[LintCode] 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

Given n = 12, return 3 because 12 = 4 + 4 + 4
Given n = 13, return 2 because 13 = 4 + 9

 

Recursion

base case f(0) = 0;

f(n) = min(f(n - i * i)) + 1,  for i * i <= n;

 public int numSquares(int n) {
    if(n == 0) {
        return 0;
    } 
    int minNum = Integer.MAX_VALUE;
    for(int i = 1; i * i <= n; i++) {
        minNum = Math.min(minNum, 1 + numSquares(n - i * i));
    }
    return minNum;
 }

 

Dynamic Programming Solution

State: dp[i]: the least number of perfect sqaure numbers that sum to i.

Function: dp[i] = 1 + min(dp[i - j * j], for all j that satisfies j * j <= i); 

Initialization: dp[i * i] = 1, for i * i <= n, the rest of dp[i] = Integer.MAX_VALUE.

Answer: dp[n]

public class Solution {
    /**
     * @param n a positive integer
     * @return an integer
     */
    public int numSquares(int n) {
        int[] dp = new int[n + 1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;
        for(int i = 1; i * i <= n; i++) {
            dp[i * i] = 1;
        }

        for (int i = 1; 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];
    }
}

 

 

Related Problems

Check Sum of Square Numbers

Ugly Number II