# 1. Introduction

In this blog post, we explore the "Perfect Squares" problem, a classic challenge in dynamic programming and number theory. The problem asks to find the minimum number of perfect square numbers that sum up to a given integer. A perfect square is an integer that is the square of another integer (e.g., 1, 4, 9, 16). This problem is often encountered in mathematical computations and algorithmic puzzles.

## Problem

Given an integer *n*, the objective is to return the least number of perfect square numbers whose sum equals *n*. The task involves identifying a set of perfect square numbers that add up to *n* with the smallest possible count.

# 2. Solution Steps

1. Use dynamic programming to solve the problem.

2. Create an array *dp* with *n + 1* elements, where *dp[i]* represents the minimum number of perfect squares that sum to *i*.

3. Initialize *dp[0]* to 0, as the sum of 0 requires 0 numbers.

4. Iterate from 1 to *n*, and for each *i*, calculate *dp[i]*:

a. Initialize *dp[i]* to a maximum value.

b. Iterate through all perfect squares less than or equal to *i*.

c. Update *dp[i]* as the minimum of its current value and *dp[i - square] + 1*.

5. Return *dp[n]*, which represents the least number of perfect squares summing to *n*.

# 3. Code Program

```
public class PerfectSquares {
public static void main(String[] args) {
int n = 12;
System.out.println(numSquares(n)); // Test the function
}
// Function to find the least number of perfect squares summing to n
public static int numSquares(int n) {
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j * j <= i; j++) {
dp[i] = Math.min(dp[i], dp[i - j * j] + 1);
}
}
return dp[n];
}
}
```

### Output:

3

### Explanation:

1. *numSquares*: This function computes the least number of perfect squares that sum to *n*.

2. It initializes a dynamic programming array *dp*, with *dp[i]* indicating the minimum number of perfect squares needed to sum to *i*.

3. The array is filled with a maximum value initially, except for *dp[0]*, which is 0.

4. For each *i* from 1 to *n*, the function iterates through all possible perfect squares *j * j* that are less than or equal to *i*. It updates *dp[i]* to the minimum of its current value and *1 + dp[i - j * j]*.

5. The function returns *dp[n]*, which gives the minimum count of perfect squares required to sum to *n*.

6. This approach effectively finds the optimal solution by considering all combinations of perfect squares that sum to each number up to *n*.

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