Check whether a number can be dialed on a phone keypad without diagonal moves
$begingroup$
I'm trying to write/come up with an algorithm for a phone keypad traversal.
Let's say I have a rook on a keypad. The rook can traverse only horizontally and vertically. My code has to input a phone number and check whether the rook is able to dial it (shouldn't go diagonally) and return a boolean based on the result.
Example: A phone number "4632871" is termed as "true" since the rook can traverse without going diagonally whereas "4853267" is termed as "false"
Below is my code implementation:
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
function rookTraversal(arr, phoneNumber) {
var arrIndex = ,
numArray = ;
var number = phoneNumber.split('');
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
arrIndex.push([i, j]);
}
}
arrIndex = arrIndex.map((item, index) => ({
[index + 1]: item,
}));
number.forEach((item) => {
numArray.push(arrIndex[item - 1]);
});
let numArrayKeys = numArray.reduce((acc, x) => [...acc, Object.values(x).map((y) => y)], );
for (let i = 1; i < numArrayKeys.length; i++) {
var x1 = numArrayKeys[i - 1][0][0];
var y1 = numArrayKeys[i - 1][0][1];
var x2 = numArrayKeys[i][0][0];
var y2 = numArrayKeys[i][0][1];
if (x1 !== x2 && y1 !== y2) {
return false;
}
}
return true;
}
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));Is there a way to optimize this piece of code?
javascript
edited 1 hour ago
200_success
129k15153415
asked 7 hours ago
a2441918a2441918
1161
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm trying to write/come up with an algorithm for a phone keypad traversal.
Let's say I have a rook on a keypad. The rook can traverse only horizontally and vertically. My code has to input a phone number and check whether the rook is able to dial it (shouldn't go diagonally) and return a boolean based on the result.
Example: A phone number "4632871" is termed as "true" since the rook can traverse without going diagonally whereas "4853267" is termed as "false"
Below is my code implementation:
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
function rookTraversal(arr, phoneNumber) {
var arrIndex = ,
numArray = ;
var number = phoneNumber.split('');
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
arrIndex.push([i, j]);
}
}
arrIndex = arrIndex.map((item, index) => ({
[index + 1]: item,
}));
number.forEach((item) => {
numArray.push(arrIndex[item - 1]);
});
let numArrayKeys = numArray.reduce((acc, x) => [...acc, Object.values(x).map((y) => y)], );
for (let i = 1; i < numArrayKeys.length; i++) {
var x1 = numArrayKeys[i - 1][0][0];
var y1 = numArrayKeys[i - 1][0][1];
var x2 = numArrayKeys[i][0][0];
var y2 = numArrayKeys[i][0][1];
if (x1 !== x2 && y1 !== y2) {
return false;
}
}
return true;
}
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));Is there a way to optimize this piece of code?
javascript
edited 1 hour ago
200_success
129k15153415
asked 7 hours ago
a2441918a2441918
1161
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm trying to write/come up with an algorithm for a phone keypad traversal.
Let's say I have a rook on a keypad. The rook can traverse only horizontally and vertically. My code has to input a phone number and check whether the rook is able to dial it (shouldn't go diagonally) and return a boolean based on the result.
Example: A phone number "4632871" is termed as "true" since the rook can traverse without going diagonally whereas "4853267" is termed as "false"
Below is my code implementation:
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
function rookTraversal(arr, phoneNumber) {
var arrIndex = ,
numArray = ;
var number = phoneNumber.split('');
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
arrIndex.push([i, j]);
}
}
arrIndex = arrIndex.map((item, index) => ({
[index + 1]: item,
}));
number.forEach((item) => {
numArray.push(arrIndex[item - 1]);
});
let numArrayKeys = numArray.reduce((acc, x) => [...acc, Object.values(x).map((y) => y)], );
for (let i = 1; i < numArrayKeys.length; i++) {
var x1 = numArrayKeys[i - 1][0][0];
var y1 = numArrayKeys[i - 1][0][1];
var x2 = numArrayKeys[i][0][0];
var y2 = numArrayKeys[i][0][1];
if (x1 !== x2 && y1 !== y2) {
return false;
}
}
return true;
}
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));Is there a way to optimize this piece of code?
javascript
edited 1 hour ago
200_success
129k15153415
asked 7 hours ago
a2441918a2441918
1161
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to write/come up with an algorithm for a phone keypad traversal.
Let's say I have a rook on a keypad. The rook can traverse only horizontally and vertically. My code has to input a phone number and check whether the rook is able to dial it (shouldn't go diagonally) and return a boolean based on the result.
Example: A phone number "4632871" is termed as "true" since the rook can traverse without going diagonally whereas "4853267" is termed as "false"
Below is my code implementation:
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
function rookTraversal(arr, phoneNumber) {
var arrIndex = ,
numArray = ;
var number = phoneNumber.split('');
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
arrIndex.push([i, j]);
}
}
arrIndex = arrIndex.map((item, index) => ({
[index + 1]: item,
}));
number.forEach((item) => {
numArray.push(arrIndex[item - 1]);
});
let numArrayKeys = numArray.reduce((acc, x) => [...acc, Object.values(x).map((y) => y)], );
for (let i = 1; i < numArrayKeys.length; i++) {
var x1 = numArrayKeys[i - 1][0][0];
var y1 = numArrayKeys[i - 1][0][1];
var x2 = numArrayKeys[i][0][0];
var y2 = numArrayKeys[i][0][1];
if (x1 !== x2 && y1 !== y2) {
return false;
}
}
return true;
}
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));Is there a way to optimize this piece of code?
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
function rookTraversal(arr, phoneNumber) {
var arrIndex = ,
numArray = ;
var number = phoneNumber.split('');
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
arrIndex.push([i, j]);
}
}
arrIndex = arrIndex.map((item, index) => ({
[index + 1]: item,
}));
number.forEach((item) => {
numArray.push(arrIndex[item - 1]);
});
let numArrayKeys = numArray.reduce((acc, x) => [...acc, Object.values(x).map((y) => y)], );
for (let i = 1; i < numArrayKeys.length; i++) {
var x1 = numArrayKeys[i - 1][0][0];
var y1 = numArrayKeys[i - 1][0][1];
var x2 = numArrayKeys[i][0][0];
var y2 = numArrayKeys[i][0][1];
if (x1 !== x2 && y1 !== y2) {
return false;
}
}
return true;
}
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
function rookTraversal(arr, phoneNumber) {
var arrIndex = ,
numArray = ;
var number = phoneNumber.split('');
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
arrIndex.push([i, j]);
}
}
arrIndex = arrIndex.map((item, index) => ({
[index + 1]: item,
}));
number.forEach((item) => {
numArray.push(arrIndex[item - 1]);
});
let numArrayKeys = numArray.reduce((acc, x) => [...acc, Object.values(x).map((y) => y)], );
for (let i = 1; i < numArrayKeys.length; i++) {
var x1 = numArrayKeys[i - 1][0][0];
var y1 = numArrayKeys[i - 1][0][1];
var x2 = numArrayKeys[i][0][0];
var y2 = numArrayKeys[i][0][1];
if (x1 !== x2 && y1 !== y2) {
return false;
}
}
return true;
}
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));javascript
javascript
edited 1 hour ago
200_success
129k15153415
asked 7 hours ago
a2441918a2441918
1161
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
200_success
129k15153415
asked 7 hours ago
a2441918a2441918
1161
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
200_success
129k15153415
edited 1 hour ago
200_success
129k15153415
edited 1 hour ago
200_success
129k15153415
129k15153415
asked 7 hours ago
a2441918a2441918
1161
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
a2441918a2441918
1161
asked 7 hours ago
a2441918a2441918
1161
1161
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
a2441918 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
There are a lot of different traits you could optimize for. I like to optimize for lines of code, so I'll try that.
You can simplify the testing by building a "graph" (2-dimensional array) of reachable numbers. Each row and column index is a phone digit. If the value at (i,j) is true, then j is reachable from i.
To do this, take all the numbers in each digit's row/column and create an array where those indexes have a value of 1.
The reachability graph looks like this:
The digit 6 can reach 3,4,5,6,9 and you can see that reachable[6][x] is true for x=3,4,5,6 or 9.
Then simply run through the list of numbers and test if reachable[current][next] is true.
I've assumed the rook can dial repeated numbers like 3333. If no, you can test for that in the graph creation or during reachability testing.
function rookTraversal(rows, phoneNumber) {
var cols=rows[0].map((col, i) => rows.map(row => row[i])),
digits=phoneNumber.match(/(d)/g),
reachable=;
for (var row = 0; row < rows.length; row++) {
for (var col = 0; col < cols.length; col++) {
reachable[ rows[row][col] ] = rows[row].concat(cols[col]).reduce( (map, value) => { map[value]=1; return map }, );
}
}
// console.log({reachable,digits});
for (var i=0; i<digits.length-1; ++i) if (! reachable[ digits[i] ][ digits[i+1] ]) return false;
return true;
}
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
$endgroup$
add a comment |
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1 Answer
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$begingroup$
There are a lot of different traits you could optimize for. I like to optimize for lines of code, so I'll try that.
You can simplify the testing by building a "graph" (2-dimensional array) of reachable numbers. Each row and column index is a phone digit. If the value at (i,j) is true, then j is reachable from i.
To do this, take all the numbers in each digit's row/column and create an array where those indexes have a value of 1.
The reachability graph looks like this:
The digit 6 can reach 3,4,5,6,9 and you can see that reachable[6][x] is true for x=3,4,5,6 or 9.
Then simply run through the list of numbers and test if reachable[current][next] is true.
I've assumed the rook can dial repeated numbers like 3333. If no, you can test for that in the graph creation or during reachability testing.
function rookTraversal(rows, phoneNumber) {
var cols=rows[0].map((col, i) => rows.map(row => row[i])),
digits=phoneNumber.match(/(d)/g),
reachable=;
for (var row = 0; row < rows.length; row++) {
for (var col = 0; col < cols.length; col++) {
reachable[ rows[row][col] ] = rows[row].concat(cols[col]).reduce( (map, value) => { map[value]=1; return map }, );
}
}
// console.log({reachable,digits});
for (var i=0; i<digits.length-1; ++i) if (! reachable[ digits[i] ][ digits[i+1] ]) return false;
return true;
}
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
$endgroup$
add a comment |
$begingroup$
There are a lot of different traits you could optimize for. I like to optimize for lines of code, so I'll try that.
You can simplify the testing by building a "graph" (2-dimensional array) of reachable numbers. Each row and column index is a phone digit. If the value at (i,j) is true, then j is reachable from i.
To do this, take all the numbers in each digit's row/column and create an array where those indexes have a value of 1.
The reachability graph looks like this:
The digit 6 can reach 3,4,5,6,9 and you can see that reachable[6][x] is true for x=3,4,5,6 or 9.
Then simply run through the list of numbers and test if reachable[current][next] is true.
I've assumed the rook can dial repeated numbers like 3333. If no, you can test for that in the graph creation or during reachability testing.
function rookTraversal(rows, phoneNumber) {
var cols=rows[0].map((col, i) => rows.map(row => row[i])),
digits=phoneNumber.match(/(d)/g),
reachable=;
for (var row = 0; row < rows.length; row++) {
for (var col = 0; col < cols.length; col++) {
reachable[ rows[row][col] ] = rows[row].concat(cols[col]).reduce( (map, value) => { map[value]=1; return map }, );
}
}
// console.log({reachable,digits});
for (var i=0; i<digits.length-1; ++i) if (! reachable[ digits[i] ][ digits[i+1] ]) return false;
return true;
}
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
$endgroup$
add a comment |
$begingroup$
There are a lot of different traits you could optimize for. I like to optimize for lines of code, so I'll try that.
You can simplify the testing by building a "graph" (2-dimensional array) of reachable numbers. Each row and column index is a phone digit. If the value at (i,j) is true, then j is reachable from i.
To do this, take all the numbers in each digit's row/column and create an array where those indexes have a value of 1.
The reachability graph looks like this:
The digit 6 can reach 3,4,5,6,9 and you can see that reachable[6][x] is true for x=3,4,5,6 or 9.
Then simply run through the list of numbers and test if reachable[current][next] is true.
I've assumed the rook can dial repeated numbers like 3333. If no, you can test for that in the graph creation or during reachability testing.
function rookTraversal(rows, phoneNumber) {
var cols=rows[0].map((col, i) => rows.map(row => row[i])),
digits=phoneNumber.match(/(d)/g),
reachable=;
for (var row = 0; row < rows.length; row++) {
for (var col = 0; col < cols.length; col++) {
reachable[ rows[row][col] ] = rows[row].concat(cols[col]).reduce( (map, value) => { map[value]=1; return map }, );
}
}
// console.log({reachable,digits});
for (var i=0; i<digits.length-1; ++i) if (! reachable[ digits[i] ][ digits[i+1] ]) return false;
return true;
}
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
$endgroup$
There are a lot of different traits you could optimize for. I like to optimize for lines of code, so I'll try that.
You can simplify the testing by building a "graph" (2-dimensional array) of reachable numbers. Each row and column index is a phone digit. If the value at (i,j) is true, then j is reachable from i.
To do this, take all the numbers in each digit's row/column and create an array where those indexes have a value of 1.
The reachability graph looks like this:
The digit 6 can reach 3,4,5,6,9 and you can see that reachable[6][x] is true for x=3,4,5,6 or 9.
Then simply run through the list of numbers and test if reachable[current][next] is true.
I've assumed the rook can dial repeated numbers like 3333. If no, you can test for that in the graph creation or during reachability testing.
function rookTraversal(rows, phoneNumber) {
var cols=rows[0].map((col, i) => rows.map(row => row[i])),
digits=phoneNumber.match(/(d)/g),
reachable=;
for (var row = 0; row < rows.length; row++) {
for (var col = 0; col < cols.length; col++) {
reachable[ rows[row][col] ] = rows[row].concat(cols[col]).reduce( (map, value) => { map[value]=1; return map }, );
}
}
// console.log({reachable,digits});
for (var i=0; i<digits.length-1; ++i) if (! reachable[ digits[i] ][ digits[i+1] ]) return false;
return true;
}
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));function rookTraversal(rows, phoneNumber) {
var cols=rows[0].map((col, i) => rows.map(row => row[i])),
digits=phoneNumber.match(/(d)/g),
reachable=;
for (var row = 0; row < rows.length; row++) {
for (var col = 0; col < cols.length; col++) {
reachable[ rows[row][col] ] = rows[row].concat(cols[col]).reduce( (map, value) => { map[value]=1; return map }, );
}
}
// console.log({reachable,digits});
for (var i=0; i<digits.length-1; ++i) if (! reachable[ digits[i] ][ digits[i+1] ]) return false;
return true;
}
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));function rookTraversal(rows, phoneNumber) {
var cols=rows[0].map((col, i) => rows.map(row => row[i])),
digits=phoneNumber.match(/(d)/g),
reachable=;
for (var row = 0; row < rows.length; row++) {
for (var col = 0; col < cols.length; col++) {
reachable[ rows[row][col] ] = rows[row].concat(cols[col]).reduce( (map, value) => { map[value]=1; return map }, );
}
}
// console.log({reachable,digits});
for (var i=0; i<digits.length-1; ++i) if (! reachable[ digits[i] ][ digits[i+1] ]) return false;
return true;
}
var data = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
];
console.log(rookTraversal(data, '4631782'));
console.log(rookTraversal(data, '4853267'));answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
edited 6 hours ago
answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
answered 6 hours ago
Oh My GoodnessOh My Goodness
39817
39817
add a comment |
add a comment |
a2441918 is a new contributor. Be nice, and check out our Code of Conduct.
a2441918 is a new contributor. Be nice, and check out our Code of Conduct.
a2441918 is a new contributor. Be nice, and check out our Code of Conduct.
a2441918 is a new contributor. Be nice, and check out our Code of Conduct.
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
