Welcome to Subscribe On Youtube

Question

Formatted question description: https://leetcode.ca/all/279.html

Given an integer n, return the least number of perfect square numbers that sum to n.

A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1, 4, 9, and 16 are perfect squares while 3 and 11 are not.

 

Example 1:

Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.

Example 2:

Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.

 

Constraints:

• 1 <= n <= 104

Algorithm

Create a one-dimensional dp array of length n+1, initialize the first value to 0, and initialize the remaining values to INT_MAX

i loops from 0 to n, j loops from 1 to the position of i+jj <= n, and then updates the value of dp[i+jj] each time, dynamically updating the dp array, where dp[i] means positive The integer i can be composed of multiple perfect squares.

Note that the wording here, i must start from 0, j must start from 1. Because our original intention is to use dp[i] to update dp[i + j * j], if i=0, j=1 , Then dp[i] and dp[i + j * j] are equal.

Code

• Java
• C++
• Python
• Go
• TypeScript

• import java.util.Arrays;
  
  public class Perfect_Squares {
  
      public static void main(String[] args) {
          Perfect_Squares out = new Perfect_Squares();
          Solution s = out.new Solution();
  
          System.out.println(s.numSquares(43));
      }
  
      class Solution_lessLoop {
          public int numSquares(int n) {
  
              if (n <= 0) {
                  return 0;
              }
  
              // count[i] means, for number i, minimum square count
              int[] dp = new int[n + 1];
              Arrays.fill(dp, Integer.MAX_VALUE);
  
              dp[0] = 0;
              for (int i = 0; i <= n; ++i) {
                  for (int j = 1; i + j * j <= n; ++j) {
                      dp[i + j * j] = Math.min(dp[i + j * j], dp[i] + 1);
                  }
              }
  
              return dp[n];
          }
      }
  
      class Solution {
          public int numSquares(int n) {
  
              if (n <= 0) {
                  return 0;
              }
  
              // count[i] means, for number i, minimum square count
              int[] count = new int[n + 1];
              Arrays.fill(count, Integer.MAX_VALUE);
              count[0] = 0;
  
              // fill in sqaures first
              int maxPossibleSqaureRoot = (int)Math.sqrt(n * 1.0);
              for (int i = 1; i <= maxPossibleSqaureRoot; i++) {
                  count[(int)Math.pow(i, 2)] = 1;
              }
  
              for (int i = 1; i <= n; i++) {
  
                  int start = 0;
                  int end = i / 2;
  
                  for (int j = start; j <= end; j++) {
                      count[i] = Math.min(count[i], count[j] + count[i - j]);
                  }
              }
  
              return count[n];
          }
      }
  }
  
  ############
  
  class Solution {
      public int numSquares(int n) {
          int[] dp = new int[n + 1];
          for (int i = 1; i <= n; ++i) {
              int mi = Integer.MAX_VALUE;
              for (int j = 1; j * j <= i; ++j) {
                  mi = Math.min(mi, dp[i - j * j]);
              }
              dp[i] = mi + 1;
          }
          return dp[n];
      }
  }
  

• // OJ: https://leetcode.com/problems/perfect-squares/
  // Time: O(NS) where S is the count of square numbers less than n.
  // Space: O(N)
  class Solution {
  public:
      int numSquares(int n) {
          vector<int> v(n + 1, INT_MAX);
          for (int i = 1; i * i <= n; ++i) v[i * i] = 1;
          for (int i = 1; i <= n; ++i) {
              if (v[i] == 1) continue;
              for (int j = 1; j * j < i; ++j) {
                  v[i] = min(v[i], 1 + v[i - j * j]);
              }
          }
          return v[n];
      }
  };
  

• class Solution:
      def numSquares(self, n: int) -> int:
          dp = [0] * (n + 1)
          for i in range(1, n + 1):
              j, mi = 1, inf
              while j * j <= i:
                  mi = min(mi, dp[i - j * j])
                  j += 1
              dp[i] = mi + 1
          return dp[-1]
  
  ############
  
  class Solution(object):
    def numSquares(self, n):
      """
      :type n: int
      :rtype: int
      """
      squares = []
      j = 1
      while j * j <= n:
        squares.append(j * j)
        j += 1
      level = 0
      queue = [n]
      visited = [False] * (n + 1)
      while queue:
        level += 1
        temp = []
        for q in queue:
          for factor in squares:
            if q - factor == 0:
              return level
            if q - factor < 0:
              break
            if visited[q - factor]:
              continue
            temp.append(q - factor)
            visited[q - factor] = True
        queue = temp
      return level
  
  

• func numSquares(n int) int {
  	dp := make([]int, n+1)
  	for i := 1; i <= n; i++ {
  		mi := 100000
  		for j := 1; j*j <= i; j++ {
  			mi = min(mi, dp[i-j*j])
  		}
  		dp[i] = mi + 1
  	}
  	return dp[n]
  }
  
  func min(a, b int) int {
  	if a < b {
  		return a
  	}
  	return b
  }
  

• function numSquares(n: number): number {
      let dp = new Array(n + 1).fill(0);
      for (let i = 1; i <= n; ++i) {
          let min = Infinity;
          for (let j = 1; j * j <= i; ++j) {
              min = Math.min(min, dp[i - j * j]);
          }
          dp[i] = min + 1;
      }
      return dp.pop();
  }
  
  

All Problems

All Solutions