I need to find missing numbers in array. There are many similar questions here & i checked it all, but can't find code who works for me. The close answer to my question was:
numbers.sort((a, b) => a - b);
miss = Array.from({
length: numbers[numbers.length - 1] - numbers[0] + 1
},
(_, idx) => numbers[0] + idx)
.filter(elem => !numbers.includes(elem));
console.log(miss.join(','));
It work correctly for this cases:
numbers = [4, 3, 2, 7, 8, 2, 3, 1] // print 5, 6
numbers = [1, 2, 3, 3, 5] // print 4
But for next case:
numbers = [1, 1, 1, 1, 1, 1, 1, 1] // returns nothing
I need to return 2, 3, 4, 5, 6, 7, 8
Edit: Need to find all the integers of [1, N] inclusive that do NOT appear in this array.
const missingValues = (numbers) => {
let size = numbers.length
let result = []
for (i = 1; i < size + 1; i++) {
if (!numbers.includes(i)) {
result.push(i)
}
}
console.log(result.join(','));
}
let numbers = [1, 1, 1, 1, 1, 1, 1, 1]
missingValues(numbers)
numbers = [4, 3, 2, 7, 8, 2, 3, 1]
missingValues(numbers)
This code will search for missing numbers, depending on the length of your given array.
function calculate(input) {
let result = list()
// We will start with i=1, since 0 isn't in our scope.
// The last value in our scope should be i <= input.length
for(i=1; i<= input.length;i++){
// Comparing if current number of for-loop is inside given array.
if (!input.includes(i)){
// If that's the case, add missing number to result-list.
result.push(i);
}
}
console.log(result)
}
You could take 1 as minimum value and check for the length of the array if the value exist or not, then add it to the missing values.
function getMissingValues(array) {
var v = 1,
i = array.length,
result = [];
while (i--) {
if (!array.includes(v)) result.push(v);
++v;
}
return result;
}
console.log(getMissingValues([4, 3, 2, 7, 8, 2, 3, 1])); // 5, 6
console.log(getMissingValues([1, 2, 3, 3, 5])); // 4
console.log(getMissingValues([1, 1, 1, 1, 1, 1, 1, 1])); // 2, 3, 4, 5, 6, 7, 8
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Checks an Array for missing numbers between 1 - 9
let ar = [1, 1, 1, 1, 1, 1, 1, 1]
let nope = []
for(i=1;i<=9;i++){
if (!ar.includes(i)) nope.push(i)
}
console.log(nope)
console.log(nope.join(", "))
Related
for learning purposes I want to get every second element of an array. I succeeded with a for loop:
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
function filterEverySecond(arr) {
let everySecondEl = [];
for (let i = 0; i < arr.length; i += 2) {
everySecondEl.push(arr[i]);
}
return everySecondEl;
}
console.log({
numbers,
result: filterEverySecond(numbers)
});
Now I want to achieve the same without a for loop, but by using array methods (forEach, filter, map or reduce). Can anyone recommend which method would be best here?
you can do it easily with filter
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const filtered = numbers.filter((_, i) => i % 2 === 0)
console.log(filtered)
you just filter out the elements that have a odd index
You can use filter and check if the index of the current array element is divisible by 2:
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
console.log({
numbers,
result: numbers.filter((_, index) => index % 2 === 0)
});
You can use filter for index.
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
console.log({
numbers,
result: numbers.filter((n,i)=>i%2==0)
});
You could use a for each loop like so to get the same desired output:
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const result = [];
numbers.forEach(number => {
if (number % 2 != 0) {
result.push(number);
}
});
console.log(numbers);
console.log(result);
The modulus operator returns the remainder of the two numbers when divided. For example: 1 % 2 will return 1 as the remainder. So in the if statement we are checking if the number is not divisible by 2.
I have two arrays (X and Y) and I need to create array Z that contains all elements from array X except those, that are present in array Y p times where p is a prime number.
I am trying to write this in JS.
For Example:
Array X:
[2, 3, 9, 2, 5, 1, 3, 7, 10]
Array Y:
[2, 1, 3, 4, 3, 10, 6, 6, 1, 7, 10, 10, 10]
Array Z:
[2, 9, 2, 5, 7, 10]
So far I have this:
const arrX = [2, 3, 9, 2, 5, 1, 3, 7, 10]
const arrY = [2, 1, 3, 4, 3, 10, 6, 6, 1, 7, 10, 10, 10]
const arrZ = []
const counts = [];
// count number occurrences in arrY
for (const num of arrY) {
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
// check if number is prime
const checkPrime = num => {
for (let i = 2; i < num; i++) if (num % i === 0) return false
return true
}
console.log(counts[10]);
// returns 4
Any hint or help appreciated. Thanks!
You're on the right track. counts should be an object mapping elements in arrY to their number of occurrences. It's easily gotten with reduce.
The prime check needs a minor edit, and the last step is to filter arrX. The filter predicate is just a prime check on the count for that element.
// produce an object who's keys are elements in the array
// and whose values are the number of times each value appears
const count = arr => {
return arr.reduce((acc, n) => {
acc[n] = acc[n] ? acc[n]+1 : 1;
return acc;
}, {})
}
// OP prime check is fine, but should handle the 0,1 and negative cases:
const checkPrime = num => {
for (let i = 2; i < num; i++) if (num % i === 0) return false
return num > 1;
}
// Now just filter with the tools you built...
const arrX = [2, 3, 9, 2, 5, 1, 3, 7, 10]
const arrY = [2, 1, 3, 4, 3, 10, 6, 6, 1, 7, 10, 10, 10]
const counts = count(arrY);
const arrZ = arrX.filter(n => checkPrime(counts[n]));
console.log(arrZ)
With an array of: [1, 2, 3, 4, 5, 6]
I would like to delete between 2 indices such as 2 and 4 to produce [1, 2, null, null, 5, 6]. What's the easiest way to do this?
Hopefully better than this:
const array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
let i = 2;
const rangeEnd = 9;
while (i < rangeEnd) {
delete array[i];
i++;
}
console.log(array)
If you want to use some native API you can actually do this with splice(). Otherwise, you should iterate a for loop through your array and change the value in each iteration.
Here is an example of how it would be done:
const array = [1, 2, 3, 4, 5, 6]
array.splice(3, 2, null, null) // the First element is beginning index and the second is count one will indicate how many indexes you need to traverse from the first one, then you should provide replace element for each of them.
console.log(array)
Note: For more info about it you can read more here.
There is a possible workaround for large scale replacement, so I will give it a touch here:
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
var anotherArr = Array(2).fill(null); // or you can simply define [null, null, ...]
Array.prototype.splice.apply(arr, [3, anotherArr.length].concat(anotherArr));
console.log(arr);
As you mean the range (2, 4] so you can follow this:
The range is: lower limit exclusive and the upper limit inclusive.
const arr = [1, 2, 3, 4, 5, 6];
const deleteRange = (arr, f, t) => {
return arr.map((item, i) => {
if (i + 1 > f && i + 1 <= t) {
return null;
}
return item;
})
}
console.log(deleteRange(arr, 2, 4));
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]))
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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);
previousValue currentValue index array return value
first call 0 1 1 [0, 1, 2, 3, 4] 1
second call 1 2 2 [0, 1, 2, 3, 4] 3
third call 3 3 3 [0, 1, 2, 3, 4] 6
fourth call 6 4 4 [0, 1, 2, 3, 4] 10
I want 1,3,6,10 in an array not the return total 10. So to return each call
You can push the return value into an array, like this. It goes against functional programming since it mutates results as a side effect. But it does meet your needs.
var array = [0, 1, 2, 3, 4];
var results = [];
array.reduce(function(previousValue, currentValue) {
var newValue = previousValue + currentValue;
results.push(newValue);
return newValue;
});
// result is 1,3,6,10
alert(results);
Don't use reduce for this. Slice the array, shift a value to start a subtotal, then use map.
var arr = [0, 1, 2, 3, 4], output = arr.slice(), subtotal = output.shift()
output = output.map(function(elem) { return subtotal += elem })
// output is [1, 3, 6, 10]
Edit - Actually, this could work fine with reduce, and even more concise than the above:
var arr = [0, 1, 2, 3, 4]
arr.reduce(function(a, b, ndx) { return a.length ? a.concat(a[ndx - 2] + b) : [a + b]})
// returns [1, 3, 6, 10]