Number of Ways of Cutting a Pizza - DP & Prefix Sum [JS]

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Description 

Solution: DP - Recursion w/ Memoization & Prefix Sum

  • Populate appleCount using prefix sum, where appleCount[i][j] = the number of apples in the piece with top left corner (i, j) and bottom right corner (n - 1, m - 1).
    The formula appleCount[i][j] = appleCount[i][j + 1] + appleCount[i + 1][j] - appleCount[i + 1][j + 1] + curr means:

    • count = right count + bottom count - bottom right count.
    • we add the right count and bottom count because we want to take the full sum.
    • we subtract the bottom right count because in the right count and bottom count, the bottom right count is added on twice.

  • Memoize each dp(i, j, k), where dp(i, j, k) = the number of ways to cut piece with top left corner (i, j) and bottom right corner (n - 1, m - 1) with k cuts left.

    • i = the row number
    • j = the column number
    • k = the amount of cuts we have left

  • For each dp(i, j, k), count the number of ways after cutting at each row and column.

    • Make sure the piece we are cutting contains at least one apple.
      • We can check this by checking whether the appleCount[newRow][newColumn] > appleCount[row][column].
      • If appleCount[newRow][newColumn] is equal to appleCount[row][column], we know the piece we are taking has no apples.

m = number of rows, n = number of columns
Time Complexity: O(mnk * (m + n)) 146ms
Space Complexity: O(mnk) 44.5MB

var ways = function(pizza, k) {
  let m = pizza.length, n = pizza[0].length, mod = 10 ** 9 + 7; 
  let appleCount = Array(m + 1).fill(0).map(() => Array(n + 1).fill(0));
  let memo = Array(m).fill(0).map(() => Array(n).fill(0).map(() => Array(k + 1).fill(-1)));
  for (let i = m - 1; i >= 0; i--) {
    for (let j = n - 1; j >= 0; j--) {
      let curr = pizza[i][j] === 'A' ? 1 : 0;
      appleCount[i][j] = appleCount[i][j + 1] + appleCount[i + 1][j] - appleCount[i + 1][j + 1] + curr;
    }
  }
  return dp(0, 0, k);
  
  function dp(i, j, k) {
    if (k === 1) return appleCount[i][j] > 0 ? 1 : 0;
    if (appleCount[i][j] === 0) return 0;
    if (memo[i][j][k] !== -1) return memo[i][j][k];
    
    let ans = 0;
    for (let newRow = i; newRow < m - 1; newRow++) {
      if (appleCount[newRow + 1][j] === appleCount[i][j]) continue; // top piece has no apples 
      ans = (ans + dp(newRow + 1, j, k - 1)) % mod;
    }
    for (let newCol = j; newCol < n - 1; newCol++) {
      if (appleCount[i][newCol + 1] === appleCount[i][j]) continue; // left piece has no apples
      ans = (ans + dp(i, newCol + 1, k - 1)) % mod;
    }
    return memo[i][j][k] = ans;
  }
};

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Location: Auckland, New Zealand

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