1183. Maximum Number of Ones

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.

• 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;
}
}

• 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;
}
};

• 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])


• 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 := range cnt[:maxOnes] {
ans += cnt[len(cnt)-i-1]
}
return ans
}

• /**
* @param {number} width
* @param {number} height
* @param {number} sideLength
* @param {number} maxOnes
* @return {number}
*/
var maximumNumberOfOnes = function (width, height, sideLength, maxOnes) {
const x = sideLength;
const cnt = new Array(x * x).fill(0);
for (let i = 0; i < width; ++i) {
for (let j = 0; j < height; ++j) {
const k = (i % x) * x + (j % x);
++cnt[k];
}
}
cnt.sort((a, b) => b - a);
return cnt.slice(0, maxOnes).reduce((a, b) => a + b, 0);
};