I am trying to count the number of permutations that do not contain consecutive letters. My code passes tests like 'aabb' (answer:8) and 'aab' (answer:2), but does not pass cases like 'abcdefa'(my answer: 2520; correct answer: 3600). Here's my code:
function permAlone(str) {
var totalPerm = 1;
var result = [];
//assign the first letter
for (var i = 0; i < str.length; i++) {
var firstNum = str[i];
var perm = firstNum;
//create an array from the remaining letters in the string
for (var k = 0; k < str.length; k++) {
if (k !== i) {
perm += str[k];
}
}
//Permutations: get the last letter and change its position by -1;
//Keep changing that letters's position by -1 until its index is 1;
//Then, take the last letter again and do the same thing;
//Keep doing the same thing until the total num of permutations of the number of items in the string -1 is reached (factorial of the number of items in the string -1 because we already established what the very first letter must be).
var permArr = perm.split("");
var j = permArr.length - 1;
var patternsLeft = totalNumPatterns(perm.length - 1);
while (patternsLeft > 0) {
var to = j - 1;
var subRes = permArr.move(j, to);
console.log(subRes);
if (noDoubleLettersPresent(subRes)) {
result.push([subRes]);
}
j -= 1;
if (j == 1) {
j = perm.length - 1;
}
patternsLeft--;
}
}
return result.length;
}
Array.prototype.move = function(from, to) {
this.splice(to, 0, (this.splice(from, 1))[0]);
return this.join("");
};
function totalNumPatterns(numOfRotatingItems) {
var iter = 1;
for (var q = numOfRotatingItems; q > 1; q--) {
iter *= q;
}
return iter;
}
function noDoubleLettersPresent(str) {
if (str.match(/(.)\1/g)) {
return false;
} else {
return true;
}
}
permAlone('abcdefa');
I think the problem was your permutation algorithm; where did you get that from? I tried it with a different one (after Filip Nguyen, adapted from his answer to this question) and it returns 3600 as expected.
function permAlone(str) {
var result = 0;
var fact = [1];
for (var i = 1; i <= str.length; i++) {
fact[i] = i * fact[i - 1];
}
for (var i = 0; i < fact[str.length]; i++) {
var perm = "";
var temp = str;
var code = i;
for (var pos = str.length; pos > 0; pos--) {
var sel = code / fact[pos - 1];
perm += temp.charAt(sel);
code = code % fact[pos - 1];
temp = temp.substring(0, sel) + temp.substring(sel + 1);
}
console.log(perm);
if (! perm.match(/(.)\1/g)) result++;
}
return result;
}
alert(permAlone('abcdefa'));
UPDATE: In response to a related question, I wrote an algorithm which doesn't just brute force all the permutations and then skips the ones with adjacent doubles, but uses a logical way to only generate the correct permutations. It's explained here: Permutations excluding repeated characters and expanded to include any number of repeats per character here: Generate all permutations of a list without adjacent equal elements
I agree with m69, the bug seems to be in how you are generating permutations. I got 3600 for 'abcdefa' by implementing a different algorithm for generating permutations. My solution is below. Since it uses recursion to generate the permutations the solution is not fast, however you may find the code easier to follow, if speed is not important.
The reason for having a separate function to generate the array index values in the permutations was to verify that the permutation code was working properly. Since there are duplicate values in the input strings it's harder to debug issues in the permutation algorithm.
// Simple helper function to compute all permutations of string indices
function permute_indices_helper(input) {
var result = [];
if (input.length == 0) {
return [[]];
}
for(var i = 0; i < input.length; i++) {
var head = input.splice(i, 1)[0];
var tails = permute_indices_helper(input);
for (var j = 0; j < tails.length; j++) {
tails[j].splice(0, 0, head);
result.push(tails[j]);
}
input.splice(i, 0, head); // check
}
return result;
};
// Given an array length, generate all permutations of possible indices
// for array of that length.
// Example: permute_indices(2) generates:
// [[0,1,2], [0,2,1], [1,0,2], ... , [2, 0, 1]]
function permute_indices(array_length) {
var result = [];
for (var i = 0; i < array_length; i++) {
result.push(i);
}
return permute_indices_helper(result);
}
// Rearrange letters of input string according to indices.
// Example: "car", [2, 1, 0]
// returns: "rac"
function rearrange_string(str, indices) {
var result = "";
for (var i = 0; i < indices.length; i++) {
var string_index = indices[i];
result += str[string_index];
}
return result;
}
function permAlone(str) {
var result = 0;
var permutation_indices = permute_indices(str.length);
for (var i = 0; i < permutation_indices.length; i++) {
var permuted_string = rearrange_string(str, permutation_indices[i]);
if (! permuted_string.match(/(.)\1/g)) result++;
}
return result;
}
You can see a working example on JSFiddle.
Related
I am trying to do one problem of hackerrank but I am not able to solve that problem
Can someone please help me with wrong logic implementation done by me?
problem
Print the length of the longest string, such that is a child of both s1 and s2.
Sample Input
HARRY
SALLY
Sample Output
2
Explanation
The longest string that can be formed by deleting zero or more characters from HARRY and SALLY is AY, whose length is 2.
Sample Input 1
AA
BB
Sample Output 1
0
Explanation 1
AA and BB have no characters in common and hence the output is 0
Sample Input 2
SHINCHAN
NOHARAAA
Sample Output 2
3
Explanation 2
The longest string that can be formed between SHINCHAN and NOHARAAA while maintaining the order is NHA.
I have written some logic which is as follows:
function commonChild(s1, s2) {
var arr = s2.split(),
currenString = '',
maxLength = 0,
index = -1;
console.log(arr);
for (var i = 0; i < s1.length; i++) {
var char = s1.charAt(i),
charIndex = arr.indexOf(char);
console.log(char)
if (index < charIndex) {
index = charIndex;
currenString +=char;
}
maxLength= Math.max(maxLength,currenString.length)
}
return maxLength;
}
commonChild('ABCDEF', 'FBDAMN');
console.log(commonChild('ABCDEF', 'FBDAMN'));
pardon me. this is an unoptimized solution.
function maxCommon(a, b, offset) {
offset = offset || 0;
if (a === b) {
return [[a, b]];
}
var possibleSolns = [];
for (var i = 0 + offset; i < a.length; i++) {
for (var j = 0 + offset; j < b.length; j++) {
if (a.charAt(i) === b.charAt(j)) {
possibleSolns.push([
a.substring(0, offset) + a.substring(i),
b.substring(0, offset) +b.substring(j)
]);
break;
}
}
}
var results = [];
possibleSolns.forEach(function(soln) {
var s = maxCommon(soln[0], soln[1], offset+1);
if (s.length === 0) {
s = [[soln[0].substring(0, offset +1), soln[1].substring(0, offset +1)]];
}
results = results.concat(s);
});
return results;
}
var maxLen = -1;
var soln = null;
maxCommon("ABCDEF", "FBDAMN").map(function(_) {
return _[0];
}).forEach(function(_) {
if (_.length > maxLen) {
soln = _;
maxLen = _.length;
}
});
console.log(soln);
I kept most of your logic in the answer:
function commonChild(s1, s2) {
var // Sets strings to arrays
arrayString1 = s1.split(""),
arrayString2 = s2.split(""),
collectedChars = "",
maxLength = 0,
max = arrayString1.length;
for (var i = 0; i < max; i++) {
var
char = arrayString1[i],
count = arrayString2.indexOf(char);
// check if char is in second string and not in collected
if (count != -1 && collectedChars.indexOf(char) == -1) {
collectedChars += char;
maxLength++;
}
}
return maxLength;
}
// expected output 4
console.log(commonChild(
'ABCDEF',
'FBDAMN'
));
// expected output 1
console.log(commonChild(
'AA',
'FBDAMN'
));
Using lodash and spread operation you can do it in this way.
const test = (first, second) => {
const stringArray1 = [...first];
const stringArray2 = [...second];
return _.intersection(stringArray1, stringArray2).length;
}
console.log(test('ABCDEF', 'FBDAMN'));
You can solve it using lcs least common subsequence
function LCS(s1,s2,x,y){
var result = 0;
if(x==0 || y==0){
result = 0
}else if(s1[x-1] == s2[y-1]){
result = 1+ LCS(s1,s2,x-1,y-1)
} else if(s1[x-1] != s2[y-1]){
result = Math.max(LCS(s1,s2,x-1,y), LCS(s1,s2,x,y-1))
}
return result;
}
// Complete the commonChild function below.
function commonChild(s1, s2) {
return LCS(s1,s2,s1.length,s2.length);
}
Based on your code before the edit.
One little change is to change var arr = s2.split() to split('').
The main change in the logic is that I added a loop to run over the string each time from another character (first loop from the first, second from the second etc).
function commonChild(s1, s2) {
var arr = s2.split(''),
currenString = '',
maxLength = 0,
index = -1,
j = -1;
for (var ii = 0; ii < s1.length; ii++) {
index = -1;
currenString = '';
for (var i = ii; i < s1.length; i++) {
var char = s1.charAt(i),
j = arr.indexOf(char);
if (index < j) {
index = j;
currenString += char;
}
maxLength = Math.max(maxLength, currenString.length)
}
}
return maxLength;
}
console.log(commonChild('ABCDEF', 'FBDAMN'));
Very new, trying to make a function that takes out and separates all negatives/positives/zeros in an array. So far Ive been able to make an acceptable for loop but only with hard coded numbers. Dont currently know how to convert it into a function. please help.
var arr=[1,3,5,-9,-3,0];
var new_arr = [];
var new_arr2 = [];
var new_arr3=[];
for(i =0; i < arr.length; i++){
if(arr[i]>0){
new_arr.push(arr[i]);
}
else if(arr[i]<0){
new_arr2.push(arr[i]);
}
else if(arr[i]===0){
new_arr3.push(arr[i]);
}
}
console.log(new_arr3.length/arr.length);
console.log(new_arr2.length/arr.length);
console.log(new_arr.length/arr.length);
How about something like this?
function division(arr) {
var new_arr = [];
var new_arr2 = [];
var new_arr3 = [];
for (i = 0; i < arr.length; i++) {
if (arr[i] > 0) {
new_arr.push(arr[i]);
} else if (arr[i] < 0) {
new_arr2.push(arr[i]);
} else if (arr[i] === 0) {
new_arr3.push(arr[i]);
}
}
console.log(new_arr3.length / arr.length);
console.log(new_arr2.length / arr.length);
console.log(new_arr.length / arr.length);
}
division([1, 3, 5, -9, -3, 0]);
This new function takes the array as a parameter, so all you need to do is call it and pass the array.
You can try this way also, here I am considering 0 as a positive number. If you want, you can tweak the condition as per requirement.
function positive_negative(array ){
positive = array.filter(function (a) { return a >= 0; });
negative = array.filter(function (a) { return a < 0; });
return [positive,negative];
}
var array = [1,3,5,-9,-3,0];
console.log(positive_negative(array));
I have written a function and called another function inside but my tests show that it is not time optimized. How can I make the following code faster?
function maxSum(arr, range) {
function sumAll(array1, myrange) {
var total = 0;
if (Array.isArray(myrange)) {
for (var i = myrange[0]; i <= myrange[1]; i++) {
total += array1[i];
}
return total;
} else return array1[myrange];
}
var mylist = [];
var l = range.length;
for (var n = 0; n < l; n++) {
mylist.push(sumAll(arr, range[n]));
}
return Math.max.apply(null, mylist);
}
Algorithmic optimization: create new array with cumulative sums from index 0 to every index
cumsum[0] = 0;
for (var i = 1; i <= arr.Length; i++) {
cumsum[i] = cumsum[i-1] + arr[i-1]
Now you don't need to calculate sums for every range - just get difference
sum for range (i..j) = cumsum[j+1] - cumsum[i];
in your terms:
function sumAll(array1, myrange) {
return cumsum[myrange[1]+1] - cumsum[myrange[0]];
}
example:
arr = [1,2,3,4]
cumsum = [0,1,3,6,10]
sum for range 1..2 = 6 - 1 = 5
P.S. If your array might be updated, consider Fenwick tree data structure
1) You can define the function sumAll outside of the function maxSum because every time you call maxSum the javascript engine is recreating a fresh new function sumAll.
2) You can define myrange[1] as a variable in the initialiser part to avoid javascript to look for myrange[1] at each iteration.
for (var i = myrange[0]; i <= myrange[1]; i++) {
total += array1[i];
}
become this:
for (var i = myrange[0], len = myrange[1]; i <= len; i++) {
total += array1[i];
}
Full working code based on #MBo's excellent optimization. This passes all the tests at https://www.codewars.com/kata/the-maximum-sum-value-of-ranges-challenge-version/train/javascript, which I gather is where this problem comes from.
function maxSum(arr, ranges) {
var max = null;
var sums = [];
var sofar = 0;
for (var i = 0; i <= arr.length; i++) {
sums[i] = sofar;
sofar += arr[i];
}
for (var i = 0; i < ranges.length; i++) {
var sum = sums[ranges[i][1]+1] - sums[ranges[i][0]];
if (max === null || sum > max) {
max = sum;
}
}
return max;
}
when I run this program I end up with NaN at the end; I'd appreciate some form of explanation, as I'm stumped! I have an odd feeling it has to do something with scope...
https://jsfiddle.net/Smuggles/evj46a23/
var array = []
var range = function(start, end) {
for (var count = start; count <= end; count++) {
array.push(start);
start += 1;
}
console.log(array);
}
var sum = function() {
var result = 0
var arrayLength = array.length
for (var count = 0; count <= arrayLength; count++) {
result += array[count]
console.log(result);
}
}
console.log(sum(range(1, 10)));
2 things:
You need to change the for loop in the sum function to be < arrayLength and not <= arrayLength. You are dealing with array lengths which start with a 0 index.
You need to return the result from the sum function
var array = [];
var range = function(start, end) {
for (var count = start; count <= end; count++) {
array.push(start);
start += 1;
}
};
var sum = function() {
var result = 0;
var arrayLength = array.length;
for (var count = 0; count < arrayLength; count++) {
result += array[count];
}
return result;
};
console.log(sum(range(1, 10)));
Given an array of [4,5,6], the indexes would be as follows:
0: 4
1: 5
2: 6
Therefore, when you use the length property of the array (3), you are referencing an index that does not exist, which returns undefined. It tries to do the math on the undefined, which causes a NaN. This is why you have to use < arrayLength.
The functional approach:
It would help to make those functions a bit more "pure". Instead of maintaining state outside of the functions (with var array = []), just return the values from the functions: See the following for example:
function range(start, end) {
var arr = [];
for (var i = start; i <= end; i++) {
arr.push(i);
}
return arr;
}
function sumArray(array) {
return array.reduce(function(a, b) {
return a + b;
});
}
console.log(sumArray(range(1, 10)));
Each function takes arguments, and simply returns the result. This way, you approach this a little more "functional".
Description in Comments of Code
var array = [];
var range = function(start, end) {
//simplified the loop to remove unnecessary variables
for (; start <= end; start++) {
array.push(start);
}
return array;
}
var sum = function() {
var result = 0;
// move length to scope of the loop
// change to < rather than <= due to zero index nature of arrays
for (var count = 0, length = array.length; count < length; count++) {
result += array[count];
}
// return the result from the function
return result;
}
// gets an array from 1-10
var arr = range(1, 10);
// print the array to the console
console.log(arr);
// print the sum to the console
console.log(sum(arr));
Here is one to get your brain going! I've not had any luck with it.
[1,2,1,1,2,1,1,1,2,2]
[1,2,1,1,2,1]
I would like to use the second array to find the values in the first, but they must be in the same order.
Once for I would like it to return the next key up from the last key in the second array.
So in this example it would use the first six digits in the first array and then return 6 as the key after the final one in the second array.
var a2 = [1,2,1,1,2,1,1,1,2,2]
var a1 = [1,2,1,1,0,1]
function find(arr1, arr2) {
var len = 1
var result = 0;
var s2 = arr2.toString();
for (len=1;len <= a1.length; len++)
{
var aa1 = arr1.slice(0, len)
var s1 = aa1.toString();
if(s2.indexOf(s1)>=0){
result = aa1.length;
}
else {
break;
}
}
return result;
}
alert(find(a1, a2));
var find = function(haystack, needle) {
var doesMatch = function(offset) {
for (var i = 0; i < needle.length; i++) {
if (haystack[i+offset] !== needle[i]) {
return false;
}
}
return true;
};
for (var j=0; j < haystack.length - needle.length; j++) {
if (doesMatch(j)) {
return j;
}
}
return -1;
};
This is quick, this is dirty, and this is correct only if your data doesn't include any comma.
var needle = [1,2,1,1,2,1];
var haystack = [1,2,1,1,2,1,1,1,2,2];
if ( needle.length <= 0 ) return 0;
var fromStr = ','+haystack.toString()+','
var findStr = ','+needle.toString()+','
// Find ',1,2,1,1,2,1,' in ',1,2,1,1,2,1,1,1,2,2,'
var pos = fromStr.indexOf(findStr);
// Count the end position requested
return pos >= 0 ? fromStr.slice(0,pos+1).match(/,/g).length + needle.length - 1 : -1;
Note: The comma at head and tail is to make sure [22,12] doesn't match [2,1].