Given the array
arr = [1,3,34,44,4,45,6,76,9,98,23]I attempted this with a solution that works well for two-digit numbers, but struggles with larger numbers. Does anyone have a more generic approach to solving this problem?
You can organize an array containing strings along with the concatenated values of a and b, as well as the value of b followed by a. By calculating the delta between them for sorting, you can determine the order in which the two strings should be arranged to yield a higher combined value when joined later.
function getLargest(array) {
return array
.map(String)
.sort((a, b) => (b + a) - (a + b))
.join('');
}
console.log(getLargest([1, 3, 34, 44, 4, 45, 6, 76, 9, 98, 23]));A comparison is also made with a simple descending sorting method using strings, but it produces incorrect results (as shown in the second line).
function getLargest(array) {
return array
.map(String)
.sort((a, b) => (b + a) - (a + b))
.join('');
}
console.log(getLargest([1, 3, 34, 44, 4, 45, 6, 76, 9, 98, 23]));
console.log([1, 3, 34, 44, 4, 45, 6, 76, 9, 98, 23].sort().reverse().join(''));Create a custom comparison function named compare() to organize numbers in ascending order. Compare two numbers, first and second, by concatenating them in different orders (firstsecond and secondfirst). Determine the correct positioning based on which concatenated number is larger.
function sortNumberArray(array){
array.sort(function compare(first,second) {
var firstsecond ='' + first + second;
var secondfirst ='' + second + first;
return firstsecond>secondfirst ? -1:1;
})
}
function findLargestNum(array){
var largestNum = array.join('')
return largestNum
}
var numArray = [1,3,34,44,4,45,6,76,9,98,23]
sortNumberArray(numArray)
var finalResult = findLargestNum(numArray)
alert(finalResult)
Why not try this:
const concatenateNumbers = (...nums) => nums.sort((a, b) => ('' + b + a) - ('' + a + b)).join('');
console.log(concatenateNumbers(1, 3, 34, 44, 4, 45, 6, 76, 9, 98, 23)); //99876645444343231
console.log(concatenateNumbers(2, 20, 24, 6, 8)); //8624220Explanation:
The ... syntax allows for spreading iterable elements in places where arguments or array elements are expected. By concatenating an empty string ('') during the sorting process, we implicitly convert values to strings before comparison.
References:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort
function switchArrayElements(inputArr, indexA, indexB) {
let temp = inputArr[indexA];
inputArr[indexA] = inputArr[indexB];
inputArr[indexB] = temp;
}
function findLargestNumber(inputArr) {
let stringArr;
stringArr = inputArr.map(num => num.toString());
for(let i = 0; i < stringArr.length - 1; i++) {
for(let j = i + 1; j < stringArr.length; j++) {
if(parseInt(stringArr[i] + stringArr[j]) < parseInt(stringArr[j] + stringArr[i])) {
switchArrayElements(stringArr, i, j);
}
}
}
return stringArr.join('');
}
let inputArr = [1,3,34,44,4,45,6,76,9,98,23];
const result = findLargestNumber(inputArr);
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