If I do a search on a page greater than 1 and if the number of items returned from the search are equal to the pagesize then the page does not flip back to the previous page and says the wrong page number.
For example if I have 5 items and the page size is set to 4. If I do a search on the second page that returns 4 items then the page won't flip back to page 1 and the grid says Page 2 of 1.
I also get the same error if I change example4 -
http://mleibman.github.io/SlickGrid/examples/example4-model.html
Replace this code -
for (var i = 0; i < 50000; i++) {
var d = (data[i] = {});
d["id"] = "id_" + i;
d["num"] = i;
d["title"] = "Task " + i;
d["duration"] = "5 days";
d["percentComplete"] = Math.round(Math.random() * 100);
d["start"] = "01/01/2009";
d["finish"] = "01/05/2009";
d["effortDriven"] = (i % 5 == 0);
}
with this -
var x;
for (var i = 0; i < 5; i++) {
var d = (data[i] = {});
if (i == 0)
x = "1";
if (i == 1)
x = "11";
if (i == 2)
x = "111";
if (i == 3)
x = "1111";
if (i == 4)
x = "2";
d["id"] = "id_" + i;
d["num"] = i;
d["title"] = "Task " + x;
d["duration"] = "5 days";
d["percentComplete"] = Math.round(Math.random() * 100);
d["start"] = "01/01/2009";
d["finish"] = "01/05/2009";
d["effortDriven"] = (i % 5 == 0);
}
Add this line -
dataView.setPagingOptions({pageSize: 4});
Go to page 2 and search for Task 1. It should say Showing Page 2 of 1.
I think your rowCount was not updated. Have you added this on your script?
dataView.onRowCountChanged.subscribe(function (e, args) {
grid.updateRowCount();
grid.render();
});
dataView.onRowsChanged.subscribe(function (e, args) {
grid.invalidateRows(args.rows);
grid.render();
});
Update: Add this code on your slick.dataview.js. Find this around line 750
var paged;
if (pagesize) {
if (filteredItems.length < pagenum * pagesize) {
pagenum = Math.floor(filteredItems.length / pagesize);
}
//.. add this code
// START
if (filteredItems.length == pagenum * pagesize && pagenum > 0)
pagenum --;
// END
paged = filteredItems.slice(pagesize * pagenum, pagesize * pagenum + pagesize);
} else {
paged = filteredItems;
}
Related
im studying Algorithm with Javascript.
It is a problem abt dijkstra algorith.
but i always meet TLE(Time Limit Exceeded) in the last case.
could i know which point makes my code slower.
i tried many ways like below
without function constructor
console output with forEach
take input with require('readline')
without recursive heapify
it is the problem link.
https://onlinejudge.u-aizu.ac.jp/courses/lesson/1/ALDS1/12/ALDS1_12_C
It is my solution.
function MinPQ() {
this.heap = [];
}
MinPQ.prototype.swap = function (i, j) {
const temp = this.heap[i];
this.heap[i] = this.heap[j];
this.heap[j] = temp;
};
MinPQ.prototype.insert = function (n) {
const parentIdx = (idx) => Math.floor((idx - 1) * 0.5);
this.heap.push(n);
let curr = this.heap.length - 1;
let parent = parentIdx(curr);
while (curr > 0 && this.compare(curr, parent)) {
this.swap(curr, parent);
curr = parent;
parent = parentIdx(curr);
}
};
MinPQ.prototype.shift = function () {
const heapify = (idx) => {
let l = idx * 2 + 1;
let r = idx * 2 + 2;
let minIdx = idx;
if (l < this.heap.length && this.compare(l, minIdx)) {
minIdx = l;
}
if (r < this.heap.length && this.compare(r, minIdx)) {
minIdx = r;
}
if (minIdx !== idx) {
this.swap(idx, minIdx);
heapify(minIdx);
}
};
this.swap(0, this.heap.length - 1);
const root = this.heap.pop();
heapify(0);
return root;
};
MinPQ.prototype.compare = function (i, j) {
return this.heap[i][1] < this.heap[j][1];
};
function solution(input) {
const n = Number(input.shift());
const G = Array(n);
for (let i = 0; i < n; i++) {
const [u, k, ...adjs] = input.shift().split(' ').map(Number);
G[u] = [];
for (let j = 0; j < k; j++) {
const v = adjs[2 * j];
const c = adjs[2 * j + 1];
G[u][v] = c;
}
}
return dijkstra(n, G)
.map((e, idx) => `${idx} ${e}`)
.join('\n');
}
function dijkstra(n, G) {
const minPQ = new MinPQ();
const d = Array(n).fill(Infinity);
let count = 0;
minPQ.insert([0, 0]);
while (count < n) {
const node = minPQ.shift();
const [u, cost] = node;
if (d[u] < Infinity) continue;
d[u] = cost;
count++;
G[u].forEach((e, idx) => {
minPQ.insert([idx, cost + e]);
});
}
return d;
}
(function (test) {
const printSolution = (input) => console.log(solution(input));
if (test) {
printSolution([
'5',
'0 3 2 3 3 1 1 2',
'1 2 0 2 3 4',
'2 3 0 3 3 1 4 1',
'3 4 2 1 0 1 1 4 4 3',
'4 2 2 1 3 3',
]);
console.log('--');
printSolution([
'9',
'0 2 1 1 3 13',
'1 3 0 1 2 1 4 11',
'2 2 5 1 1 1',
'3 3 4 1 0 13 6 1',
'4 4 1 11 5 1 3 1 7 4',
'5 3 2 1 8 7 4 1',
'6 2 3 1 7 1',
'7 3 4 4 6 1 8 1',
'8 2 5 7 7 1',
]);
return;
}
printSolution(
require('fs').readFileSync('/dev/stdin', 'utf-8').split('\n')
);
})(0);
It is the solution what passed last case.
var h = [],hs = 0;
h[0] = new Array(2);
h[0][0] = -1;h[0][1] = -1;
function insert(key){
h[++hs] = key;
var i = hs;
while(h[i][1] < h[Math.floor(i / 2)][1]){
var ex = h[i];
h[i] = h[Math.floor(i / 2)];
h[Math.floor(i / 2)] = ex;
i = Math.floor(i / 2);
}
}
function extract(){
if(hs <= 0)
return;
var ret = h[1];
h[1] = h[hs--];
i = 1;
while((i * 2 <= hs && h[i][1] > h[i * 2][1]) || (i * 2 + 1 <= hs && h[i][1] > h[i * 2 + 1][1])){
var l = i * 2;var r = i * 2;
if(i * 2 + 1 <= hs)
r++;
var m = h[l][1] <= h[r][1] ? l : r;
var ex = h[i];
h[i] = h[m];
h[m] = ex;
i = m;
}
return ret;
}
function Main(input){
input = input.split("\n");
var n = parseInt(input[0],10);
var graph = new Array(n);
for(var i = 0;i < n;i++){
input[i + 1] = input[i + 1].split(" ");
var u = parseInt(input[i + 1][0],10);
var k = parseInt(input[i + 1][1],10);
graph[u] = new Array(k);
for(var j = 0;j < k;j++)
graph[u][j] = new Array(2);
for(var j = 0;j < k;j++){
graph[u][j][0] = parseInt(input[i + 1][j * 2 + 2],10);
graph[u][j][1] = parseInt(input[i + 1][j * 2 + 3],10);
}
}
var count = 1;
var sum = Array(n);
for(var i = 0;i < n;i++)
sum[i] = 1000000000;
sum[0] = 0;
for(var i = 0;i < graph[0].length;i++){
insert(graph[0][i]);
}
while(count < n){
var aa = extract();
if(sum[aa[0]] < 1000000000)
continue;
sum[aa[0]] = aa[1];
count++;
for(var i = 0;i < graph[aa[0]].length;i++){
graph[aa[0]][i][1] += sum[aa[0]];
insert(graph[aa[0]][i]);
}
}
for(var i = 0;i < n;i++){
console.log(i + " " + sum[i]);
}
}
Main(require("fs").readFileSync("/dev/stdin","utf8"));
We have new entries being pulled into a Google sheet daily using XML (Sheet 1). We are using the code below to pull these entries into a separate Google sheet for end users (Sheet 2). The end users can then edit the data. We are trying to compare the ID numbers in Sheet 1 and Sheet 2. Each time we run the code, ID numbers from Sheet 1 that do not have a match in Sheet 2 should be added to Sheet 2. If the ID number in Sheet 1 already exists in Sheet 2, that ID number should be skipped. It isn't skipping the matches. Instead it is adding everything to Sheet 2 every time we run the code and Sheet 2 now contains duplicates.
for(var i = 1; i < slateDT.length; i ++) {
var bannerid = slateDT[i][0];
var match = "No Match";
var j = 1;
while(j < gradingDT.length && match == "No Match") {
var matchID = gradingDT[j][1].trim();
if(bannerid.trim() != matchID){
j++;
} else {
match = "Match";
}
}
if(match == "No Match"){
additions.push(moveColumns(slateDT[i]));
}
}
if(additions.length > 0) {
gradingSS.getRange(gradingDT.length + 1, 2, additions.length, additions[0].length).setValues(additions);
gradingDT = getDataValues(gradingSS.getName());
var sortRng = gradingSS.getRange(2, 1, gradingDT.length, gradingDT[0].length);
sortRng.sort(3);
}
function moveColumns(studentRow) {
studentRow.splice(17, 3);
var v = checkDefined(studentRow.splice(20, 1));
studentRow.splice(10, 0, v.join());
v = checkDefined(studentRow.splice(18, 1));
studentRow.splice(13, 0, v.join());
v = checkDefined(studentRow.splice(20));
studentRow.splice(14, 0, v.join());
return studentRow;
}
Ok, I'm assuming that your weird moveColumns function does what you want and that the column numbers mismatch per my question above. Replace your for loop with this:
for (var i = 0; i < slateDT.length; i++) {
var oldID = slateDT[i][0].trim();
var matchID = 0;
for (var j = 1; j < gradingDT.length; j++) {
var newID = gradingDT[j][1].trim();
if (oldID == newID) {
matchID = j;
break; //ends the j loop when it meets the match
}
} //for [j] loop
if (matchID == 0) {
additions.push(moveColumns(slateDT[i]));
Logger.log("No match was found for " + i);
} else {
Logger.log("A match was found for " + i + " at " + j);
}
} //for [i] loop
This is very similar to what you are trying to do with the while loop but without managing to never increment J under certain circumstances.
Once you are sure it is working comment out the two logger lines for performance.
Given a number x, insert elements 1 to x^2 in a matrix spirally.
e.g. For x = 3, matrix looks like [[1,2,3],[8,9,4],[7,6,5]].
For this I've written following snippet. However, I'm getting o/p as [[7,9,5],[7,9,5],[7,9,5]]
while(t<=b && l<=r){
System.out.print(t+" "+b+" "+l+" "+r+"\n");
if(dir==0){
for(int i = l;i<=r;i++){
arr.get(t).set(i,x);
x++;
}
t++;
}else if(dir==1){
for(int i = t;i<=b;i++){
arr.get(i).set(r,x);
x++;
}
r--;
}else if(dir==2){
for(int i = r;i>=l;i--){
arr.get(b).set(i,x);
x++;
}
b--;
}else if(dir==3){
for(int i = b;i>=t;i--){
arr.get(l).set(i,x);
x++;
}
l++;
}
dir = (dir+1)%4;
}
You can use the next code (which I developed for some implementation that handles huge martrix sizes). It will use width (columns) and height (rows) of any matrix size and produce the output you need
List<rec> BuildSpiralIndexList(long w, long h)
{
List<rec> result = new List<rec>();
long count = 0,dir = 1,phase = 0,pos = 0;
long length = 0,totallength = 0;
bool isVertical = false;
if ((w * h)<1) return null;
do
{
isVertical = (count % 2) != 0;
length = (isVertical ? h : w) - count / 2 - count % 2;
phase = (count / 4);
pos = (count % 4);
dir = pos > 1 ? -1 : 1;
for (int t = 0; t < length; t++)
// you can replace the next code with printing or any other action you need
result.Add(new rec()
{
X = ((pos == 2 || pos == 1) ? (w - 1 - phase - (pos == 2 ? 1 : 0)) : phase) + dir * (isVertical ? 0 : t),
Y = ((pos <= 1 ? phase + pos : (h - 1) - phase - pos / 3)) + dir * (isVertical ? t : 0),
Index = totallength + t
});
totallength += length;
count++;
} while (totallength < (w*h));
return result;
}
This solution walks from the top left to the top right, the top right to the bottom right, the bottom right to the bottom left and the bottom left up to the top left.
It is a tricky problem, hopefully my comments below assist in explaining.
Below is a codepen link to see it added to a table.
https://codepen.io/mitchell-boland/pen/rqdWPO
const n = 3; // Set this to a number
matrixSpiral(n);
function matrixSpiral(number){
// Will populate the outer array with n-times inner arrays
var outerArray = [];
for(var i = 0; i < number; i++){
outerArray.push([]);
}
var leftColumn = 0;
var rightColumn = number - 1;
var topRow = 0;
var bottomRow = number-1;
var counter = 1; // Used to track the number we are up to.
while(leftColumn <= rightColumn && topRow <=bottomRow){
// populate the top row
for(var i = leftColumn; i <= rightColumn; i++){
outerArray[leftColumn][i] = counter;
counter++;
}
// Top row is now populated
topRow ++;
// Populate the right column
for(var i = topRow ; i <= bottomRow; i++){
outerArray[i][rightColumn] = counter;
counter++;
}
// Right column now populated.
rightColumn--;
// Populate the bottom row
// We are going from the bottom right, to the bottom left
for(var i = rightColumn; i >= leftColumn; i--){
outerArray[bottomRow][i] = counter;
counter++;
}
// Bottom Row now populated
bottomRow--;
// Populate the left column
// We are going from bottom left, to top left
for(var i = bottomRow; i >= topRow ; i--){
outerArray[i][leftColumn] = counter;
counter++;
}
// Left column now populated.
leftColumn++;
// While loop will now repeat the above process, but a step in.
}
// Console log the results.
for(var i = 0; i < number; i++){
console.log(outerArray[i]);
}
}
Taking the example from here
I want to make the events look like this:
instead of how it is on the page:
Notice how there is a space between the event and the bounding box.
It seems the element style is auto generated by the scheduler and calculates a width. How would I go about widening the event to fit exactly in the bounding box?
I found the answer on another stackedoverflow question.
This is her code diving by columns.length instead of 2
//Override kendo function for deciding events with
kendo.ui.MultiDayView.fn._arrangeColumns = function (element, top, height, slotRange) {
var startSlot = slotRange.start;
element = { element: element, slotIndex: startSlot.index, start: top, end: top + height };
var columns,
slotWidth = startSlot.clientWidth,
eventRightOffset = slotWidth * 0.10,
columnEvents,
eventElements = slotRange.events(),
slotEvents = kendo.ui.SchedulerView.collidingEvents(eventElements, element.start, element.end);
slotRange.addEvent(element);
slotEvents.push(element);
columns = kendo.ui.SchedulerView.createColumns(slotEvents);
//This is where the magic happens
var columnWidth = slotWidth / columns.length;
//Original code: var columnWidth = (slotWidth - eventRightOffset) / columns.length;
//This is where the magic ends
for (var idx = 0, length = columns.length; idx < length; idx++) {
columnEvents = columns[idx].events;
for (var j = 0, eventLength = columnEvents.length; j < eventLength; j++) {
columnEvents[j].element[0].style.width = columnWidth - 4 + "px";
columnEvents[j].element[0].style.left = (this._isRtl ? this._scrollbarOffset(eventRightOffset) : 0) + startSlot.offsetLeft + idx * columnWidth + 2 + "px";
}
}
};
Can anyone give an algorithm to find the number of times a number repeats in pascal's triangle? For example
num - No of times
1 - infinite
2 - 1
3 - 2
4 - 2
. .
6 - 3
. .
10 - 4
. .
for image Link
Or in other way, how many nCr 's are possible for nCr = x , where x is any given integer?
Just count. You know n > 1 can only appear in the first n+1 rows of Pascal's triangle. And that each row is symmetric, and increasing (for the first half). That saves time.
See http://oeis.org/A003016 for more about the sequence
I had to write something similar for a hackathon challenge. This code will find all the numbers 1 to MAX_NUMBER_TO_SEARCH that have a count of more than MINIMUM_COUNT in the Pascal Triangle of size PASC_SIZE. You can obviously alter it to only count for a single number. Not super efficient obviously.
function pasc(n) {
var xx = [];
var d = 0;
var result = [];
result[0] = [1];
result[1] = [1, 1];
for (var row = 2; row < n; row++) {
result[row] = [1];
for (var col = 1; col <= row - 1; col++) {
result[row][col] = result[row - 1][col] + result[row - 1][col - 1];
result[row].push(1);
}
for (var ff = 0; ff < result[row].length; ff++) {
xx[d++] = (result[row][ff]);
}
}
return xx;
}
function countInArray(array, what) {
var count = 0;
for (var i = 0; i < array.length; i++) {
if (array[i] === what) {
count++;
}
}
return count;
}
var MAX_NUMBER_TO_SEARCH = 5000;
var MINIMUM_COUNT = 5;
var PASC_SIZE = 1000;
var dataset = pasc(PASC_SIZE);
for (var i = 0; i < MAX_NUMBER_TO_SEARCH; i++) {
if (countInArray(dataset, i) >= MINIMUM_COUNT) {
console.log(i + " Count:" + countInArray(dataset, i) + "\n");
}
}