FastPrep
Problem · Dynamic Programming

Matrix Traversal

● EasyFlexportFULLTIMEOA

Given a 4 x 4 matrix mat, the initial energy is 100. The task is to reach the last row of the matrix with the maximum possible energy left.

The matrix can be traversed in the following way:

  • Start with any cell in the first row.
  • In each move, traverse from cell (i, j) of the ith row and jth column to any existing cell out of (i + 1, j - 1), (i + 1, j), or (i + 1, j + 1).
  • Finish the traversal in the last row.

    • After stepping on a cell (i, j), energy decreases by mat[i][j] units. Find the maximum possible energy left at the end of the traversal.

    Note: The final energy can be negative.

    Function Description

    Complete the function maxEnergy in the editor below.

    maxEnergy has the following parameter:

    • int mat[4][4]: a matrix of integers

    Returns

    int: the maximum possible energy at the end of the traversal

    Examples
    01 · Example 1
    mat = [[10, 20, 30, 40], [60, 50, 20, 80], [10, 10, 10, 10], [60, 50, 60, 50]]
return = 0
    Possible paths (0-based indexing is used):
    • (0, 0) - (1, 1) - (2, 2) - (3, 3)
    • (0, 1) - (1, 2) - (2, 2) - (3, 2)
    For the first path, energy left = 100 - 10 - 50 - 10 - 50 = -20 For the second path, energy left = 100 - 20 - 20 - 10 - 50 = 0 It can be proven that 0 is the maximum energy possible at the end of the traversal so return 0.
    Constraints
    0 <= mat[i][j] < 100
    More Flexport problems
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    public int maxEnergy(int[][] mat) {
  // write your code here
}
    mat[[10, 20, 30, 40], [60, 50, 20, 80], [10, 10, 10, 10], [60, 50, 60, 50]]
    expected0
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