = min(f[i][j], f[i][j - i * i] + 1)  # Option 2: Use i²

State Transition Logic: For each state f[i][j], we have two choices:

1. Don't use the current perfect square i²: Copy the result from f[i-1][j] (minimum squares needed using only 1² to (i-1)²)
2. Use the current perfect square i²: Take the result from f[i][j - i*i] and add 1 (we can reuse i² multiple times, hence we look at f[i][...] not f[i-1][...])

The key insight is that we check f[i][j - i*i] instead of f[i-1][j - i*i], allowing unlimited use of each perfect square (unbounded knapsack property).

Final Result: Return f[m][n], which contains the minimum number of perfect squares needed to sum to n using all available perfect squares from 1² to m².

Example Walkthrough

Let's walk through the solution with n = 12.

Step 1: Setup

• Calculate m = int(sqrt(12)) = 3, so we'll consider perfect squares: 1², 2², 3² (which are 1, 4, 9)
• Create DP table f[4][13] (dimensions are (m+1) × (n+1))
• Initialize all values to infinity, except f[0][0] = 0

Step 2: Build DP Table

Let me show the key iterations:

When i = 1 (considering square 1² = 1):

• For j = 0: f[1][0] = f[0][0] = 0 (no squares needed for sum 0)
• For j = 1:
  • Don't use 1²: f[1][1] = f[0][1] = ∞
  • Use 1²: f[1][1] = f[1][0] + 1 = 0 + 1 = 1
  • Take minimum: f[1][1] = 1
• For j = 2: f[1][2] = min(∞, f[1][1] + 1) = 2 (need two 1's)
• ...continuing this pattern...
• For j = 12: f[1][12] = 12 (need twelve 1's)

When i = 2 (considering squares 1² and 2² = 4):

• For j = 0 to 3: Same as row i = 1 (can't use 4 for these)
• For j = 4:
  • Don't use 4: f[2][4] = f[1][4] = 4 (four 1's)
  • Use 4: f[2][4] = f[2][0] + 1 = 0 + 1 = 1 (one 4)
  • Take minimum: f[2][4] = 1
• For j = 8:
  • Don't use 4: f[2][8] = f[1][8] = 8
  • Use 4: f[2][8] = f[2][4] + 1 = 1 + 1 = 2 (two 4's)
  • Take minimum: f[2][8] = 2
• For j = 12:
  • Don't use 4: f[2][12] = f[1][12] = 12
  • Use 4: f[2][12] = f[2][8] + 1 = 2 + 1 = 3 (three 4's)
  • Take minimum: f[2][12] = 3

When i = 3 (considering squares 1², 2², and 3² = 9):

• For j = 0 to 8: Same as row i = 2 (can't use 9 for these)
• For j = 9:
  • Don't use 9: f[3][9] = f[2][9] = 3 (e.g., 4+4+1)
  • Use 9: f[3][9] = f[3][0] + 1 = 0 + 1 = 1 (one 9)
  • Take minimum: f[3][9] = 1
• For j = 12:
  • Don't use 9: f[3][12] = f[2][12] = 3 (three 4's)
  • Use 9: f[3][12] = f[3][3] + 1 = 3 + 1 = 4 (9+1+1+1)
  • Take minimum: f[3][12] = 3

Step 3: Final Answer Return f[3][12] = 3

The minimum number of perfect squares that sum to 12 is 3, achieved by using 4 + 4 + 4.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * sqrt(n))

The algorithm uses dynamic programming with a 2D table. The outer loop runs from 1 to m where m = int(sqrt(n)), giving us O(sqrt(n)) iterations. The inner loop runs from 0 to n, giving us O(n) iterations. Since these loops are nested, the total time complexity is O(sqrt(n) * n) = O(n * sqrt(n)).

Space Complexity: O(n * sqrt(n))

The algorithm creates a 2D array f with dimensions (m + 1) × (n + 1) where m = int(sqrt(n)). This results in a space requirement of O((sqrt(n) + 1) * (n + 1)) = O(n * sqrt(n)) for storing the DP table. No additional significant space is used beyond this table, so the overall space complexity remains O(n * sqrt(n)).

Common Pitfalls

1. Memory Inefficiency with Large Values of n

The current solution uses O(m × n) space where m = √n, resulting in O(n√n) space complexity. For large values of n (e.g., n = 10,000), this creates a table with millions of entries, which is unnecessary since we only need the previous row to compute the current row.

Solution: Optimize to 1D DP array since we only need the current state:

def numSquares(self, n: int) -> int:
    from math import sqrt, inf
  
    # Use 1D DP array instead of 2D
    dp = [inf] * (n + 1)
    dp[0] = 0
  
    max_square_root = int(sqrt(n))
  
    for i in range(1, max_square_root + 1):
        square = i * i
        for j in range(square, n + 1):
            dp[j] = min(dp[j], dp[j - square] + 1)
  
    return dp[n]

2. Initialization Error with Base Cases

A common mistake is forgetting to properly initialize all base cases. The code only sets dp[0][0] = 0, but doesn't handle dp[i][0] for i > 0, relying on the iteration to propagate the value. If the loop logic changes, this could break.

Solution: Explicitly initialize all dp[i][0] = 0 values:

# Initialize all dp[i][0] = 0 (0 squares needed for sum 0)
for i in range(max_square_root + 1):
    dp[i][0] = 0

3. Integer Overflow in Square Calculation

Computing i * i repeatedly in the inner loop is inefficient and could theoretically cause issues with very large numbers (though Python handles big integers well).

Solution: Pre-compute the square value:

for i in range(1, max_square_root + 1):
    square = i * i  # Compute once per outer loop
    for j in range(n + 1):
        dp[i][j] = dp[i - 1][j]
        if j >= square:
            dp[i][j] = min(dp[i][j], dp[i][j - square] + 1)

4. Missing Edge Case: n = 0

While the code handles n = 0 correctly, it's worth noting that some implementations might not account for this edge case where the answer should be 0.

Solution: Add explicit edge case handling if needed:

if n == 0:
    return 0

5. Suboptimal Algorithm Choice

While the 2D DP approach works, this problem can be solved more efficiently using BFS or mathematical approaches (Lagrange's four-square theorem). The DP solution has O(n√n) time complexity, while BFS can potentially find the answer faster for certain inputs.

Alternative BFS Solution:

def numSquares(self, n: int) -> int:
    from collections import deque
  
    squares = []
    i = 1
    while i * i <= n:
        squares.append(i * i)
        i += 1
  
    queue = deque([n])
    level = 0
    visited = {n}
  
    while queue:
        level += 1
        for _ in range(len(queue)):
            remainder = queue.popleft()
            for square in squares:
                if remainder == square:
                    return level
                if remainder < square:
                    break
                next_val = remainder - square
                if next_val not in visited:
                    visited.add(next_val)
                    queue.append(next_val)
  
    return level