LeetCode //C - 279. Perfect Squares

279. Perfect Squares

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 < = 1 0 4 1 <= n <= 10^4 1<=n<=104

From: LeetCode

Link: 279. Perfect Squares

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Solution:

Ideas：

1. Initialize an array dp with size n+1 and fill it with INT_MAX to represent infinity, since we are looking for the minimum value. This array will store the least number of perfect squares that sum to every number up to n.
2. Set dp[0] = 0 because there are 0 perfect squares that sum to 0.
3. Use nested loops to populate the dp array. The outer loop iterates through each number from 1 to n, and the inner loop iterates through each square number jj that could be used to form i. It updates dp[i] to the minimum between its current value and dp[i - jj] + 1.
4. After filling the dp array, dp[n] contains the least number of perfect squares that sum to n.

Code:

int numSquares(int n) {
    int dp[n+1];
    for(int i = 0; i <= n; i++) {
        dp[i] = INT_MAX;
    }
    dp[0] = 0;
    
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j*j <= i; j++) {
            if(dp[i - j*j] != INT_MAX) {
                dp[i] = dp[i] < dp[i - j*j] + 1 ? dp[i] : dp[i - j*j] + 1;
            }
        }
    }
    
    return dp[n];
}