(Javascript - Arrays) Find the leftmost and rightmost connected characters

Question

(Javascript - Arrays) Find the leftmost and rightmost connected characters

Looking for the leftmost and topmost connected '1' character in a 2D matrix?

Find the most left & top connected '1' character
Find the most right & bottom connected '1' character

EDIT 2.0:

To start, provide the coordinates for the initial '1' character.

10001
00001
11111
01110   --> (x: 1, y: 3)

The function will iterate through adjacent columns to find connected '1' characters for x or y value calculation.

Your function initiates at a specific point (e.g. y: 2, x: 0)

var array = [
 '00000',
 '01111',       --> get position of the most top & right '1'-character (x: 4, y: 1)
 '11000',
 '00000'
]

This function retrieves the top-right end of '1' characters:

var array = [
 '00000',
 '01111',      
 '11000',
 '00000'
]

Array.prototype.get_top_right = function(x_pos, y_pos) {
    var matrix = this, y1= y_pos;
    for (var x1 = x_pos; x1 < this[0].length; x1++) {
      try {
        if (matrix[(y1-1)][x1] == '1') y1--;
        else if (matrix[y1][(x1+1)] != '1') break;
      } catch(e) {}
    }; return [x1,y1]
 }
 
 var result=array.get_top_right(0,2)
 
 console.log(result)

Want to find the most bottom-left connected '1' character now?

var array = [
  '00000',
  '01111',      
  '11000', --> get position of the most bottom & left '1'-character (x: 0, y: 2)
  '00000'
]

Check out the updated function below to achieve this:

Array.prototype.get_bottom_left = function(x_pos, y_pos) {
    var matrix = this, y2= y_pos;
    for (var x2 = x_pos; x2 > 0; x2--) {
      try {
        if (matrix[(y2+1)][x2] == '1') y2++;
        if (matrix[y2][(x2-1)] != '1') break;
      } catch(e) {}
    }; return [x2,y2]
 }

Don't forget to test your updated function with the error_array provided below.

var error_array = [
    '000000',
    '000011',
    '111110',
    '111111'
]

If you encounter any issues, seek help from the community. Good luck!

Best regards,
- hans

javascript arrays algorithm

Answer 1

Answer №1

Two variations of the get_bottom_left function were created:

bottom_left_up which moves from the point (x, y) to the left and up
bottom_left_down that moves from the point (x, y) to the right and down.

Implementation details are as follows:

Array.prototype.bottom_left_up = function(x, y) {
  if(this[y][x] === '0') {
    return;
  }
  while(y >= 0) {
    while(--x >= 0) {
      if(this[y][x] === '0') {
        return [x + 1, y];
      }
    }
    if(--y === -1 || this[y][this[y].length - 1] === '0') {
      return [0, y + 1];
    }
    x = this[y].length;
  }
};

Array.prototype.bottom_left_down = function(x, y) {
  if(this[y][x] === '0') {
    return;
  }
  while(y < this.length) {
    if(this[y].indexOf('0', x) !== -1) {
      return [x, y];
    }
    if(++y === this.length || this[y][0] === '0') {
      return [x, y - 1];
    }
    x = 0;
  }
};

No out-of-range protections are present in the code, but they can be easily added without any issues. Let's validate the logic with some tests:

var array = [
 '00000',
 '01111',
 '11000',
 '00000'
];
console.log(array.bottom_left_up(2, 1)); // [1, 1]
console.log(array.bottom_left_down(2, 1)); // [0, 2]

var array2 = [
  '000000',
  '000011',
  '111110',
  '111111'
];
console.log(array2.bottom_left_up(3, 3)); // [0, 3]
console.log(array2.bottom_left_down(3, 3)); // [3, 3]

When it comes to method protection, using try-catch is not recommended. Instead, consider a solution like the following:

function(x, y) {
  x = parseInt(x, 10);
  y = parseInt(y, 10);
  if(!this.length || isNaN(x) || isNaN(y) || x < 0 || y < 0 || x >= this.length || y >= this[0].length) {
    return;
  }
  // ...
}

This way, 'undefined' will be returned in cases of an empty array, incorrect parameters, or no match found.

Answer 2

Two variations of the get_bottom_left function were created:

bottom_left_up which moves from the point (x, y) to the left and up
bottom_left_down that moves from the point (x, y) to the right and down.

Implementation details are as follows:

Array.prototype.bottom_left_up = function(x, y) {
  if(this[y][x] === '0') {
    return;
  }
  while(y >= 0) {
    while(--x >= 0) {
      if(this[y][x] === '0') {
        return [x + 1, y];
      }
    }
    if(--y === -1 || this[y][this[y].length - 1] === '0') {
      return [0, y + 1];
    }
    x = this[y].length;
  }
};

Array.prototype.bottom_left_down = function(x, y) {
  if(this[y][x] === '0') {
    return;
  }
  while(y < this.length) {
    if(this[y].indexOf('0', x) !== -1) {
      return [x, y];
    }
    if(++y === this.length || this[y][0] === '0') {
      return [x, y - 1];
    }
    x = 0;
  }
};

No out-of-range protections are present in the code, but they can be easily added without any issues. Let's validate the logic with some tests:

var array = [
 '00000',
 '01111',
 '11000',
 '00000'
];
console.log(array.bottom_left_up(2, 1)); // [1, 1]
console.log(array.bottom_left_down(2, 1)); // [0, 2]

var array2 = [
  '000000',
  '000011',
  '111110',
  '111111'
];
console.log(array2.bottom_left_up(3, 3)); // [0, 3]
console.log(array2.bottom_left_down(3, 3)); // [3, 3]

When it comes to method protection, using try-catch is not recommended. Instead, consider a solution like the following:

function(x, y) {
  x = parseInt(x, 10);
  y = parseInt(y, 10);
  if(!this.length || isNaN(x) || isNaN(y) || x < 0 || y < 0 || x >= this.length || y >= this[0].length) {
    return;
  }
  // ...
}

This way, 'undefined' will be returned in cases of an empty array, incorrect parameters, or no match found.

Answer 3

Answer №2

Check out this function that can handle matrices of any size, although I'm unsure if you always want to locate the first occurrence of "1" in the matrix, even if it's isolated on a line and to the right side.

    var array = [
    '00000',
    '01111',       
    '11000',
    '00100'];

var error_array = [
    '000000',
    '000011',
    '111111',
    '111111',
    '000000',
    '000010',
    '100000'
    ] 


getLeftBottom(error_array);

function getLeftBottom(testArray)
{
    var firstFound;
    for (var index = testArray.length; index > 0; index--)
    {
        console.log(testArray[index-1]);
        str = testArray[index-1];
        firstFound = str.indexOf("1");
        if (firstFound !== -1)
        {
            console.log("Coordinates: x = {0}, y = {1}", firstFound, index)   // This part behaves differently than expected in JavaScript, but still useful for locating the coordinates
            break;
        }

    }
}

Answer 4

Check out this function that can handle matrices of any size, although I'm unsure if you always want to locate the first occurrence of "1" in the matrix, even if it's isolated on a line and to the right side.

    var array = [
    '00000',
    '01111',       
    '11000',
    '00100'];

var error_array = [
    '000000',
    '000011',
    '111111',
    '111111',
    '000000',
    '000010',
    '100000'
    ] 


getLeftBottom(error_array);

function getLeftBottom(testArray)
{
    var firstFound;
    for (var index = testArray.length; index > 0; index--)
    {
        console.log(testArray[index-1]);
        str = testArray[index-1];
        firstFound = str.indexOf("1");
        if (firstFound !== -1)
        {
            console.log("Coordinates: x = {0}, y = {1}", firstFound, index)   // This part behaves differently than expected in JavaScript, but still useful for locating the coordinates
            break;
        }

    }
}

(Javascript - Arrays) Find the leftmost and rightmost connected characters

Answer №1

Answer №2

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