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
-
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(); }