I'm learning algorithms and doing JavaScript exercise questions, and I don't understand how one reaches the correct answer for a particular algorithm.
The question provided in the exercise is:
Have the function CorrectPath(str) read the str parameter being
passed, which will represent the movements made in a 5x5 grid of cells
starting from the top left position. The characters in the input
string will be entirely composed of: r, l, u, d, ?. Each of the
characters stand for the direction to take within the grid, for
example: r = right, l = left, u = up, d = down. Your goal is to
determine what characters the question marks should be in order for a
path to be created to go from the top left of the grid all the way to
the bottom right without touching previously travelled on cells in the
grid.
For example, the input drdr??rrddd? should ouptut drdruurrdddd
I've not found a solution on my own. I'm taking a look at a solution provided, and I'm bothered because:
A. pure functions are not used to manipulate values within the CorrectPath function (note the addX() and addY() methods contained within). I'm not convinced the solution provided is using best practices, especially coming from a functional programming background.
B. I don't understand how the steps taken, specifically in the while block and the succeeding for block, are taken to reach the correct answer and why sometimes the missingLetters array has letters remaining and other times not
The working solution provided is below
function CorrectPath(str) {
let x = 0, //start x coord
y = 0, //start y coord
missingLetters = []
const unknowns = str.match(/\W/g)
function addX() {
while(x !== 4) {
if (x > 4) {
x--;
missingLetters.push('l');
} else {
x++;
missingLetters.push('r');
}
}
}
function addY() {
while (y !== 4) {
if (y > 4) {
y--;
missingLetters.push('u');
} else {
y++;
missingLetters.push('d');
}
}
}
//tallies current number of x and y movements
for (let i=0; i<str.length; i++) {
switch (str[i]) {
case 'd':
y += 1;
break;
case 'u':
y -= 1;
break;
case 'l':
x -= 1;
break;
case 'r':
x += 1;
break;
}
}
if (x > y) { addX(); addY(); }
if (y >= x) { addY(); addX(); }
while (missingLetters.length < unknowns.length) {
var pos = missingLetters.length - 1;
if (missingLetters[pos] === 'r') {x += 1; missingLetters.push('r'); addX()}
if (missingLetters[pos] === 'l') {x -= 1; missingLetters.push('l'); addX()}
if (missingLetters[pos] === 'd') {y += 1; missingLetters.push('d'); addY()}
if (missingLetters[pos] === 'u') {y -= 1; missingLetters.push('u'); addY()}
}
var newStr = str.split('');
for (var j=0; j<str.length; j++) {
if (newStr[j] === '?') {
newStr[j] = missingLetters.shift()
}
}
return newStr.join('');
}
CorrectPath(readline());
Here's a solution I found
const dirMap = {
u: { x: 0, y: -1 },
r: { x: 1, y: 0 },
d: { x: 0, y: 1 },
l: { x: -1, y: 0 }
}
function CorrectPath(pathString) {
const map = Array(5*5)
return trace(pathString, map)
}
function trace(path, [...map], x = 0, y = 0, newPath = "") {
const steps = path.split(""),
nextMove = steps.shift()
if (nextMove === undefined) {
if (5 * y + x === (5*5-1)) return newPath
return "Bad move"
}
if (nextMove === "?") {
const moves = availableMoves(x,y,map)
if (!moves.length) return "Bad move"
for(let i = 0; i<moves.length; i++) {
let move = moves[i],
trySteps = [move,...steps].join("")
res = trace(trySteps,map,x,y,newPath)
if (!res || res === "Bad move") continue
else return res
}
return "Bad move"
} else {
if (!canMove(nextMove, x, y, map)) return "Bad move"
const pos = dirMap[nextMove],
newX = pos.x + x,
newY = pos.y + y
newPath += nextMove
map[5*newY+newX] = nextMove
return trace(steps.join(""),map,newX,newY,newPath)
}
}
function availableMoves(x,y,map) {
const steps = []
Object.keys(dirMap).forEach(z => {
if (canMove(z,x,y,map)) steps.push(z)
})
return steps
}
function canMove(dir, xPath, yPath, map) {
const pos = dirMap[dir],
x = pos.x + xPath,
y = pos.y + yPath
if (x > 4 || x < 0 || y > 4 || y < 0) return false
if (map[5*y+x] !== undefined) return false
return true
}
CorrectPath(readline());
What I'd like to do is take an SVG shape drawn by an enclosed path (in this case, a region of a map) and reduce the number of points to create a simpler shape.
I've tried implementing the Ramer-Douglas-Peucker algorithm to reduce the number of points. For example, here's a fiddle using the simplify.js library:
https://jsfiddle.net/0t3n8762/
After reading about the issue, if I understood it correctly, it seems the algorithm isn't really designed to work on enclosed shapes but open ended paths. I tried splitting each path into two (so there are two lines that together make the entire shape) and running the algorithm on each before recombining them, though the results seem essentially identical:
https://jsfiddle.net/caqwL3t7/
It may be (and indeed is quite likely) that I'm just not grasping how the algorithm is supposed to work and am implementing it incorrectly. Or perhaps that I should be trying a completely different method altogether.
coords = pathToCoords($("#JP-01").attr("d"));
for (var i = 0; i < coords.length; i++) {
newCoords[i] = simplify(coords[i], 2);
}
newPath = coordsToPath(newCoords);
$("#JP-01").attr("d", newPath);
What I would want to produce is a simpler path that still retains the overall shape of the original, drawn with fewer points. The actual result is a distorted shape that shares little in common with the original.
As Paul pointed out in the comments, you haven't considered about l and v command in a path.
I made a snippet to show how to achieve your goal but it won't work in all the cases (I guess) and it still needs to be improved.
Kindly find the snippet and comments below.
$("#simplify-1").on("click", function(){
var coords;
var newCoords = [];
var newPath;
var coordsObject;
// get the coordinates from a given path
function pathToCoords(path) {
// save each path individually (.i.e islands are stored separately)
var data = path.match(/(?<=m).*?(?=z)/igs);
var coordsArray = [];
var objectArray = [];
var objectArrayA = [];
var objectArrayB = [];
var objectContainer = [];
// split each pair of coordinates into their own arrays
for (var i = 0; i < data.length; i++) {
// should remove anything between h or v and l instead?
data[i] = data[i].split(/[LlHhVv]/);
coordsArray[i] = [];
for (var j = 0; j < data[i].length; j++) {
coordsArray[i].push(data[i][j].split(",").map(Number));
}
}
// convert each pair of coordinates into an object of x and y
for (var i = 0; i < coordsArray.length; i++) {
objectArray[i] = [];
for (var j = 0; j < coordsArray[i].length; j++) {
objectArray[i].push({
x: coordsArray[i][j][0],
y: coordsArray[i][j][1]
});
}
// split each array of coordinates in half
var halfway = Math.floor(objectArray[i].length / 2);
objectArrayB[i] = JSON.parse(JSON.stringify(objectArray[i]));;
objectArrayA[i] = objectArrayB[i].splice(0, halfway);
}
objectContainer = [objectArrayA, objectArrayB];
return objectContainer;
}
// convert the coordinates back into a string for the path
function coordsToPath(objectContainer) {
var objectArray = [];
var coordsArray = [];
var data;
// recombine the two objectArrays
for (var i = 0; i < objectContainer[0].length; i++) {
objectArray[i] = objectContainer[0][i].concat(objectContainer[1][i])
}
for (var i = 0; i < objectArray.length; i++) {
coordsArray[i] = [];
// take the X and Y values from the objectArray and strip the unwanted information
for (var j = 0; j < objectArray[i].length; j++) {
if (j == 0) {
// add 'M' in front of the first entry
coordsArray[i].push("M" + Object.values(objectArray[i][j]));
} else if (j == objectArray[i].length - 1) {
// add 'z' to the end of the last entry
coordsArray[i].push("l" + Object.values(objectArray[i][j]) + "z");
} else {
// add 'l' in front of each coordinate pair
coordsArray[i].push("l" + Object.values(objectArray[i][j]));
}
}
coordsArray[i] = coordsArray[i].toString();
}
// put everything back into a single valid SVG path string
data = coordsArray.join("");
return data;
}
// -- simplify.js -- //
/*
(c) 2017, Vladimir Agafonkin
Simplify.js, a high-performance JS polyline simplification library
mourner.github.io/simplify-js
*/
(function() {
'use strict';
// to suit your point format, run search/replace for '.x' and '.y';
// for 3D version, see 3d branch (configurability would draw significant performance overhead)
// square distance between 2 points
function getSqDist(p1, p2) {
var dx = p1.x - p2.x,
dy = p1.y - p2.y;
return dx * dx + dy * dy;
}
// square distance from a point to a segment
function getSqSegDist(p, p1, p2) {
var x = p1.x,
y = p1.y,
dx = p2.x - x,
dy = p2.y - y;
if (dx !== 0 || dy !== 0) {
var t = ((p.x - x) * dx + (p.y - y) * dy) / (dx * dx + dy * dy);
if (t > 1) {
x = p2.x;
y = p2.y;
} else if (t > 0) {
x += dx * t;
y += dy * t;
}
}
dx = p.x - x;
dy = p.y - y;
return dx * dx + dy * dy;
}
// rest of the code doesn't care about point format
// basic distance-based simplification
function simplifyRadialDist(points, sqTolerance) {
var prevPoint = points[0],
newPoints = [prevPoint],
point;
for (var i = 1, len = points.length; i < len; i++) {
point = points[i];
if (getSqDist(point, prevPoint) > sqTolerance) {
newPoints.push(point);
prevPoint = point;
}
}
if (prevPoint !== point) newPoints.push(point);
return newPoints;
}
function simplifyDPStep(points, first, last, sqTolerance, simplified) {
var maxSqDist = sqTolerance,
index;
for (var i = first + 1; i < last; i++) {
var sqDist = getSqSegDist(points[i], points[first], points[last]);
if (sqDist > maxSqDist) {
index = i;
maxSqDist = sqDist;
}
}
if (maxSqDist > sqTolerance) {
if (index - first > 1) simplifyDPStep(points, first, index, sqTolerance, simplified);
simplified.push(points[index]);
if (last - index > 1) simplifyDPStep(points, index, last, sqTolerance, simplified);
}
}
// simplification using Ramer-Douglas-Peucker algorithm
function simplifyDouglasPeucker(points, sqTolerance) {
var last = points.length - 1;
var simplified = [points[0]];
simplifyDPStep(points, 0, last, sqTolerance, simplified);
simplified.push(points[last]);
return simplified;
}
// both algorithms combined for awesome performance
function simplify(points, tolerance, highestQuality) {
if (points.length <= 2) return points;
var sqTolerance = tolerance !== undefined ? tolerance * tolerance : 1;
points = highestQuality ? points : simplifyRadialDist(points, sqTolerance);
points = simplifyDouglasPeucker(points, sqTolerance);
return points;
}
// export as AMD module / Node module / browser or worker variable
if (typeof define === 'function' && define.amd) define(function() {
return simplify;
});
else if (typeof module !== 'undefined') {
module.exports = simplify;
module.exports.default = simplify;
} else if (typeof self !== 'undefined') self.simplify = simplify;
else window.simplify = simplify;
})();
// -- end simplify.js -- //
coords = pathToCoords($("#OriginalJP-01").attr("d"));
for (var i = 0; i < coords.length; i++) {
newCoords[i] = [];
for (var j = 0; j < coords[i].length; j++) {
newCoords[i][j] = simplify(coords[i][j], 1);
}
}
newPath = coordsToPath(newCoords);
$("#JP-01").attr("d", newPath);
});
$("#simplify-2").on("click", function(){
let d = $("#OriginalJP-01").attr("d");
let coordsArray = [];
let data = d.match(/(?<=m).*?(?=z)/igs);
// split each pair of coordinates into the array as an object {x, y}
for (var i = 0; i < data.length; i++) {
let ca = coordsArray[i] = [];
// split data[i] into each coordinate text
let matches = data[i].match(/((\w?-?\d+(\.\d+)?)(,-?\d+(\.\d+)?)?)/g);
for(let j=0;j<matches.length;j++){
let x, y,
text = matches[j],
// split with comma and convert it to a number
temp = text.split(",").map(v=>+v.replace(/^[^\-\d]/g,""));
switch(text[0]){
default:
case "L": // absolute
x = temp[0];
y = temp[1];
break;
case "l": // relative
x = ca[j-1].x + temp[0];
y = ca[j-1].y + temp[1];
break;
case "V": // absolute
x = ca[j-1].x;
y = temp[0];
break;
case "v": // relative
x = ca[j-1].x;
y = ca[j-1].y + temp[0];
break;
case "H": // absolute
x = temp[0];
y = ca[j-1].y;
break;
case "h": // relative
x = ca[j-1].x + temp[0];
y = ca[j-1].y;
break;
}
x = +x.toFixed(2);
y = +y.toFixed(2);
ca.push({x, y});
}
}
let mArray = [];
// calculate the slopes
for(let i=0;i<coordsArray.length;i++){
mArray[i] = [];
for(let j=0;j<coordsArray[i].length-1;j++){
let {x, y} = coordsArray[i][j], // current point's x and y
{x: nx, y: ny} = coordsArray[i][j+1], // next point's x and y
dy = (ny - y);
if(dy === 0) // to check if the denominator is legal or not
// in your case, it would not enter here
mArray[i].push(Infinity);
else
mArray[i].push((nx - x) / dy);
}
}
let abandonFactor = +$("#abandonFactor").val();
let newCoordsArray = [];
for(let i=0;i<mArray.length;i++){
let na = newCoordsArray[i] = [];
// calculate the abandonRate base on the amount of the original points
let abandonRate = coordsArray[i].length * abandonFactor;
for(let j=0;j<mArray[i].length-1;j++){
let m = mArray[i][j], // this slope
nm = mArray[i][j+1]; // next slope
let diffRate = Math.abs((m - nm) / m); // calculate the changes of the slope
// check if the diffRate is greater than abandonRate
// or the sign of m not equals the sign of nm
// you can try out removing the "sign check part" and see what would happen ;)
if(diffRate >= abandonRate || (Math.sign(m) !== Math.sign(nm))){
na.push(coordsArray[i][j]);
}
}
}
let newPath = [];
// create new path
for(let i=0;i<newCoordsArray.length;i++){
let temp = [];
for(let j=0;j<newCoordsArray[i].length;j++){
let {x, y} = newCoordsArray[i][j];
let p = `${x},${y}`;
temp.push(p);
}
newPath.push("M" + temp.join("L") + "z");
}
$("#JP-01").attr("d", newPath.join(""));
}).click();
$("#abandonFactor").on("change", function(){
$("#abandonFactor_text").text(`Abandon factor: ${this.value}`);
$("#simplify-2").click();
});
div {
width: 50%;
}
.original {
float: left;
}
.simplified {
float: right;
}
.map {
box-sizing: border-box;
width: 100%;
padding: 0.5rem;
text-shadow: 1px 1px white;
cursor: grab;
}
.land {
fill: lightgreen;
fill-opacity: 1;
stroke: white;
stroke-opacity: 1;
stroke-width: 0.5;
cursor: pointer;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="original">
<svg id="OriginalMap" viewBox="324.55999755859375 0 126.44989013671875 111.65999603271484" class="map" xmlns="http://www.w3.org/2000/svg" xmlns:amcharts="http://amcharts.com/ammap">
<g>
<path id="OriginalJP-01" title="Hokkaido" class="land" d="
M344.04,71.29l2.71,-2.13l0.04,-0.59l-1.96,-3.17l-1.38,-0.99l-0.7,-1.16l0.68,-2.54l-0.27,-0.48l1.64,0.11l0.6,-0.42l0.21,-0.54l0.53,-0.21l2.75,2.66l1.93,0.87l1.13,1.28l2.48,-0.47l0.17,-0.42l1.02,-0.32l0.72,0.02l-0.14,1.17l3.04,1.39l2.2,-1.22l1.97,-2.15l0.99,-1.6l0.1,-2.35l-1.38,-2.76l0.54,-1.94L363,51.89l-0.45,-2.27l1.02,-2.1l1.47,-0.89l0.76,-0.05l2.03,-1.32l0.93,-1.89l0.41,-1.95l-0.35,-7.86l0.5,-0.32l1.64,-3.44l0.21,-2.86l0.38,-0.79l0.13,-1.6l-0.3,-3.27l-0.77,-3.84l-3.04,-7.32l-0.31,-1.16l0.07,-0.99l0.86,-1.27l0.79,-1.95l-0.41,-1.4l0.15,-1.28l0.75,0.77l0.13,0.52l0.39,0.16l1.76,-0.28l1.35,-0.94l0.55,-1.99L374.52,0l0.47,0.33l0.56,1.29l0.59,0.34l0.63,1.47l2.32,1.7l2.21,3.12l4.95,5.56l3.91,7.23l2.58,2.55l2.44,3.1l5.58,4.4l1.57,0.89l0.33,0.95l1.26,0.96l3.7,1.94l3.24,1.28l5.88,1.69l2.79,0.53l0.44,-0.31l0.29,0.77l-0.06,1.28l0.87,1.55l0.58,0.48l3.21,1.01l3.41,0.21l2.57,-0.32l2.28,-1.99l1.54,-1.92l2.18,-2.04l1.83,-1.26l0.66,-1.45l1.1,-1.16l0.78,-1.38l0.46,0.03l0.45,2.23l-1.21,2.49l-0.91,0.92l-0.6,2.4l-1.73,2.13l-0.94,2.12l0.1,0.83L436.06,48l0.22,1.06l1.35,2.58l1.49,1.97l1.52,5.23l1.49,1.96l-1.16,-0.53l-0.19,-0.53l-1.14,-0.22l-0.24,0.37l0.21,1.26l0.86,-0.67l0.47,0.4l-0.29,0.55l0.37,0.39l0.4,-0.28l2.55,0.62l2.09,-1.7l0.32,-0.6l1.11,-0.71l0.24,-0.5l3.28,0.2l-0.19,0.5l-1.17,0.91l-1.73,0.31l-1.87,1.03l-0.8,2.93l-0.77,-0.36l-1.24,-0.05l-2.71,0.51l-1.51,0.9l-1.26,-0.21l-0.62,0.6l-0.14,1.15l-1.6,2.15l-1.02,0.49l-1.9,-0.1l-0.9,-0.79l0.03,-0.53l0.33,-0.31l-0.55,-0.37l-0.73,0.1l-1.19,1.73l-0.02,0.41l1.15,1.13l-2.01,-0.12l-1.15,-0.35l-3.58,-0.02l-0.69,-0.15l-1.58,-1.27l-3.06,0.54l-2.43,1.18l-3.4,2.53l-5.27,5.27l-4.08,5.8l-1.53,3.31l-0.28,1.67l0.55,1.44l-0.82,3.88l-0.91,1.53l0,1.61l-3.7,-3.72l-4.85,-2.02l-8.09,-4.45l-2.38,-1.67l-1.84,-2.05l-3.21,-0.94l-2.21,-2.05l-1.82,-1.2l-1.83,-0.53l-2.38,-0.04l-3.17,1.04l-2.08,1.29l-2.68,2.28l-1.59,0.89l-2.12,2.45l-0.34,0.75l-1.07,-0.4l-0.33,-0.55l0.38,-0.29l0.59,0.64l0.12,-0.81l-1.35,-0.41l-0.87,-2.37l-0.79,-0.4l-1.07,-1.12l-0.12,-0.58l-1.27,-1.23l-1.24,-0.09l-1.25,0.53l-0.97,-0.59l-1.21,0.03l-1.68,1.81l-1.44,2.75l-0.71,2.55l0.28,1.94l2.32,1.2l2.74,2.35l0.75,0.14l1.37,-0.53l2.04,0.31l1.23,2.21l1.63,1.22l1.08,1.88l3.15,1.36l0.55,0.89l0.79,0.59l-0.61,0.61l-0.95,0.01l-1.1,1.3l-1.66,0.66l-0.86,-0.79l-2.78,-0.91l-0.93,0.16l-0.28,0.62l-0.41,0.1l-0.18,-0.61l0.68,-0.45l-0.2,-0.67l-0.59,-0.37l-0.86,0.08l-0.41,0.33l-0.43,1.68l-1.44,1.06l-1.29,0.1l-0.35,0.33l-0.31,3.94l-0.4,0.48l-0.94,0.03l-1.96,0.87l-0.9,1.89l-0.39,0.32l-1.02,-0.8l-1.14,0.22l-0.9,-0.71l-1.08,-2.72l-0.09,-0.86l0.41,-1.77l1.35,-3.03v-1.05l0.77,0.08l0.29,-0.59l0.31,-2.7l-0.4,-1.98l-1.9,-2.94l-0.67,-0.36l-1.3,-0.12L334.29,91l-0.85,-0.88l-1.14,-0.41l-0.39,-0.73l-0.19,-1.39l0.31,-1.12l0.94,-1.03l0.33,-0.95l0.03,-2.65l-0.49,-2.23l0.92,-1.45l1.12,-0.62l2.28,-0.06l0.61,-0.91l1.39,-0.73l0.85,-1.92l0.82,0.57l0.56,0.97l0.89,-0.23l0.09,-1.39l0.82,-0.84L344.04,71.29z
M358.76,9.86l-0.09,-1.23l1.28,-1.25l2.07,1.21l0.51,1.81l-1.81,1.49l-1.37,-1.07L358.76,9.86z
M326.88,91.03l-0.29,2.22l-0.39,0.63l-0.65,0.23l-0.02,0.36l-0.61,-0.54l0.06,-1.71l-0.42,-1.11l0.74,-1.09l1.91,-0.38l0.46,-0.52l0.03,0.64L326.88,91.03z
M357.23,4.25l-0.26,2.59l-0.39,0.13L355.7,4l0.22,-1.31l-0.61,-0.34l0.09,-0.73l0.65,0.85l0.66,-0.15l-0.02,-0.59L357.03,2l0.35,0.78L357.23,4.25z"/>
</g>
</svg> Original
</div>
<div class="simplified">
<svg id="Map" viewBox="324.55999755859375 0 126.44989013671875 111.65999603271484" class="map" xmlns="http://www.w3.org/2000/svg" xmlns:amcharts="http://amcharts.com/ammap">
<g>
<path id="JP-01" title="Hokkaido" class="land" d=""/>
</g>
</svg> Simplified
<button id="simplify-1">Your method</button>
<button id="simplify-2">New method</button>
<input type="range" min="0.0001" max="0.1" value="0.01" step="0.0001" class="slider" id="abandonFactor">
<div id="abandonFactor_text">Abandon factor: 0.01</div>
</div>
In Mars, there are only two denominations of currency ,x and y. A
Marsian goes to a bar and the bill is "z". Using x and y he has to pay
the bill. But the bar doesn't tender change, any extra money will be
taken as tips.
So write a function in JavaScript that helps the marsian to reduce the
tips.
The function takes in x, y, z and returns the amount of tip he has to
pay.
Example 1
Input: 2, 5, 109
Output: 0
Explanation: 21 coins of 5, and 2 coins of 2
Example 2
Input: 5, 7, 43
Output: 0
Explanation: 4 coins of 7, and 3 coins of 5
Example 3
Input: 15, 19, 33
Output: 1
Explanation: 1 coin of 15 and 1 coin of 19
Solution: I think this is level one DP problem, something like subset sum. Like for finding the optimal tip for the larger number, knowing the optimal tip for all the below numbers would help.
const coinA = 2
const coinB = 5
const sum = 13
var arr = [];
arr[0] =0;
console.log(getMyTip(coinA, coinB, sum));
function getMyTip(){
for(var i=1; i<= sum; i++){
var minA, minB;
if( i < coinA){
minA = coinA - i;
}else{
minA = arr[i - coinA];
}
if( i < coinB){
minB = coinB - i;
}else{
minB = arr [i - coinB]
}
arr[i] = Math.min(minA, minB);
}
return arr[sum];
}
Jsfiddle: https://jsfiddle.net/7c4sbe46/
But I'm not sure why it is not getting accepted. Please let me know if I'm missing something with the logic here.
It is more related to diophantine equations, i.e. is there a solution to a.x+b.y=z ? The answer is yes if z is a multiple of the greatest common divisor of x and y (called it gcd). If not, your tip will be the difference between 1. the smaller number divisible by gcd and greater than z
and 2. z.
Once you know the value of the tip, you can even easily know the number of x and y that you need by slightly modifying the value of z to (z+tip).
#include <stdio.h>
int main()
{
int curr1, curr2, bill;
scanf("%d %d %d",&curr1,&curr2,&bill);
int gcd, tip=0;
int x=curr1;
int y=curr2;
while(x!=y)
{
if(x > y)
x -= y;
else
y -= x;
}
gcd=x;
if((bill%curr1==0) || (bill%curr2==0) || (bill%(curr1 + curr2)==0)){
tip = 0;
} else if(bill>(curr1 + curr2) && (bill % gcd==0)) {
tip = 0;
} else if((curr1 + curr2) > bill){
if(curr2 > curr1){
tip = (bill % (curr2-curr1));
}else{
tip = (bill % (curr1-curr2));
}
}
printf("%d",tip);
return 0;
}
There is no need to use dp for this. Here is the simple solution -
// x -> first currency denomination
// y -> second currency denomination
// z -> total bill
var calculateTip = function(x,y,z) {
var xMax = Math.floor(z/x);
var tip = y;
if(xMax == 0) {
tip = (x-z) < (Math.ceil(z/y)*y - z) ? (x-z) : (Math.ceil(z/y)*y - z);
}
while (xMax>=0) {
var tempTip = xMax*x + Math.ceil((z-xMax*x)/y)*y - z;
if(tempTip < tip) {
tip = tempTip;
}
xMax--;
}
return tip;
}
var minimumTip = function(x,y,z) {
if(x>y) {
return calculateTip(x,y,z);
} else {
return calculateTip(y,x,z);
}
}
console.log(minimumTip(2, 5, 109));
var findTip = function(x=2, y=5, z=13){
var x = x;
var y = y;
var z = z;
var tip ;
var temp1 = x;
var temp2 = y
function findNumber(num,total){
if(num > total){
return num-total;
}
else{
var q = Math.floor(total/num);
return ((q+1)*num)-total;
}
}
function findMin(a,b,c){
var min ;
if(a<b && a<c){
min = a
}else{
if(b<c){
min = b;
}else{
min = c;
}
}
return min;
}
while(temp1!=temp2)
{
if(temp1 > temp2)
temp1 -= temp2;
else
temp2 -= temp1;
}
var factor =temp1;
if(z%x == 0 || z%y == 0 || z%(x+y) == 0) {
tip = 0;
}else if(z%factor == 0 && z>=x*y - x -y){
tip = 0;
}
else {
var minX= findNumber(x,z);
var minY = findNumber(y,z);
var minXY = findNumber(x+y,z);
console.log(minX,minY,minXY)
tip = findMin(minX,minY,minXY);
}
alert('the tip is '+ tip.toString());
return tip;
}
findTip(21, 11, 109);
So I wrote this code for a simple game. The code runs at 60 fps in both Chrome and Safari but Firefox barely manages 30-40 fps. The code looks simple enough to me. What could be causing the delay?
I checked in firebug and found out that only one function "follow" is taking up all the time. Here is the code:
function checkCollision (ball0, ball1) {
var dx = ball1.X - ball0.X,
dy = ball1.Y - ball0.Y,
dist = Math.sqrt(dx * dx + dy * dy);
if (dist < ball0.rad + ball1.rad) {
var angle = Math.atan2(dy, dx),
sin = Math.sin(angle),
cos = Math.cos(angle);
var pos0 = {x: 0, y: 0},
pos1 = rotate(dx, dy, sin, cos, true),
vel0 = rotate(ball0.spdX, ball0.spdY, sin, cos, true),
vel1 = rotate(ball1.spdX, ball1.spdY, sin, cos, true),
vxTotal = vel0.x - vel1.x;
vel0.x = ((ball0.mass - ball1.mass) * vel0.x + 2 * ball1.mass * vel1.x) /
(ball0.mass + ball1.mass);
vel1.x = vxTotal + vel0.x;
var absV = Math.abs(vel0.x) + Math.abs(vel1.x),
overlap = (ball0.rad + ball1.rad) - Math.abs(pos0.x - pos1.x);
pos0.x += vel0.x / absV * overlap;
pos1.x += vel1.x / absV * overlap;
//rotate positions back
var pos0F = rotate(pos0.x, pos0.y, sin, cos, false),
pos1F = rotate(pos1.x, pos1.y, sin, cos, false);
ball1.X = ball0.X + pos1F.x;
ball1.Y = ball0.Y + pos1F.y;
ball0.X = ball0.X + pos0F.x;
ball0.Y = ball0.Y + pos0F.y;
var vel0F = rotate(vel0.x, vel0.y, sin, cos, false),
vel1F = rotate(vel1.x, vel1.y, sin, cos, false);
ball0.spdX = vel0F.x;
ball0.spdY = vel0F.y;
ball1.spdX = vel1F.x;
ball1.spdY = vel1F.y;
}
}
function move()
{
var side,i;
for (i=0 ; i < balls.length; i++)
{
var obj = balls[i];
if (side=obj.edgeX())
{
if (side === 'l')
obj.X = obj.rad;
else if (side === 'r')
obj.X = canvas.width() - obj.rad;
obj.spdX*=-1;
}
if (side=obj.edgeY())
{
if (side === 't')
obj.Y = obj.rad;
else if (side === 'b')
obj.Y = canvas.height() - obj.rad;
obj.spdY*=-1;
}
if (leash == true && i === 0)
{
if (mouse.X>obj.X && (obj.spdX<10))
obj.spdX+=obj.accX;
else if (mouse.X<obj.X && (obj.spdX>-10))
obj.spdX-=obj.accX;
if (mouse.Y>obj.Y && (obj.spdY<10))
obj.spdY+=obj.accY;
else if (mouse.Y<obj.Y && (obj.spdY>-10))
obj.spdY-=obj.accY;
}
obj.X+=obj.spdX;
obj.Y+=obj.spdY;
if (Math.abs(obj.spdX)>0.1)
obj.spdX*=0.98;
else obj.spdX=0;
if (Math.abs(obj.spdY)>0.1)
obj.spdY*=0.98;
else obj.spdY = 0;
};
}
function follow()
{
var ballA, i, ballB,j;
requestAnimationFrame(follow);
//stats.begin();
context.clearRect(0,0,canvas.width(),canvas.height());
move();
for (i = 0, len = balls.length - 1; i < len; i++) {
ballA = balls[i];
for (j = i + 1; j < balls.length; j++) {
ballB = balls[j];
checkCollision(ballA, ballB);
}
}
balls.forEach(function(obj){
drawCircle(obj.X,obj.Y,obj.rad, obj.color);
if (leash == true && obj.control === true)
{drawLeash(mouse.X,mouse.Y,obj.X,obj.Y,obj.color);}
});
//stats.end();
};
Here is the animation: http://ipsumturpis.xtreemhost.com/follower/index.html
I vaguely remembered there used to be a problem in FF regarding with canvas drawing performance, so I have commented out drawCircle(obj.X,obj.Y,obj.rad, obj.color); and poof, magic happened - my frame rate went up from 11 FPS to 60.
Try caching balls length in a variable. Unless it’s absolutely necessary for reasons I don’t see, running balls.length (or any function) in every iteration of a loop is naturally going to be time consuming.
So try something like this;
ballslen = balls.length;
for (j = i + 1; j < ballslen; j++)