The code is to return the lowest index in an array when the value in the array is the same as the index. If there are no matches i should return -1. For example:
indexEqualsValue([-8,0,2,5])
output: 2 //array[2] == 2
indexEqualsValue([-1,0,3,6])
output: -1 //no matches
The code works when there are no matches or if the length of the array is zero, but not at other times. I think the problem is the first condition in my if statement. I don't necessarily want an answer, more tips on what I should check/rewrite.
Thanks!
function indexEqualsValue(a) {
return a.reduce((acc, currV, currI) => {
if (currI === currV) {
return currV;
}
return -1;
}, 0);
}
You could just find the index with Array#findIndex.
const indexEqualsValue = array => array.findIndex((v, i) => v === i);
console.log(indexEqualsValue([-8, 0, 2, 5])); // 2
console.log(indexEqualsValue([-1, 0, 3, 6])); // -1
some exits when it matches, so you can use it to quickly find what you need:
const indexEqualsValue = array => {
let match;
const didMatch = array.some((v, i) => {
match = i;
return v === i;
})
return didMatch ? match : -1;
}
console.log(indexEqualsValue([-8,0,2,5]))
console.log(indexEqualsValue([-8,0,2,5,0]))
console.log(indexEqualsValue([-1,0,3,6]))
nina-scholz's answer is better, the only advantage of using some over findIndex is that some is supported in ie whereas it seems findIndex is not.
for(i = 0,c=0;i < arr.length; i++ ) {
if(arr[i] == i) {
c = 1;
break;
}
}
if( c == 0 ) {
print(c);
} else {
print (i);
}
Related
I am trying to compare two arrays( containing 3 integers) and return a hint message array that conform to the logic
-Push “Almost” when 1 digit and position match in array
-push “not quite” when 1 digit matches but different position
-push “incorrect “ when no digits match
push “correct” When exact match
Example of arrays :
Array1 = [2,7,6]
ReferenceArray= [2,9,7]
Hint= [“Almost”, “Not Quite”];
Code I have so far:
function check( array1, referenceArray ) {
let hint=[];
for(i=0;i<referenceArray.length;i++){
for (j=0;j<Array1.length;j++){
//value and position match
if ((referenceArray[i] && reference.indexOf[i]) === (Array1[j] && Array1.indexOf[j])) {
return hint.push('almost');
}
//value matches but not position
else if(( referenceArray[i] ===Array1[j]) && !(referenceArray.indexOf[i]===Array1.indexOf[j] )){
return hint.push('not quite');
}
}// end of Array1 iteration
} // end of reference interation
// if all values and position match
if(referenceArray===Array1){
return hint.push("correct");
}
//if no values match
else if (referenceArray!==Array1){
return hintArray.push("incorrect");
}
I would use some built in Array methods to help achieve this: every(), map() and findIndex().
I generally avoid using .push() because it mutates the array. Immutable code is nice to read 😉
const check = (array, referenceArray) => {
if (array.every((val, index) => val === referenceArray[index] )) {
return ['Correct']
}
const allHints = array.map((val, index) => {
const refArrayIndex = referenceArray.findIndex(refVal => val === refVal);
if (refArrayIndex === index) {
return 'Almost'
}
if (refArrayIndex !== -1) {
return 'Not Quite'
}
return undefined
});
const hints = allHints.filter((hint) => hint !== undefined);
if (hints.length > 0) {
return hints;
}
return ['Incorrect']
};
const hints = check([2,7,6],[2,9,7]);
console.log('hints', hints)
I did this code, tell me if it works or not 😁
const array1 = [2,7,6]
const ReferenceArray = [2,9,7]
function compareArrays(arr){
let perfect = true
for(let i = 0; i < ReferenceArray.length; i++){
if(ReferenceArray[i] != arr[i]) {
perfect = false
break
}
}
if(perfect) return 'correct'
let hint = []
for(let i = 0; i < ReferenceArray.length; i++){
if(arr[i] == ReferenceArray[i]) hint.push('Almost')
else if(ReferenceArray.includes(arr[i])) hint.push('Not Quite')
}
if(hint.length > 0) return hint
return 'incorrect'
}
console.log(compareArrays(array1))
Have to create a function that return the sum of the element in the array but if the array is
["a","b","c"] // output : abc
So far I have
function calculateSumRecursion(array) {
//your code
if (array.length === 0 ) {
return 0
}
return array[0] + calculateSumRecursion(array.slice(1))
}
I found out how to calculate the sum of all numbers using recursion but when it's an array of string like
array = ["a","b","c"]
it returns me
// abc0
because of the if statement.. is there any way to say
if (array.length === 0) return nothing instead of a 0 (that work only when it's an array of number?)
You just need to return the only value in the array when the length is 1, rather than waiting until you get a length of 0. That way you are always summing compatible types (numbers or strings). Note that you still need a test for a 0 array length in case the function gets called with an empty array. In this case you need to choose what to return; as requested, it is 0.
function calculateSumRecursion(array) {
if (array.length === 0) {
return 0;
}
if (array.length === 1) {
return array[0];
}
return array[0] + calculateSumRecursion(array.slice(1))
}
console.log(calculateSumRecursion([1, 2, 3, 4, 5]));
console.log(calculateSumRecursion(['a', 'b', 'c']));
console.log(calculateSumRecursion([]));
let arr = [1,2,3,4,5] // output : abc
let sum = calculateSumRecursion(arr);
function calculateSumRecursion (arr) {
return arr.length ? arr.pop() + calculateSumRecursion(arr) : 0;
}
Slice version
let arr = [1,2,3,4,5] // output : abc
let sum = calculateSumRecursion(arr);
function calculateSumRecursion (arr) {
return arr.length ? arr[0] + calculateSumRecursion(arr.slice(1)) : 0;
}
Change return 0 to return "" which will add an empty string to the sum.
You have returned 0 when the array is empty.
Now, you are doing string operations so it is needed to return empty value (not zero) so it will be affordable to return "".
function calculateSumRecursion(array) {
return array.length === 0 ? "" : array[0] + calculateSumRecursion(array.slice(1));
}
There's a way easier way to do this:
function calculateSumRecursion(array) {
var out = array[0];
for (let i = 1; i < array.length; i++) {
out = out + array[i];
}
return out;
}
Return empty string on recursion base case. Just replace your return 0 to return ''.
const array = ['a', 'b', 'c'];
function calculateSumRecursion(array) {
if (array.length === 0) {
return '';
}
return array[0] + calculateSumRecursion(array.slice(1));
}
console.log(calculateSumRecursion(array));
If you are want to work with number also then check array length for zero as well as one.
const array = ['a', 'b', 'c', 'e'];
const array2 = [];
const array3 = [1, 2, 3];
function calculateSumRecursion(array) {
const rec =
array.length === 1
? array[0]
: array.length >= 1 && array[0] + calculateSumRecursion(array.slice(1));
return array.length === 0 ? 0 : rec;
}
console.log(calculateSumRecursion(array));
console.log(calculateSumRecursion(array2));
console.log(calculateSumRecursion(array3));
I'm just beginning with programming using javascript and I need to practice some questions to get EXP with the logic of code build.
I got this question for homework but I can't make it work for some reason, even though it seems "logic" to me.
Check if an array is descending, ascending or not sorted using loops.
I'm just a noob so please try and help me figure this out as I only got to loops in my studies (:
this is the code I wrote:
var array = [1, 2, 3, 7 ];
var d = 0;
var c =0 ;
var b = 1;
var a = 0;
for (var i = 1; i <= array.length; i++)
{
if (array[c]<array[b] && a!== -1 ){
d = -1;
c =c+1;
b = b+1;
if(c==array.length){
console.log("asc");
break;
}else{
continue;
}
} else if (array[c]>array[b] && d!==-1 ){
a = -1;
d= d+1;
b = b+1;
if(i=array.length){
console.log("dsc");
break;
}else{continue;}
} else{
console.log("unsorted array");
break;
}
}
Array.prototype.every passes its predicate an index, which you can use to get an element’s predecessor:
function isAscending(arr) {
return arr.every(function (x, i) {
return i === 0 || x >= arr[i - 1];
});
}
Here, we’re checking that every item (x) is greater than or equal to the item before it (arr[i - 1]) or has no item before it (i === 0).
Flip >= to <= for isDescending.
"Check if an array is descending, ascending or not sorted using loops"
// define the array
var array = [1,2,3,7];
// keep track of things
var isDescending = true;
var isAscending = true;
// we're looking ahead; loop from the first element to one before the last element
for (var i=0, l=array.length-1; i<l; i++)
{
////////////////////////////////////////////////////////////
// log to the console to show what's happening for each loop iteration
// this is the ith iteration
console.log("loop iteration %s", i);
// breaking isDescending down:
// is this value greater than the next value?
console.log("A: (%s > %s) = %s", array[i], array[i+1], (array[i] > array[i+1]));
// have all values been descending so far?
console.log("B: isDescending: %s", isDescending);
// if this value is greater than the next and all values have been descending so far, isDescending remains true. Otherwise, it's set to false.
console.log("are A and B both true? %s", (isDescending && (array[i] > array[i+1])));
// add a line break for clarity
console.log("");
////////////////////////////////////////////////////////////
// true if this is greater than the next and all other so far have been true
isDescending = isDescending && (array[i] > array[i+1]);
// true if this is less than the next and all others so far have been true
isAscending = isAscending && (array[i] < array[i+1]);
}
if (isAscending)
{
console.log('Ascending');
}
else if (isDescending)
{
console.log('Descending');
}
else
{
console.log('Not Sorted');
}
This is a question that requires some sort of loop, with several if statements because there are several cases you need to tackle:
Array is empty or has only one element.
All items in the array are equal
Array is ascending - delta between 2 elements > 0, but some deltas may be 0
Array is descending - delta between 2 elements < 0, but some deltas may be 0
Not sorted - some deltas are > 0, and some are < 0
Depending on how the sorted is defined in the question, cases 1 & 2 might be regarded as unsorted as well.
function findSortOrder(arr) {
if(arr.length < 2) { // case 1
return 'not enough items'; // can also be 'unsorted'
}
var ascending = null;
var nextArr = arr.slice(1); // create an array that starts from the 2nd element of the original array
for(var i = 0; i < nextArr.length; i++) {
if (nextArr[i] === arr[i]) { // neutral - do nothing
} else if(ascending === null) { // define the the direction by the 1st delta encountered
ascending = nextArr[i] > arr[i];
} else if (ascending !== nextArr[i] > arr[i]) { // case 5
return 'unsorted';
}
}
if(ascending === null) { // case 2
return 'all items are equal'; // can also be 'unsorted'
}
return ascending ? 'ascending' : 'descending'; // cases 3 & 4
}
console.log(findSortOrder([1])); // case 1
console.log(findSortOrder([1, 1, 1, 1])); // case 2
console.log(findSortOrder([1, 1, 2, 3, 7, 7])); // case 3
console.log(findSortOrder([7, 2, 2, 1])); // case 4
console.log(findSortOrder([7, 2, 1, 3, 2, 1])); // case 5
You could use a copy from the second element and check the predecessor for the wanted sort order.
function checkArray(array) {
var aa = array.slice(1);
if (!aa.length) {
return "Just one element";
}
if (aa.every((a, i) => array[i] > a)) {
return "Ascending";
}
if (aa.every((a, i) => array[i] < a)) {
return "Descending";
}
return "Unsorted";
}
console.log(checkArray([1, 2, 3, 4, 5]));
console.log(checkArray([5, 4, 3, 2, 1]));
console.log(checkArray([3, 1, 4, 2, 5]));
console.log(checkArray([42]));
function isAscending(arr = []) {
for (let i = 0; i < arr.length; i++) {
if (arr[i + 1] <= arr[i]) {
return false;
}
}
return true;
}
function isAscending(arr) {
return arr
.slice(1)
.every((num,i) => num >= arr[i]);
}
console.log(isAscending([1,2,3])); // true
console.log(isAscending([12,38,25])); // false
console.log(isAscending([103,398,52,629])); // false
arr.slice(1) --> allows us to start iteration at index 1 instead of 0
we'll iterate over "arr" with "every" & compare the current "num" against the previous "num" with "arr[i]"
function findOrder(array) {
var asc = true;
var desc = true;
if(array.length < 2){
return 'array is too small'
}
for(var i=1, len=array.length;i<len;i++){
//if current element is bigger than previous array is not descending
if(array[i]>array[i-1]){
desc = false;
//if current element is smaller than previous array is not ascending
}else if(array[i]<array[i-1]){
asc = false;
}
if(!asc && !desc){
return 'not sorted'
}
}
if(asc && desc){
return 'array values are equal'
}else if (asc){
return 'array is ascending'
}else {
return 'array is descending'
}
}
Actually we may do even more by creating an Array method like natureOf which can tell us more about the nature of the array than just ascending, descendig or flat. Array.natureOf() shall give us a value between -1 and 1. If it is -1 the array is fully descending and 1 would mean fully ascending of course. Any value inbetween would give us the inclination of the array. As you would guess 0 would mean totally random or flat. (it's fairly easy to insert the logic to distinguish random from flat if that's also needed)
Array.natureOf = function(a){
var nature = a.reduce((n,e,i,a) => i && a[i-1] !== e ? a[i-1] < e ? (n.asc++, n)
: (n.dsc++, n)
: n, {asc:0, dsc:0});
return (nature.asc - nature.dsc) / (a.length-1);
};
var arr = [1,2,3,4,5,6,7,8,9,10,7,11,13,14,15],
brr = Array.from({length:2000000}, _ => ~~(Math.random()*1000000000));
console.log(`The nature of "arr" array is ${Array.natureOf(arr)}`);
console.log(`The nature of "brr" array is ${Array.natureOf(brr)}`);
console.log(checkSort([1, 2, 3, 3, 4, 5]));
console.log(checkSort([5, 4, 3, 2, 1, 1]));
console.log(checkSort([2, 5, 8, 9, 4, 6]));
function checkSort(arr){
var isDescending, isAscending;
isDescending = isAscending = true;
const len = arr.length - 1;
for (var index = 0 ; index < len ; index++){
if(isAscending)
isAscending = arr[index] <= arr[index + 1];// '<=' so as to check for same elements
if(isDescending)
isDescending = arr[index] >= arr[index + 1];//'<=' so as to check for same elements
}
var result = "Array is ";
if (isAscending)
return result.concat("sorted in ascending order");
if (isDescending)
return result.concat("sorted in descending order");
return result.concat("not sorted");
const isAscending=(arr)=>{
for(let i=0;i<arr.length;i++){
if(arr[i+1]<arr[i]){ return false } } return true }
I'm trying to use the reduce() method to find one skipped (missing) number in an array of (sometimes almost) consecutive numbers. Only one number will be missing at most.
This is my codepen: http://codepen.io/PiotrBerebecki/pen/zBrRVd
For example,
findMissing([1,2,3,5]) should return 4
findMissing([1,2,3,4]) should return undefined
findMissing([2,3,4,6]) should return 5
findMissing([2,4,5,6]) should return 3
The code that I developed seems to work fine if there is indeed a number that was skipped. But it returns an undesired value if all numbers are present. Would you know how to fix it?
My JS code:
function findMissing(arr) {
return arr.reduce(function(prev, curr) {
if (curr - prev !== 1) {
return curr - 1;
}
});
}
// This should return 4, and it indeed returns 4
console.log( findMissing([1,2,3,5]) );
// This should return 'undefined', but it returns 3
console.log( findMissing([1,2,3,4]) );
// This should return 5, and it indeed returns 5
console.log( findMissing([2,3,4,6]) );
UPDATE 1:
Based on the answers below, the following code delivers the desired outcome using the reduce() method:
// ****SOLUTION:****
function findMissing2(arr) {
return arr.reduce(function(prev, curr, index, array) {
if (curr === index + array[0]) {
return prev;
} else {
return index + array[0]++;
}
}, void 0);
}
console.log( findMissing2([1,2,3,4]) ); // Undefined
console.log( findMissing2([1,2,3,5]) ); // 4
console.log( findMissing3([2,3,4,6]) ); // 5
console.log( findMissing2([2,3,4,5]) ); // Undefined
console.log( findMissing2([2,4,5,6]) ); // 3
I would do this job as follows;
var a1 = [1,2,3,5],
a2 = [2,3,4,5],
a3 = [2,4,5,6],
res1 = a1.reduce((p,c,i,a) => c == i+a[0] ? p : i + a[0]++, void 0),
res2 = a2.reduce((p,c,i,a) => c == i+a[0] ? p : i + a[0]++, void 0),
res3 = a3.reduce((p,c,i,a) => c == i+a[0] ? p : i + a[0]++, void 0);
console.log(res1);
console.log(res2);
console.log(res3);
Note: void 0 is a very safe undefined value in JS. The above code will mutate the tested array. You might prefer to call like a1.slice().reduce... if you want to keep the tested array as it is.
Instead of reduce you should use for loop here
function findMissing(arr) {
var r = [];
for (var i = arr[0]; i <= arr[arr.length - 1]; i++) {
if (arr.indexOf(i) == -1) r.push(i);
}
return r;
}
console.log(findMissing([1, 2, 3, 5]));
console.log(findMissing([1, 3, 6]));
console.log(findMissing([10, 13, 16]));
Your reduce closure/callback function needs to return the value to be used as the next prev in the next iteration.
Because of this, in the second example, the first iteration returns undefined, as it does not enter the if block. The second iteration is passed the parameters of undefined, 3, where undefined - 3 !== 1 so it returns 2.
This propagates the way up your iterations, until it returns 3.
As such, I'm struggling to think of a way your reduce function could be adapted to correct this.
Perhaps using a simple for loop might be a bit more robust?
function findMissing(arr) {
for(var i = 1; i < arr.length; i++) {
if(arr[i] - arr[i-1] !== 1) {
return arr[i]-1;
}
}
}
As I said in comments, if you are looking for efficiency you could do it with recursion:
function findMissing(arr) {
if (arr.length === 1) return;
if(arr[1] - arr[0] !== 1){
return arr[0];
}
else{
return findMissing(arr.slice(1,arr.length));
}
}
Or even with a while loop:
function findMissing(arr) {
var i = 0;
while (arr[i+1] - arr[i] === 1) {
i++;
}
if (i < arr.length-1) return arr[i];
}
var findMissing = function (list) {
var expected_sum = (list[0] + list[list.length - 1]) * (list.length + 1) / 2;
var sum = list.reduce((a,b)=>a+b);
return expected_sum - sum;
}
console.log(findMissing([-5,-1,1,3,5,7,9,11]))
I need to test whether each item in an array is identical to each other. For example:
var list = ["l","r","b"]
Should evaluate as false, because each item is not identical. On the other hand this:
var list = ["b", "b", "b"]
Should evaluate as true because they are all identical. What would be the most efficient (in speed/resources) way of achieving this?
In ES5, you could do:
arr.every(function(v, i, a) {
// first item: nothing to compare with (and, single element arrays should return true)
// otherwise: compare current value to previous value
return i === 0 || v === a[i - 1];
});
.every does short-circuit as well.
function identical(array) {
for(var i = 0; i < array.length - 1; i++) {
if(array[i] !== array[i+1]) {
return false;
}
}
return true;
}
You could always do a new Set, and check the length.
var set1 = [...new Set(list)].length === 1;
The one line answer is:
arr.every((val, ind, arr) => val === arr[0]);
You can look into Array.every for more details.
Note:
Array.every is available ES5 onwards.
This method returns true for any condition put on an empty array.
Syntax: arr.every(callback[, thisArg]) or array.every(function(currentValue, index, arr), thisValue)
It does not change the original array
The execution of every() is short-circuited. As soon as every() finds an array element that doesn't match the predicate, it immediately returns false and doesn't iterate over the remaining elements
arr.every(i=>i==arr[0]) //will return true if all items in arr are identical
function matchList(list) {
var listItem = list[0];
for (index in list) {
if(list[index] != listItem {
return false;
}
}
return true;
}
var list = ["b", "b", "b"];
var checkItem = list[0];
var isSame = true;
for (var i = 0; i < list.length; i++) {
if (list[i] != checkItem) {
isSame = false;
break;
}
}
return isSame;
function identical(array) {
// a variable holding standard value
//against this standard value we are examining the array
var standard = array[1];
for (var i = 0; i < array.length; i++) {
if (array[i] !== standard) {
return false;
}
}
return true;
}
identical([1, 1, 1, 1, 1]); //return true
identical(['a', 'a', 'a']); //return true
identical(['a', 'a', 'b'])
function identical(array) {
// a variable holding standard value
//against this standard value we are examining the array
var standard = array[1];
for (var i = 0; i < array.length; i++) {
if (array[i] !== standard) {
return false;
}
}
return true;
}
identical([1, 1, 1, 1, 1]); //return true
identical(['a', 'a', 'a']); //return true
identical(['a', 'a', 'b'])
My suggestion would be to remove duplicates (check out Easiest way to find duplicate values in a JavaScript array), and then check to see if the length == 1. That would mean that all items were the same.
function allEqual(list)
{
if(list.length == 0 || list.length == 1)
{
return true;
}
for (index in list) {
if(list[index] != list[index+1] {
return false;
}
}
return true;
}