Question

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

 279	Perfect Squares

 Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

 Example 1:

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

 Example 2:

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

 @tag-dp

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

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];
        }
    }
}

All Problems

All Solutions