Minimum Number of Squares

Logical Breakdown

• [ ] Subproblem Identification:
• [ ] Optimal Substructure:
• [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
• [ ] Optimization: (Matrix Exponentiation if 'A' is huge)

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Mathematical Rigor

State Definition

Let \(dp[i]\) be the state for the \(i\)-th subproblem.

Recurrence Relation

\[ dp[i] = ... \]

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Visualization

graph TD A["Current State"] --> B["Previous State 1"] A --> C["Previous State 2"]

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Complexity Analysis

Approach	Time Complexity	Space Complexity
Iterative DP	\(O(N)\)	\(O(1)\)

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Code Reference

package com.dsa.dynamic_programming;

import java.util.*;

/ * Problem: Minimum Number of Squares * Group: 11. Dynamic Programming */ / * ============================================================================================ * PROBLEM: Minimum Number of Squares * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given an integer A. Return the minimum number of perfect squares that sum to A. * Example: A = 12 -> 4 + 4 + 4 (Count = 3). * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * Let dp[i] be the minimum squares needed to sum to 'i'. * * To calculate dp[i], we can try subtracting every possible square \(x^2\) (where \(x^2 \le i\)). * If we pick \(x^2\), the remaining sum is \(i - x^2\). * The cost becomes: 1 (for the current square) + dp[i - x^2]. * * Recurrence Relation: * dp[i] = 1 + min(dp[i - x^2]) for all x where xx <= i * * Base Case: * dp[0] = 0 * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= A <= 100,000 * * Time Complexity: * O(A * sqrt(A)) * - Outer loop runs A times. * - Inner loop runs sqrt(A) times. * * Space Complexity: * O(A) - DP array. * * --------------------- * 4. Examples * --------------------- * Input: 12 * Options: * - 12 - 1^2 = 11 -> 1 + dp[11] * - 12 - 2^2 = 8 -> 1 + dp[8] * - 12 - 3^2 = 3 -> 1 + dp[3] * Result: 3 (4+4+4) * * ============================================================================================ / public class MinimumNumberOfSquares { public int solve(int A) { // TODO: Implement solution return 0; }

public int countMinSquares(int i) {
    return 0;
}

}