I need to find the maximum slice of the array which contains no more than two different numbers.
Here is my array [1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8]
My thought process on this is to find the numbers that are not repeated and return their index within a new array.
Here is what I have so far:
function goThroughInteger(number) {
var array = [];
//iterate the array and check if number is not repeated
number.filter(function (element, index, number) {
if(element != number[index-1] && element != number[index+1]) {
array.push(index);
return element;
}
})
console.log(array);
}
goThroughInteger([1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8]);
I'm unsure where to go next, I'm struggling to understand the question that being - find the maximum slice which contains no more than two different numbers - that confuses me.
A solution with a single loop, which checks the last values and increments a counter.
function getLongestSlice(array) {
var count = 0,
max = 0,
temp = [];
array.forEach(function (a) {
var last = temp[temp.length - 1];
if (temp.length < 2 || temp[0].value === a || temp[1].value === a) {
++count;
} else {
count = last.count + 1;
}
if (last && last.value === a) {
last.count++;
} else {
temp.push({ value: a, count: 1 });
temp = temp.slice(-2);
}
if (count > max) {
max = count;
}
});
return max;
}
console.log(getLongestSlice([58, 800, 0, 0, 0, 356, 8988, 1, 1])); // 4
console.log(getLongestSlice([58, 800, 0, 0, 0, 356, 356, 8988, 1, 1])); // 5
console.log(getLongestSlice([1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8])); // 10
function goThroughInteger(array) {
var solutionArray = [];
var max = 0;
for (var i = 0; i <= array.length; i++) {
for (var j = i + 1; j <= array.length; j++) {
var currentSlice= array.slice(i,j);
var uniqSet = [...new Set(currentSlice)];
if(uniqSet.length <3) {
if(currentSlice.length>max) {
max= currentSlice.length;
}
}
}
}
console.log(max);
}
goThroughInteger([1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8]);
This solution checks every possible slice of the array, checks if it has not more than 2 different numbers and finally prints out the length of the longest slice.
This is a possible solution, with complexity O(n²) (as pointed out by #le_m in the comments)
goThroughInteger = (list) => {
let scores = list.reduce((slices, num, pos) => {
let valid = [num];
let count = 0;
for (let i = pos; i < list.length; i++) {
if (valid.indexOf(list[i]) == -1) {
if (valid.length < 2) {
valid.push(list[i]);
count++;
} else {
break;
}
} else {
count++;
}
}
slices[pos] = { pos, count };
return slices;
}, []);
scores.sort((a, b) => b.count - a.count);
let max = scores[0];
return list.slice(max.pos, max.pos + max.count);
};
console.log(goThroughInteger([1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8]));
console.log(goThroughInteger([58, 800, 0, 0, 0, 356, 8988, 1, 1]));
```
The solution calculates the 'score' at every position of the input list, counting the length of a sequence of no more than 2 different values, then takes the result with the highest score and extracts a slice from the original list based on that information.
It can definitely be cleaned and optimized but I think it's a good starting point.
Using the sliding window algorithm in O(n) time:
const arr = [1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8, 1, 1 ,1 ,1, 8, 1, 1, 8, 8];
const map = {
length: 0
};
let required = [];
for(start = 0, end = 0; end <= arr.length; ){
if(map.length > 2){
if(map[arr[start]] === 1){
delete map[arr[start]];
map.length --;
}else{
map[arr[start]]--;
};
start++;
}else{
if(end - start > required.length){
required = arr.slice(start, end);
};
if(map[arr[end]]){
map[arr[end]]++;
}else{
map[arr[end]] = 1;
map.length++;
}
end++;
}
}
console.log(required);
Related
I need to check if a number repeats itself at least three times in an array. How can I refactor it to decrease the Cognitive Complexity that Lint keeps complaining about.
Heres my code:
let array11 = [1, 3, 2, 3, 5, 6, 7, 8, 9, 0, 1];
function checkDuplicateNumber (array11) {
for (let i = 0; i < array11.length; i += 1) {
let sameNumberLoop = 0;
for (let i2 = i; i2 < array11.length; i2 += 1) {
if (array11[i] === array11[i2]) {
sameNumberLoop += 1;
if (sameNumberLoop >= 3) {
return true;
}
}
}
}
}
Instead of iterating multiple times, iterate just once, while counting up the number of occurrences in an object or Map:
let array11 = [1, 3, 2, 3, 5, 6, 7, 8, 9, 0, 1];
function checkDuplicateNumber (array) {
const counts = {};
for (const num of array) {
counts[num] = (counts[num] || 0) + 1;
if (counts[num] === 3) return true;
}
return false;
};
console.log(checkDuplicateNumber(array11));
console.log(checkDuplicateNumber([3, 1, 3, 5, 3]));
let array11 = [1, 3, 2, 3, 5, 6, 7, 8, 9, 0, 1]
let array22 = [1, 3, 2, 3, 5, 6, 7, 1, 9, 0, 1]
function checkDuplicateNumber(arr) {
const map = new Map()
return arr.some((v) => (map.has(v) ? (++map.get(v).count === 3) : (map.set(v, { count: 1 }), false)))
}
console.log(checkDuplicateNumber(array11))
console.log(checkDuplicateNumber(array22))
I found a great solution to get maximum Bi-valued slice of an array. And I want to apply it in Swift. However, I'm not familiar with JavaScript, I search it via Google, but no result yet.
Well, I simply want to know what does last.count & last.value means in below code. What they do with element last?
function getLongestSlice(array) {
var count = 0,
max = 0,
temp = [];
array.forEach(function (a) {
var last = temp[temp.length - 1];
if (temp.length < 2 || temp[0].value === a || temp[1].value === a) {
++count;
} else {
count = last.count + 1;
}
if (last && last.value === a) {
last.count++;
} else {
temp.push({ value: a, count: 1 });
temp = temp.slice(-2);
}
if (count > max) {
max = count;
}
});
return max;
}
console.log(getLongestSlice([58, 800, 0, 0, 0, 356, 8988, 1, 1])); // 4
console.log(getLongestSlice([58, 800, 0, 0, 0, 356, 356, 8988, 1, 1])); // 5
console.log(getLongestSlice([1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8])); // 10
Update:
Thanks. Now it is working in Swift.
import Foundation
struct Tmp {
var value: Int
var count: Int
}
func getLongestSlice(A: [Int]) -> Int {
var count = 0,
max = 0,
temp: [Tmp] = []
var last: Tmp = Tmp(value: 0, count: 0)
for a in A {
if temp.count != 0 {
last = temp[temp.count - 1]
}
print("last: \(last)")
if temp.count < 2 || temp[0].value == a || temp[1].value == a {
count += 1
} else {
count = last.count + 1
}
if last.value == a {
last.count += 1
// assign last.count to last element's count of temp array
temp[temp.count - 1].count = last.count
} else {
temp.append(Tmp(value: a, count: 1))
temp = Array(temp.suffix(2))
}
if count > max {
max = count
}
}
return max
}
getLongestSlice(A: [58, 800, 0, 0, 0, 356, 8988, 1, 1]) // return 4
getLongestSlice(A: [58, 800, 0, 0, 0, 356, 356, 8988, 1, 1]) // return 5
getLongestSlice(A: [1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 6, 2, 1, 8]) // return 10
getLongestSlice(A: [777]) // return 1
getLongestSlice(A: []) // return 0
From thread: Find Max Slice Of Array | Javascript
The temp array holds objects of the following shape:
{ count: ..., value: ... }
So the last refers to the last element of that array, while count and value are the properties of that element.
This is not related to JavaScript.
I have an array with numbers in the range of 0 - 100. I need to find all the same numbers and add 1 to them.
my code worked well with arrays like [100, 2, 1, 1, 0]
const findAndChangeDuplicates = (arr: any) => {
for (let i = arr.length - 1; i >= 0; i--) {
if (arr[i + 1] === arr[i] && arr[i] <= 5) {
arr[i] += 1;
} else if (arr[i - 1] === arr[i] && arr[i] >= 5) {
arr[i] -= 1;
findAndChangeDuplicates(arr);
}
}
return arr;
};
but when I came across this
[100, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0]
my code let me down.
Expected Result:
[100, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
Have any ideas?
An approach by using at least one loop from the end to adjust the values and if necessary another loop from the beginning to set the largest value to 100.
Both loops feature a value variable v. In the first loop, it starts with the last value of the array and increments its value and check is the item is smaller than this value.
If smaller, then the value is assigned, otherwise the actual value is taken for the next item.
if necessary, the other loop works in opposite direction and with a start value of 100 and checks if the item is greater than wanted and takes the smaller value, or the value is taken from the item.
The result is an array which has a gereatest value of 100 at start and goes until zero or greater to the end of the array.
function update(array) {
var i = array.length,
v = array[--i];
while (i--) if (array[i] < ++v) array[i] = v; else v = array[i];
if (array[0] > 100) {
v = 100;
for (i = 0; i < array.length; i++) {
if (array[i] > v) array[i] = v; else v = array[i];
v--;
}
}
return array;
}
console.log(update([100, 2, 1, 1, 0]));
console.log(update( [100, 100, 99, 86, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0]))
.as-console-wrapper { max-height: 100% !important; top: 0; }
The following assumes you want them ordered from highest to lowest, if not this might ba as well as useless to you.
The idea is to first create an Object to keep track of how many of each number exist. We then map each value by first checking whether it's unique and if not increasing it until we can't find any value inside the Object anymore. This will not neatly order the numbers by itself so we will have to sort afterwards.
let arr1 = [100, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0],
arr2 = [100, 2, 1, 1, 0];
const f = (arr) => arr.reduce((a,c) => (a[c] = (a[c] || 0) + 1, a),{}),
g = (arr, obj) => arr.map(v => {
if (obj[v] > 1) {
let i = 1;
obj[v] = obj[v] - 1;
while (obj[v + i]) {
i++;
}
obj[v + i] = (obj[v + i] || 0) + 1;
return v + i;
} else {
return v;
}
}).sort((a,b) => +b - +a);
console.log(g(arr1, f(arr1)))
console.log(g(arr2, f(arr2)))
Here is a verbose solution that will work with unordered arrays as well.
It's not efficient, neither brilliant, but it takes care of unordered arrays as well.
Basically, it takes advantage of reduce to collect all the occurrences of each element. Each time it finds more than one, it increases all the occurrences by 1 except the last one.
Next, it checks whether there still are duplicates. If there are, it repeats the process until none is found. Of course, it's not the cleverest approach, but it works.
// Increases all duplicates until there are no more duplicates.
const increaseDuplicates = (arr, index) => {
// Repeat the code until no duplicate is found
while (!noDuplicates(arr)) {
// Acquire all the occurrences of each item, keeping track of the index.
Object.entries(arr.reduce((acc, next, i) => {
acc[next] = acc[next] || [];
return acc[next].push(i), acc;
}, {})).forEach(([n, indexes]) => {
// for each value found, check whether it appears at least twice.
if (indexes.length > 1) {
// if it does, increase the value of every item but the last one.
for (var i = 0; i < indexes.length - 1; i++) {
arr[indexes[i]]++;
}
}
});
}
return arr;
};
// Asserts an array has no duplicates.
const noDuplicates = (arr) => [...new Set(arr)].length === arr.length;
const input = [100, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0];
console.log(increaseDuplicates(input));
const unorderedInput = [6,4,5,6,6,6,6,5,6,3,1,2,3,99,403,100, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0];
console.log(increaseDuplicates(unorderedInput));
You can use a forEach on your array to do this, using the 3rd parameter of the callback, the array itself, and a bit of recursivity
const increment_to_unicity = (value, index, self) => {
if (self.indexOf(value) !== index) {
self[index]++
increment_to_unicity(self[index], index, self)
}
return self[index];
}
arr = arr.map(increment_to_unicity).sort((a, b) => b - a);
Given this array:
[1, 1, 2, 1, 1, 1, 2, 3, 4, 4, 4, 6, 4, 4]
How can I efficently count the nearest same elements in array, the result I would expect is:
1 => 2,
2 => 1,
1 => 3,
2 => 1,
3 => 1,
4 => 3,
6 => 1,
4 => 2
I don't know how to formulate correctly the question but I think the example is pretty clear.
I tried using reduce to make more compact and elegant but I always get a value with total number of same value in array.
let result = testArray.reduce((allValues, value) => {
if(value in allValues){
allValues[value]++;
} else {
allValues[value] = 1;
}
return allValues;
}, {});
You could check the last element and if equal, increment count, if not push a new object to the result set.
var array = [1, 1, 2, 1, 1, 1, 2, 3, 4, 4, 4, 6, 4, 4],
count = array.reduce((r, a, i, aa) => {
if (aa[i - 1] === a) {
r[r.length - 1].count++;
} else {
r.push({ value: a, count: 1 });
}
return r;
}, []);
console.log(count);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Another solution using Array.prototype.reduce and a hash table - see demo below:
var array = [1, 1, 2, 1, 1, 1, 2, 3, 4, 4, 4, 6, 4, 4];
var result = array.reduce(function(hash){
return function(p,c,i){
hash[c] = (hash[c] || 0) + 1;
if(hash.prev && (hash.prev !== c || i == array.length - 1)) {
let obj= {};
obj[hash.prev] = hash[hash.prev];
delete hash[hash.prev];
p.push(obj);
}
hash.prev = c;
return p;
}
}(Object.create(null)),[]);
console.log(result);
.as-console-wrapper{top:0;max-height:100%!important;}
While Array.prototype.reduce is a nice function, it does have some compatibility issues.
Here is a solution using old school for loops:
var arr = [1, 1, 2, 1, 1, 1, 2, 3, 4, 4, 4, 6, 4, 4];
var results = [];
for (var indexA = 0; indexA < arr.length; indexA++) {
var a = arr[indexA];
if (results.length > 0) {
if (a == results[results.length - 1]["value"]) {
continue;
}
}
var r = { value: a, index: indexA, count: 0 };
for (var indexB = indexA; indexB < arr.length; indexB++) {
var b = arr[indexB];
if (a != b) {
break;
}
r.count++;
}
results.push(r);
}
console.log(results);
The posted solutions are fine but as noted by #Emil S. Jørgensen there might be compatibility issues using Array.reduce. Also, the suggested old school solution by #Emil S. Jørgensen uses two loops. If you want to have a slightly more efficient,simple and straightforward solution which will work in all browsers then use:
var arr = [1, 1, 2, 1, 1, 1, 2, 3, 4, 4, 4, 6, 4, 4];
var result = [];
var current = arr[0];
var count = 1;
for (var i = 1; i <= arr.length; i++)
{
if(arr[i] === current)
{
count+= 1;
}
else
{
var newObj = {};
newObj[current] = count;
result.push(newObj);
current = arr[i];
count = 1;
}
}
console.log(result); //prints the solution
Okay, I've only figured out how to get one mode out of the array..
But I want to get 2, 3 or more if they occur the same amount of times.
This is the code:
var frequency = {}; // array of frequency.
var maxFreq = 0; // holds the max frequency.
for (var i in array) {
frequency[array[i]] = (frequency[array[i]] || 0) + 1; // increment frequency.
if (frequency[array[i]] > maxFreq) { // is this frequency > max so far ?
maxFreq = frequency[array[i]]; // update max.
mode = array[i]; // update result.
}
}
So right now, if I've got a array = [3, 8, 3, 6, 1, 2, 9];
I get mode = 3;
But what I'm looking for is if array = [3, 6, 1, 9, 2, 3, 6, 6, 3, 1, -8, 7];
I want to get the mode = 3, 6;
The question doesn't state how to get the modes, but if we want them in an array, we could change the code like this:
function getModes(array) {
var frequency = []; // array of frequency.
var maxFreq = 0; // holds the max frequency.
var modes = [];
for (var i in array) {
frequency[array[i]] = (frequency[array[i]] || 0) + 1; // increment frequency.
if (frequency[array[i]] > maxFreq) { // is this frequency > max so far ?
maxFreq = frequency[array[i]]; // update max.
}
}
for (var k in frequency) {
if (frequency[k] == maxFreq) {
modes.push(k);
}
}
return modes;
}
alert(getModes([3, 6, 1, 9, 2, 3, 6, 6, 3, 1, -8, 7]));
function modeCount(data) {
let modecount = [];
let valueArr = [];
let dataSet = new Set(data);
for (const iterator of dataSet) {
const filteredNum = data.filter((num) => iterator === num);
modecount.push({
mode: iterator,
count: filteredNum.length
});
}
modecount.sort((a, b) => {
return b.count - a.count;
});
modecount.forEach(value => {
if (value.count === modecount[0].count) {
valueArr.push(value.mode);
}
});
return valueArr;
}
let ages = [3, 6, 1, 9, 2, 3, 6, 6, 3, 1, -8, 7]
console.log(modeCount(ages));