#leetcode #hard 3548. Equal Sum Grid Partition II Дейлик литкода 26.03.2026 You are given an m x n matrix grid of positive integers. Your task is to determine if it is possible to make either one horizontal or one vertical cut on the grid such that: Each of the two resulting sections formed by the cut is non-empty. The sum of elements in both sections is equal, or can be made equal by discounting at most one single cell in total (from either section). If a cell is discounted, the rest of the section must remain connected. Return true if such a partition exists; otherwise, return false. Note: A section is connected if every cell in it can be reached from any other cell by moving up, down, left, or right through other cells in the section. Тут чет многовато. Прикладываю едиториал. А так в целом скип задача, так как никто не спрашивает. class Solution { public boolean canPartitionGrid(int[][] grid) { long total = 0; int m = grid.length; int n = grid[0].length; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { total += grid[i][j]; } } for (int k = 0; k < 4; k++) { Set<Long> exist = new HashSet<>(); exist.add(0L); long sum = 0; m = grid.length; n = grid[0].length; if (m < 2) { grid = rotation(grid); continue; } if (n == 1) { for (int i = 0; i < m - 1; i++) { sum += grid[i][0]; long tag = sum 2 - total; if (tag == 0 || tag == grid[0][0] || tag == grid[i][0]) { return true; } } grid = rotation(grid); continue; } for (int i = 0; i < m - 1; i++) { for (int j = 0; j < n; j++) { exist.add((long) grid[i][j]); sum += grid[i][j]; } long tag = sum 2 - total; if (i == 0) { if ( tag == 0 || tag == grid[0][0] || tag == grid[0][n - 1] ) { return true; } continue; } if (exist.contains(tag)) { return true; } } grid = rotation(grid); } return false; } public int[][] rotation(int[][] grid) { int m = grid.length; int n = grid[0].length; int[][] tmp = new int[n][m]; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { tmp[j][m - 1 - i] = grid[i][j]; } } return tmp; } }