Game-Hit-the-target

Description

given a matrix n x n , determine if the arrow is directed to the target (x). There will be only 1 arrow ‘>’ and 1 target ‘x’ An empty spot will be denoted by a space ” “, the target with a cross”x”, and the scope “>”

Examples:

given matrix 4x4:

[

  [' ', ' ', ' ', ' '],

  [' ', ' ', ' ', ' '], --> return true

  [' ', '>', ' ', 'x'],

  [' ', ' ', ' ', ' ']

] 

given matrix 4x4:

[

  [' ', ' ', ' ', ' '],

  [' ', '>', ' ', ' '], --> return false

  [' ', ' ', ' ', 'x'],

  [' ', ' ', ' ', ' ']

] 

given matrix 4x4:

[

  [' ', ' ', ' ', ' '],

  [' ', 'x', '>', ' '], --> return false

  [' ', '', ' ', ' '],

  [' ', ' ', ' ', ' ']

] 

In this example, only a 4x4 matrix was used, the problem will have matrices of dimensions from 2 to 7 Happy hacking as they say!

Solutions

Julia

using Test


"""
只有 `>` 和 `x` 在同一行且 `>` 在 `x` 前，才返回 true
"""

#> "只有 `>` 和 `x` 在同一行且 `>` 在 `x` 前，才返回 true\n"

function solution(A::Matrix{Char})::Bool
    arrow_indices = findfirst(==('>'), A) # 返回CartesianIndex{2}
    arrowᵢ, arrowⱼ = arrow_indices[1], arrow_indices[2]
    target_indices = findfirst(==('x'), A)
    targetᵢ, targetⱼ = target_indices[1], target_indices[2]

    return arrowᵢ == targetᵢ && arrowⱼ < targetⱼ
end

#> solution (generic function with 1 method)

A = [
    ' ' ' ' ' ' ' ' ' '
    ' ' ' ' ' ' ' ' ' '
    ' ' ' ' ' ' ' ' ' '
    ' ' ' ' '>' ' ' 'x'
    ' ' ' ' ' ' ' ' ' '
]

#> 5×5 Matrix{Char}:
#>  ' '  ' '  ' '  ' '  ' '
#>  ' '  ' '  ' '  ' '  ' '
#>  ' '  ' '  ' '  ' '  ' '
#>  ' '  ' '  '>'  ' '  'x'
#>  ' '  ' '  ' '  ' '  ' '

@test solution(A) == true

#> Test Passed
#>   Expression: solution(A) == true
#>    Evaluated: true == true

R

library(tidyverse)


solution <- function(M) {
    # 将M沿列展开为向量的一维坐标
    arrow_index <- match(">", M)
    target_index <- match("x", M)

    # 二维坐标中的行
    arrow_i <- arrow_index%%nrow(M)
    target_i <- target_index%%nrow(M)

    arrow_i == target_i && arrow_index < target_index
}


M <- c(rep(" ", 17), ">", " ", "x", rep(" ", 5)) %>%
    matrix(nrow = 5, ncol = 5, byrow = TRUE)

solution(M)

#> [1] TRUE

JavaScript

const solution = function (mtrx) {
  for (let row of mtrx) {
    let scope = row.indexOf('>');
    let target = row.indexOf('x');
    if (target > -1 && scope > -1 && scope < target) return true;
  }
  return false;
};


console.log(solution([
  [' ', ' ', ' ', ' ', ' '],
  [' ', ' ', ' ', ' ', ' '],
  [' ', ' ', ' ', ' ', ' '],
  [' ', ' ', '>', ' ', 'x'],
  [' ', ' ', '', ' ', ' ']
]));

#> true