Question

1074. Number of Submatrices That Sum to Target

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Description

Given a matrix and a target, return the number of non-empty submatrices that sum to target.

A submatrix x1, y1, x2, y2 is the set of all cells matrix[x][y] with x1 <= x <= x2 and y1 <= y <= y2.

Two submatrices (x1, y1, x2, y2) and (x1', y1', x2', y2') are different if they have some coordinate that is different: for example, if x1 != x1'.

 

Example 1:

Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
Output: 4
Explanation: The four 1x1 submatrices that only contain 0.

Example 2:

Input: matrix = [[1,-1],[-1,1]], target = 0
Output: 5
Explanation: The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.

Example 3:

Input: matrix = [[904]], target = 0
Output: 0

 

Constraints:

• 1 <= matrix.length <= 100
• 1 <= matrix[0].length <= 100
• -1000 <= matrix[i][j] <= 1000
• -10^8 <= target <= 10^8

Solutions

Solution 1

class Solution:
  def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
    def f(nums: List[int]) -> int:
      d = defaultdict(int)
      d[0] = 1
      cnt = s = 0
      for x in nums:
        s += x
        cnt += d[s - target]
        d[s] += 1
      return cnt

    m, n = len(matrix), len(matrix[0])
    ans = 0
    for i in range(m):
      col = [0] * n
      for j in range(i, m):
        for k in range(n):
          col[k] += matrix[j][k]
        ans += f(col)
    return ans

class Solution {
  public int numSubmatrixSumTarget(int[][] matrix, int target) {
    int m = matrix.length, n = matrix[0].length;
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      int[] col = new int[n];
      for (int j = i; j < m; ++j) {
        for (int k = 0; k < n; ++k) {
          col[k] += matrix[j][k];
        }
        ans += f(col, target);
      }
    }
    return ans;
  }

  private int f(int[] nums, int target) {
    Map<Integer, Integer> d = new HashMap<>();
    d.put(0, 1);
    int s = 0, cnt = 0;
    for (int x : nums) {
      s += x;
      cnt += d.getOrDefault(s - target, 0);
      d.merge(s, 1, Integer::sum);
    }
    return cnt;
  }
}

class Solution {
public:
  int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
    int m = matrix.size(), n = matrix[0].size();
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      vector<int> col(n);
      for (int j = i; j < m; ++j) {
        for (int k = 0; k < n; ++k) {
          col[k] += matrix[j][k];
        }
        ans += f(col, target);
      }
    }
    return ans;
  }

  int f(vector<int>& nums, int target) {
    unordered_map<int, int> d{{0, 1}};
    int cnt = 0, s = 0;
    for (int& x : nums) {
      s += x;
      if (d.count(s - target)) {
        cnt += d[s - target];
      }
      ++d[s];
    }
    return cnt;
  }
};

func numSubmatrixSumTarget(matrix [][]int, target int) (ans int) {
  m, n := len(matrix), len(matrix[0])
  for i := 0; i < m; i++ {
    col := make([]int, n)
    for j := i; j < m; j++ {
      for k := 0; k < n; k++ {
        col[k] += matrix[j][k]
      }
      ans += f(col, target)
    }
  }
  return
}

func f(nums []int, target int) (cnt int) {
  d := map[int]int{0: 1}
  s := 0
  for _, x := range nums {
    s += x
    if v, ok := d[s-target]; ok {
      cnt += v
    }
    d[s]++
  }
  return
}

function numSubmatrixSumTarget(matrix: number[][], target: number): number {
  const m = matrix.length;
  const n = matrix[0].length;
  let ans = 0;
  for (let i = 0; i < m; ++i) {
    const col: number[] = new Array(n).fill(0);
    for (let j = i; j < m; ++j) {
      for (let k = 0; k < n; ++k) {
        col[k] += matrix[j][k];
      }
      ans += f(col, target);
    }
  }
  return ans;
}

function f(nums: number[], target: number): number {
  const d: Map<number, number> = new Map();
  d.set(0, 1);
  let cnt = 0;
  let s = 0;
  for (const x of nums) {
    s += x;
    if (d.has(s - target)) {
      cnt += d.get(s - target)!;
    }
    d.set(s, (d.get(s) || 0) + 1);
  }
  return cnt;
}