1074. Number of Submatrices That Sum to Target LeetCode Solution

In this guide, you will get 1074. Number of Submatrices That Sum to Target LeetCode Solution with the best time and space complexity. The solution to Number of Submatrices That Sum to Target problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Number of Submatrices That Sum to Target solution in C++
4. Number of Submatrices That Sum to Target solution in Java
5. Number of Submatrices That Sum to Target solution in Python
6. Additional Resources

Problem Statement of Number of Submatrices That Sum to Target

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 1×1 submatrices that only contain 0.

Example 2:

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

Complexity Analysis

• Time Complexity: O(mn^2)
• Space Complexity: O(n)

1074. Number of Submatrices That Sum to Target LeetCode Solution in C++

class Solution {
 public:
  int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
    const int m = matrix.size();
    const int n = matrix[0].size();
    int ans = 0;

    // Transfer each row in the matrix to the prefix sum.
    for (auto& row : matrix)
      for (int i = 1; i < n; ++i)
        row[i] += row[i - 1];

    for (int baseCol = 0; baseCol < n; ++baseCol)
      for (int j = baseCol; j < n; ++j) {
        unordered_map<int, int> prefixCount{{0, 1}};
        int sum = 0;
        for (int i = 0; i < m; ++i) {
          if (baseCol > 0)
            sum -= matrix[i][baseCol - 1];
          sum += matrix[i][j];
          if (const auto it = prefixCount.find(sum - target);
              it != prefixCount.cend())
            ans += it->second;
          ++prefixCount[sum];
        }
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

1074. Number of Submatrices That Sum to Target LeetCode Solution in Java

class Solution {
  public int numSubmatrixSumTarget(int[][] matrix, int target) {
    final int m = matrix.length;
    final int n = matrix[0].length;
    int ans = 0;

    // Transfer each row in the matrix to the prefix sum.
    for (int[] row : matrix)
      for (int i = 1; i < n; ++i)
        row[i] += row[i - 1];

    for (int baseCol = 0; baseCol < n; ++baseCol)
      for (int j = baseCol; j < n; ++j) {
        Map<Integer, Integer> prefixCount = new HashMap<>();
        prefixCount.put(0, 1);
        int sum = 0;
        for (int i = 0; i < m; ++i) {
          if (baseCol > 0)
            sum -= matrix[i][baseCol - 1];
          sum += matrix[i][j];
          ans += prefixCount.getOrDefault(sum - target, 0);
          prefixCount.merge(sum, 1, Integer::sum);
        }
      }

    return ans;
  }
}
// code provided by PROGIEZ

1074. Number of Submatrices That Sum to Target LeetCode Solution in Python

class Solution:
  def numSubmatrixSumTarget(self, matrix: list[list[int]], target: int) -> int:
    m = len(matrix)
    n = len(matrix[0])
    ans = 0

    # Transfer each row in the matrix to the prefix sum.
    for row in matrix:
      for i in range(1, n):
        row[i] += row[i - 1]

    for baseCol in range(n):
      for j in range(baseCol, n):
        prefixCount = collections.Counter({0: 1})
        summ = 0
        for i in range(m):
          if baseCol > 0:
            summ -= matrix[i][baseCol - 1]
          summ += matrix[i][j]
          ans += prefixCount[summ - target]
          prefixCount[summ] += 1

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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