Leetcode

Perfect Squares

  • Time:O(\log n)
  • Space:O(n\log n)

C++

class Solution {
 public:
  int numSquares(int n) {
    vector<int> dp(n + 1, n);  // 1^2 x n

    dp[0] = 0;  // no way
    dp[1] = 1;  // 1^2

    for (int i = 2; i <= n; ++i)
      for (int j = 1; j * j <= i; ++j)
        dp[i] = min(dp[i], dp[i - j * j] + 1);

    return dp[n];
  }
};

JAVA

class Solution {
  public int numSquares(int n) {
    int[] dp = new int[n + 1];
    Arrays.fill(dp, n); // 1^2 x n

    dp[0] = 0; // no way
    dp[1] = 1; // 1^2

    for (int i = 2; 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];
  }
}

Python

class Solution:
  def numSquares(self, n: int) -> int:
    dp = [n] * (n + 1)

    dp[0] = 0
    dp[1] = 1

    for i in range(2, n + 1):
      j = 1
      while j * j <= i:
        dp[i] = min(dp[i], dp[i - j * j] + 1)
        j += 1

    return dp[n]