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UniquePaths2.js
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UniquePaths2.js
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/*
* Unique Paths 2
*
* There is a robot on an `m x n` grid.
* The robot is initially located at the top-left corner
* The robot tries to move to the bottom-right corner.
* The robot can only move either down or right at any point in time.
*
* Given grid with obstacles
* An obstacle and space are marked as 1 or 0 respectively in grid.
* A path that the robot takes cannot include any square that is an obstacle.
* Return the number of possible unique paths that the robot can take to reach the bottom-right corner.
*
* More info: https://leetcode.com/problems/unique-paths-ii/
*/
/**
* @description Return 'rows x columns' grid with cells filled by 'filler'
* @param {Number} rows Number of rows in the grid
* @param {Number} columns Number of columns in the grid
* @param {String | Number | Boolean} filler The value to fill cells
* @returns {Array [][]}
*/
const generateMatrix = (rows, columns, filler = 0) => {
const matrix = []
for (let i = 0; i < rows; i++) {
const submatrix = []
for (let k = 0; k < columns; k++) {
submatrix[k] = filler
}
matrix[i] = submatrix
}
return matrix
}
/**
* @description Return number of unique paths
* @param {Array [][]} obstacles Obstacles grid
* @returns {Number}
*/
const uniquePaths2 = (obstacles) => {
if (!Array.isArray(obstacles)) {
throw new Error('Input data must be type of Array')
}
// Create grid for calculating number of unique ways
const rows = obstacles.length
const columns = obstacles[0].length
const grid = generateMatrix(rows, columns)
// Fill the outermost cell with 1 b/c it has
// the only way to reach neighbor
for (let i = 0; i < rows; i++) {
// If robot encounters an obstacle in these cells,
// he cannot continue moving in that direction
if (obstacles[i][0]) {
break
}
grid[i][0] = 1
}
for (let j = 0; j < columns; j++) {
if (obstacles[0][j]) {
break
}
grid[0][j] = 1
}
// Fill the rest of grid by dynamic programming
// using following recurrent formula:
// K[i][j] = K[i - 1][j] + K[i][j - 1]
for (let i = 1; i < rows; i++) {
for (let j = 1; j < columns; j++) {
grid[i][j] = obstacles[i][j] ? 0 : grid[i - 1][j] + grid[i][j - 1]
}
}
return grid[rows - 1][columns - 1]
}
export { uniquePaths2 }