Minimum Operations to Write the Letter Y on a Grid - Counting & Enumeration [JS]

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Description 

Solution: Counting & Enumeration

  1. Count the occurances of each value in the grid, inside the Y and outside the Y.
  2. Enumerate through all combinations of the values inside the Y and outside the Y.
    • For a combination of values (i, j),
    • calculate the number of operations to change all values inside the Y to i, and all values outside the Y to j.
    • Record and return the minimum operations out of all combinations.

How to check if (row, col) is inside the Y:

  • Separate the checks into three segments, the left part (\), the right part (/), and the middle part (|).
  • The left part (\): row <= floor(n / 2) and row === col
  • The right part (/): row <= floor(n / 2) and row + col === n - 1
  • The middle part (|): row >= floor(n / 2) and col === floor(n / 2)

Time Complexity: O(n^2)
Space Complexity: O(1)

var minimumOperationsToWriteY = function(grid) {
  let countY = Array(3).fill(0);
  let countOutside = Array(3).fill(0);
  let n = grid.length;
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < n; j++) {
      if (isPartOfY(i, j, n)) {
        countY[grid[i][j]]++;
      } else {
        countOutside[grid[i][j]]++;
      }
    }
  }
  let ans = Infinity;
  for (let i = 0; i < 3; i++) {
    for (let j = 0; j < 3; j++) {
      if (i === j) continue;
      let costY = countY[0] + countY[1] + countY[2] - countY[i];
      let costOutsideY = countOutside[0] + countOutside[1] + countOutside[2] - countOutside[j];
      ans = Math.min(ans, costY + costOutsideY);
    }
  }
  return ans;
};

function isPartOfY(row, col, n) {
  let isLeftPart = row <= Math.floor(n / 2) && row === col;
  let isRightPart = row <= Math.floor(n / 2) && row + col === n - 1;
  let isMiddlePart = row >= Math.floor(n / 2) && col === Math.floor(n / 2);
  return isLeftPart || isRightPart || isMiddlePart;
}
Location: Auckland, New Zealand

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