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1183. Maximum Number of Ones πŸ”’

Description

Consider a matrix M with dimensions width * height, such that every cell has value 0 or 1, and any square sub-matrix of M of size sideLength * sideLength has at most maxOnes ones.

Return the maximum possible number of ones that the matrix M can have.

 

Example 1:

Input: width = 3, height = 3, sideLength = 2, maxOnes = 1
Output: 4
Explanation:
In a 3*3 matrix, no 2*2 sub-matrix can have more than 1 one.
The best solution that has 4 ones is:
[1,0,1]
[0,0,0]
[1,0,1]

Example 2:

Input: width = 3, height = 3, sideLength = 2, maxOnes = 2
Output: 6
Explanation:
[1,0,1]
[1,0,1]
[1,0,1]

 

Constraints:

  • 1 <= width, height <= 100
  • 1 <= sideLength <= width, height
  • 0 <= maxOnes <= sideLength * sideLength

Solutions

Solution 1: Count Equivalent Positions

For convenience, let's denote $x = sideLength$.

Consider a $x \times x$ square, we need to select at most $maxOnes$ points inside the square and set them to 1. Note that when the point at coordinate $(i, j)$ is selected, all points at coordinates $(i\pm k_1 \times x, j\pm k_2 \times x)$ can be equivalently set to 1. Therefore, we calculate the number of equivalent positions of the coordinate $(i, j)$ in the matrix, and select the top $maxOnes$ with the most quantities.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively.

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class Solution:
    def maximumNumberOfOnes(
        self, width: int, height: int, sideLength: int, maxOnes: int
    ) -> int:
        x = sideLength
        cnt = [0] * (x * x)
        for i in range(width):
            for j in range(height):
                k = (i % x) * x + (j % x)
                cnt[k] += 1
        cnt.sort(reverse=True)
        return sum(cnt[:maxOnes])

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class Solution {
    public int maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
        int x = sideLength;
        int[] cnt = new int[x * x];
        for (int i = 0; i < width; ++i) {
            for (int j = 0; j < height; ++j) {
                int k = (i % x) * x + (j % x);
                ++cnt[k];
            }
        }
        Arrays.sort(cnt);
        int ans = 0;
        for (int i = 0; i < maxOnes; ++i) {
            ans += cnt[cnt.length - i - 1];
        }
        return ans;
    }
}

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class Solution {
public:
    int maximumNumberOfOnes(int width, int height, int sideLength, int maxOnes) {
        int x = sideLength;
        vector<int> cnt(x * x);
        for (int i = 0; i < width; ++i) {
            for (int j = 0; j < height; ++j) {
                int k = (i % x) * x + (j % x);
                ++cnt[k];
            }
        }
        sort(cnt.rbegin(), cnt.rend());
        int ans = 0;
        for (int i = 0; i < maxOnes; ++i) {
            ans += cnt[i];
        }
        return ans;
    }
};

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func maximumNumberOfOnes(width int, height int, sideLength int, maxOnes int) int {
    x := sideLength
    cnt := make([]int, x*x)
    for i := 0; i < width; i++ {
        for j := 0; j < height; j++ {
            k := (i%x)*x + (j % x)
            cnt[k]++
        }
    }
    sort.Ints(cnt)
    ans := 0
    for i :=