Develop Reference

java.lang.StackOverflowError on factorial function being called by another recursive function

I have a factorial function on my program that works fine until i try to execute the function deleteRepeated(), the console is telling me that the error is in the return of the factorial function, maybe it's being called by a single function too many times in a short period of time? I've been stuck for hours.
import java.util.Scanner;
public class ex9 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
int[] newArr = new int[n - repeated(arr)];
int[] finalArr = deleteRepeated(arr, newArr);
for (int a : finalArr) {
System.out.println(a);
}
}
public static long factorial(int n) {
if (n == 0)
return 1;
return (n * factorial(n - 1));
}
public static int repeated(int arr[]) {
int n = arr.length;
int mix = (int) (factorial(n) / (2 * factorial(n - 2)));
int i = 0;
int k = 0;
int rep = 0;
int a = -100;
while (i < mix) {
for (int j = k + 1; j < n; j++) {
if (arr[k] == arr[j] && a != j) {
a = j;
rep += 1;
}
i++;
}
k++;
}
return rep;
}
public static int[] deleteRepeated(int arr[], int newArr[]) {
int n = arr.length;
int rep = repeated(arr);
int i = 0;
int k = 0;
int a = -100;
while (i < newArr.length) {
for (int j = k + 1; j < n; j++) {
if (arr[k] == arr[j] && a != arr[k]) {
a = arr[j];
newArr[k] = arr[k];
}
i++;
}
k++;
}
rep = repeated(newArr);
if (rep > 0) {
int[] newArr2 = new int[newArr.length - rep];
deleteRepeated(newArr, newArr2);
}
return newArr;
}
}
Only thing i could do to avoid the error was stopping the function from executing :/, maybe it has to do with how i'm re-calling it at the end of each execution...? is what i did allowed?

So, deleteRepeated is all messed up. The issue is deleteRepeated does not actually remove duplicate elements, so the check for the base case of recursion always fails. I'm not sure why you're using recursion here anyway; if the while loop worked properly, it could remove all duplicates without need for recursion.
It appears that you copy-pasted the implementation of repeated into deleteRepeated, and you replaced the logic for handling repeated elements with logic that handles non-repeated elements.
Here is how I would implement the method:
public static int deleteRepeated(int arr[], int newArr[]) {
int n = 0;
for(int i = 0; i < arr.length; i++) {
boolean unique = true;
for(int j = 0; j < n; j++)
unique = unique && newArr[j] != arr[i];
if(unique)
newArr[n++] = arr[i];
if(n >= newArr.length) break;
}
return n;
}

Two Sum II - Input array is sorted

Leetcode #167 is almost same as #1, but why I cannot only add a if condition?
Q: Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.
Note:
Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
The sum of 2 and 7 is 9.
Therefore index1 = 1, index2 = 2.
My code:
class Solution {
public int[] twoSum(int[] numbers, int target) {
for (int i = 1; i < numbers.length; i++) {
for (int j = i + 1; j < numbers.length; j++) {
if (numbers[j] == target - numbers[i]) {
if(numbers[i] < numbers[j])
return new int[] { i, j };
}
}
}
return null;
}
}
Why I always return null? where is my mistake? How to fix it?

Because the question says array starts from 1 does not mean array starts from 1 in java.If you want to return i,j as non-zero you should go from 1 to length+1 and then inside the conditions you should check indexes as i-1,j-1 or just start from 0 and return i+1,j+1.
class Solution {
public int[] twoSum(int[] numbers, int target) {
for (int i = 1; i < numbers.length+1; i++) {
for (int j = i + 1; j < numbers.length+1; j++) {
if (numbers[j-1] == target - numbers[i-1]) {
if(numbers[i-1] < numbers[j-1])
return new int[] { i, j };
}
}
}
return null;
}
}
or you can do,
class Solution {
public int[] twoSum(int[] numbers, int target) {
for (int i = 0; i < numbers.length; i++) {
for (int j = i + 1; j < numbers.length; j++) {
if (numbers[j] == target - numbers[i]) {
if(numbers[i] < numbers[j])
return new int[] { i+1, j+1 };
}
}
}
return null;
}
}

https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
[question]: 167. Two Sum II - Input array is sorted
Using the two-pointer technique:-
class Solution {
public int[] twoSum(int[] numbers, int target) {
if (numbers == null || numbers.length == 0)
return null;
int i = 0;
int j = numbers.length - 1;
while (i < j) {
int x = numbers[i] + numbers[j];
if (x < target) {
++i;
} else if (x > target) {
j--;
} else {
return new int[] { i + 1, j + 1 };
}
}
return null;
}
}

I have modified your code and added code comments on why your previous code has errors. Refer to code below for details.
public class Main {
public static void main(String[] args) {
int target = 9;
int[] numbers = new int[] { 2, 7, 11, 15 };
int[] result = twoSum(numbers, target);
if (result != null) {
System.out
.println("The sum of " + numbers[result[0]] + " and " + numbers[result[1]] + " is " + target + ".");
System.out.println("Therefore index1 = " + (result[0] + 1) + ", index2 = " + (result[1] + 1));
} else {
System.out.println("No Solution found!");
}
}
public static int[] twoSum(int[] numbers, int target) {
for (int i = 0; i < numbers.length; i++) { // array index starts at 0
for (int j = i + 1; j < numbers.length; j++) {
if (numbers[j] + numbers[i] == target) { // add the current numbers
// if (numbers[i] < numbers[j]) // not needed
return new int[] { i, j };
}
}
}
return null;
}
}
Sample input:
numbers = [2, 7, 11, 15];
Sample output:
The sum of 2 and 7 is 9.
Therefore index1 = 1, index2 = 2

You are starting first for-loop with i = 0, what you should do is start it with i = 1.
Working code:
public class Solution
{
public static void main(String[] args)
{
int[] num = {2,7,11,5};
int n = 13;
int[] answer = new int[2];
answer = twoSum(num,n);
if(answer != null)
for(int i=0;i<2;i++)
System.out.printf( answer[i] +" ");
}
public static int[] twoSum(int[] numbers, int target)
{
for (int i = 0; i < numbers.length; i++)
{
for (int j = i + 1; j < numbers.length; j++)
{
if (numbers[j] == target - numbers[i])
{
if(numbers[i] < numbers[j])
return new int[] { i+1, j+1};
}
}
}
return null;
}
}
Note: I have placed an IF before FOR in main() so that if we find no such integers that adds up to give target integer, it'll not throw a NullPointerException.

This is a better solution as it's much faster and covers all test cases as well:
class Solution {
public int[] twoSum(int[] numbers, int target) {
int l = 0, r = numbers.length - 1;
while (numbers[l] + numbers[r] != target) {
if (numbers[l] + numbers[r] > target)
r--;
else
l++;
if (r == l) return new int[]{};
}
return new int[]{l + 1, r + 1};
}
}

public int[] twoSum(int[] nums, int target) {
int start = 0, end = nums.length -1;
while (start < end){
if (nums[start]+ nums[end]== target)
return new int []{start+1, end+1};
if (nums[start]+ nums[end]> target){
end--;}
else if (nums[start]+ nums[end]< target){
start++;
}
}
return null;
}

Counting number of duplicates in a given array

Input: 1,4,2,6,7,5,1,2
Output:2
Counting the number of duplicated numbers in a given array for java. I first sorted the array and then counted duplicates. It's showing me error that variable c is not used and that this method should return value of int.
public class Duplicates
public static void main(String[] args) {
int[]list;
int[]c;
int[] c = new int[list.length];
int temp;
for (int i = 0; i < list.length - 1; i++) {
for (int j = i + 1; j < list; j++) {
if (list[I] > list[j]) {
temp = list[i];
list[i] = list[j];
list[j] = temp;
c = list;
}
}
}
int n = 0;
int counter = 0;
int a = -1;
for (int i = 0; i < c.length; ++i) {
if (c[i] == a) {
++n;
if (n == 1) {
++counter;
if (counter == 1) {
System.out.print(c[i]);
} else {
System.out.print("," + c[i]);
}
}
} else {
a = c[i];
n = 0;
}
}
System.out.println("\nNumber of Duplicated Numbers in array:" + counter);
}
}

It's showing me error that variable c is not used
This should be a warning. So the code should still run correctly even with this is showing.
this method should return value of int
This is a compilation error and since you are not returning any int array at the end of the method, your method's return type should be void. You should change your method signature as below,
public static void c(int[] list)
Otherwise you will need to return an int array at the end of your method.
After fixing your code,
public class Duplicates {
public static void main(String[] args) {
int[] list = new int[]{1, 4, 2, 6, 7, 5, 1, 2};
int temp;
for (int i = 0; i < list.length; ++i) {
for (int j = 1; j < (list.length - i); ++j) {
if (list[j - 1] > list[j]) {
temp = list[j - 1];
list[j - 1] = list[j];
list[j] = temp;
}
}
}
int n = 0, counter = 0;
int previous = -1;
for (int i = 0; i < list.length; ++i) {
if (list[i] == previous) {
++n;
if (n == 1) {
++counter;
if (counter == 1) {
System.out.print(list[i]);
} else {
System.out.print(", " + list[i]);
}
}
} else {
previous = list[i];
n = 0;
}
}
System.out.println("\nNumber of Duplicated Numbers in array: " + counter);
}
}

How can I modify the boyer moore algorithm so that it runs in 0(n+m) time?

So I'm using the Boyer-Moore function and trying to integrate the failure function from the KMP and I don't know how to do that per se. I added the calling method to the failure function but I don't know how to make use of it. The two methods are given below.
public static int findBoyerMoore(char[] text, char[] pattern) {
int n = text.length;
int m = pattern.length;
int[] fail = computeFailKMP(pattern); ---> I did this but how can I use fail?
if(m == 0){
return 0;
}
Map<Character, Integer> last = new HashMap<>();
for(int i = 0; i< n; i++) {
last.put(text[i], -1);
}
for(int k = 0; k < m; k++) {
last.put(pattern[k], k);
}
int i = m - 1;
int k = m - 1;
while(i < n) {
if(text[i] == pattern[k]) {
if(k == 0) {
return i;
}
i--;
k--;
} else {
i+= m - Math.min(k, 1+ last.get(text[i]));
k = m-1;
}
}
return -1;
}
public static int[] computeFailKMP(char[] pattern) {
int m = pattern.length;
int[] fail = new int[m];
int j = 0;
int k = 0;
while(j < m) {
if(pattern[j] == pattern[k]) {
fail[j] = k + 1;
j++;
k++;
}
else if(k > 0) {
k = fail[k - 1];
}
else {
j++;
}
}
return fail;
}

Find the largest span between the same number in an array

Merry Christmas and hope you are in great Spirits,I have a Question in Java-Arrays as shown below.Im stuck up with this struggling to get it rite.
Consider the leftmost and righmost appearances of some value in an array. We'll say that the "span" is the number of elements between the two inclusive. A single value has a span of 1. Write a **Java Function** that returns the largest span found in the given array.
**Example:
maxSpan({1, 2, 1, 1, 3}) → 4,answer is 4 coz MaxSpan between 1 to 1 is 4
maxSpan({1, 4, 2, 1, 4, 1, 4}) → 6,answer is 6 coz MaxSpan between 4 to 4 is 6
maxSpan({1, 4, 2, 1, 4, 4, 4}) → 6,answer is 6 coz Maxspan between 4 to 4 is 6 which is greater than MaxSpan between 1 and 1 which is 4,Hence 6>4 answer is 6.
I have the code which is not working,it includes all the Spans for a given element,im unable to find the MaxSpan for a given element.
Please help me out.
Results of the above Program are as shown below
Expected This Run
maxSpan({1, 2, 1, 1, 3}) → 4 5 X
maxSpan({1, 4, 2, 1, 4, 1, 4}) → 6 8 X
maxSpan({1, 4, 2, 1, 4, 4, 4}) → 6 9 X
maxSpan({3, 3, 3}) → 3 5 X
maxSpan({3, 9, 3}) → 3 3 OK
maxSpan({3, 9, 9}) → 2 3 X
maxSpan({3, 9}) → 1 1 OK
maxSpan({3, 3}) → 2 3 X
maxSpan({}) → 0 1 X
maxSpan({1}) → 1 1 OK
::Code::
public int maxSpan(int[] nums) {
int count=1;//keep an intial count of maxspan=1
int maxspan=0;//initialize maxspan=0
for(int i=0;i<nums.length;i++){
for(int j=i+1;j<nums.length;j++){
if(nums[i] == nums[j]){
//check to see if "i" index contents == "j" index contents
count++; //increment count
maxspan=count; //make maxspan as your final count
int number = nums[i]; //number=actual number for maxspan
}
}
}
return maxspan+1; //return maxspan
}

Since a solution has been given, here is a more efficient solution which uses one pass.
public static void main(String... args) {
int maxspan = maxspan(3, 3, 3, 2, 1, 4, 3, 5, 3, 1, 1, 1, 1, 1);
System.out.println(maxspan);
}
private static int maxspan(int... ints) {
Map<Integer, Integer> first = new LinkedHashMap<Integer, Integer>(); // use TIntIntHashMap for efficiency.
int maxspan = 0; // max span so far.
for (int i = 0; i < ints.length; i++) {
int num = ints[i];
if (first.containsKey(num)) { // have we seen this number before?
int span = i - first.get(num) + 1; // num has been found so what is the span
if (span > maxspan) maxspan = span; // if the span is greater, update the maximum.
} else {
first.put(num, i); // first occurrence of number num at location i.
}
}
return maxspan;
}

I see the following problems with your attempt:
Your count is completely wrong. You can instead calculate count from i and j: j - i + 1
You're overriding maxcount as soon as you get any span, so you're going to end up with the last span, not the maximum span. Fix it by going maxspan = Math.max(maxspan, count);.
You can remove the line int number = nums[i]; as you never use number.
Remove the +1 in the returnmaxspan+1;` if you follow my tips above.
Initial maxspan should be 1 if there are any values in the array, but 0 if the array is empty.
That should help you get it working. Note that you can do this in a single pass of the array, but that's probably stretching it too far for you. Concentrate on getting your code to work before considering efficiency.

Here is the solution of this problem:
public int maxSpan(int[] nums) {
int maxSpan=0;
int tempSpan=0;
if(nums.length==0){
return 0;
}
for(int i=0;i<nums.length;i++){
for(int j=nums.length-1;j>i;j--){
if(nums[i]==nums[j]){
tempSpan=j-i;
break;
}
}
if(tempSpan>maxSpan){
maxSpan=tempSpan;
}
}
return maxSpan+1;
}

I did it with a List. Easier way to do it.
The only problem is if the Array is too big, maybe it's gonna take a while..
import java.util.ArrayList;
import java.util.List;
public class StackOverflow {
public static void main(String[] args) {
List<Integer> listNumbers = new ArrayList<Integer>();
listNumbers.add(3);
listNumbers.add(3);
listNumbers.add(3);
listNumbers.add(2);
listNumbers.add(1);
listNumbers.add(4);
listNumbers.add(3);
listNumbers.add(5);
listNumbers.add(1);
listNumbers.add(1);
listNumbers.add(1);
listNumbers.add(1);
listNumbers.add(1);
listNumbers.add(3);
int result = 0;
Integer key = null;
for(Integer i : listNumbers){
int resultDistance = returnDistance(listNumbers, i);
if (resultDistance > result){
result = resultDistance;
key = i;
}
}
System.out.println("MaxSpan of key " + key + " is: " + result);
}
private static int returnDistance(List<Integer> listNumbers, Integer term){
Integer startPosition = null;
Integer endPosition = null;
boolean bolStartPosition = false;
boolean bolResult = false;
int count = 1;
int result = 0;
for (Integer i : listNumbers){
if (i == term && !bolStartPosition){
startPosition = count;
bolStartPosition = true;
continue;
}
if (i == term && bolStartPosition){
endPosition = count;
}
count++;
}
if (endPosition != null){
// because it's inclusive from both sides
result = endPosition - startPosition + 2;
bolResult = true;
}
return (bolResult?result:-1);
}
}

public int maxSpan(int[] nums) {
int b = 0;
if (nums.length > 0) {
for (int i = 0; i < nums.length; i++) {
int a = nums[0];
if (nums[i] != a) {
b = nums.length - 1;
} else {
b = nums.length;
}
}
} else {
b = 0;
}
return b;
}

One brute force solution may like this, take one item from the array, and find the first occurance of item from the left most, and calculate the span, and then compare with the previous result.
public int maxSpan(int[] nums) {
int result = 0;
for(int i = 0; i < nums.length; i++) {
int item = nums[i];
int span = 0;
for(int j = 0; j<= i; j++) {//find first occurance of item from the left
if(nums[j]==item) {
span = i -j+1;
break;
}
}
if(span>result) {
result = span;
}
}
return result;
}

Here is the solution -
public int maxSpan(int[] nums) {
int span = 0;
for (int i = 0; i < nums.length; i++) {
for(int j = i; j < nums.length; j++) {
if(nums[i] == nums[j]) {
if(span < (j - i + 1)) {
span = j -i + 1;
}
}
}
}
return span;
}

public int maxSpan(int[] nums) {
int span = 0;
//Given the values at position i 0..length-1
//find the rightmost position of that value nums[i]
for (int i = 0; i < nums.length; i++) {
// find the rightmost of nums[i]
int j =nums.length -1;
while(nums[i]!=nums[j])
j--;
// j is at the rightmost posititon of nums[i]
span = Math.max(span,j-i+1);
}
return span;
}

public int maxSpan(int[] nums) {
//convert the numnber to a string
String numbers = "";
if (nums.length == 0)
return 0;
for(int ndx = 0; ndx < nums.length;ndx++){
numbers += nums[ndx];
}
//check beginning and end of string
int first = numbers.indexOf(numbers.charAt(0));
int last = numbers.lastIndexOf(numbers.charAt(0));
int max = last - first + 1;
int efirst = numbers.indexOf(numbers.charAt(numbers.length()-1));
int elast = numbers.lastIndexOf(numbers.charAt(numbers.length()-1));
int emax = elast - efirst + 1;
//return the max span.
return (max > emax)?max:emax;
}

public int maxSpan(int[] nums) {
int current = 0;
int currentcompare = 0;
int counter = 0;
int internalcounter = 0;
if(nums.length == 0)
return 0;
for(int i = 0; i < nums.length; i++) {
internalcounter = 0;
current = nums[i];
for(int x = i; x < nums.length; x++) {
currentcompare = nums[x];
if(current == currentcompare) {
internalcounter = x - i;
}
if(internalcounter > counter) {
counter = internalcounter;
}
}
}
return counter + 1;
}

public int maxSpan(int[] nums) {
if(nums.length<1){
return 0;
}
int compare=1;
for (int i=0; i<nums.length; i++){
for (int l=1; l<nums.length; l++){
if((nums[l]==nums[i])&&(Math.abs(l)-Math.abs(i))>=compare){
compare = Math.abs(l)-Math.abs(i)+1;
}
}
}
return compare;
}

public static int MaxSpan(int[] input1, int key)
{
int Span = 0;
int length = input1.length;
int i,j,k = 0;
int start = 0, end = 0 ;
k = key;
for (int l = 0; l < length; l++) {
if(input1[l] == k) { start = l;
System.out.println("\nStart = " + start);
break;
}
}
if(start == 0) { Span = 0; System.out.println("Key not found"); return Span;}
for (j = length-1; j> start; j--) {
if(input1[j] == k) { end = j;
System.out.println("\nEnd = " + end);
break;
}
}
Span = end - start;
System.out.println("\nStart = " + start + "End = " + end + "Span = " + Span);
return Span;
}

public int maxSpan(int[] nums) {
int length = nums.length;
if(length <= 0)
return 0;
int left = nums[0];
int rigth = nums[length - 1];
int value = 1;
//If these values are the same, then the max span is length
if(left == rigth)
return length;
// the last match is the largest span for any value
for(int x = 1; x < length - 1; x++)
{
if(nums[x] == left || nums[x] == rigth)
value = x + 1;
}
return value;
}

public int maxSpan(int[] nums) {
int count, largest=0;
for (int x=0; x< nums.length; x++)
{
for (int y=0; y< nums.length; y++)
{
if (nums[x]==nums[y])
{
count= y-x+1;
if (count > largest)
{
largest= count;
}
}
}
}
return largest;
}

import java.io.*;
public class maxspan {
public static void main(String args[])throws java.io.IOException{
int A[],span=0,pos=0;
DataInputStream in=new DataInputStream(System.in);
System.out.println("enter the number of elements");
A=new int[Integer.parseInt(in.readLine())];
int i,j;
for(i=0;i<A.length;i++)
{
A[i]=Integer.parseInt(in.readLine());
}
for(i=0;i<A.length;i++)
{
for(j=A.length-1;j>=0;j--)
if(A[i]==A[j]&&(j-i)>span){span=j-i;pos=i;}
}
System.out.println("maximum span => "+(span+1)+" that is of "+A[pos]);
}
}

public static int maxSpan(int[] nums) {
int left = 0;
int right = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[0] == nums[nums.length - 1 - i]) {
left = nums.length - i;
break;
} else if (nums[nums.length - 1] == nums[i]) {
right = nums.length - i;
break;
}
}
return Math.max(left, right);
}

The above solutions are great, if your goal is to avoid using Arrays.asList and indexOf and LastIndexOf, the code below does the job as lazy as possible, while still being clear and concise.
public int maxSpan(int[] nums) {
if(nums.length < 2){ //weed out length 0 and 1 cases
return nums.length;
}
int maxSpan = 1; //start out as 1
for(int a = 0; a < nums.length; a++){
for(int b = nums.length - 1; b > a; b--){
if(nums[a] == nums[b]){
maxSpan = Math.max(maxSpan, (b + 1 - a));
//A little tricky getting those indices together.
break; //there's no reason to continue,
//at least for this single loop execution inside another loop
}
}
}
return maxSpan; //the maxSpan is here!
}
The Math.max method returns the larger of 2 values, one of them if they are equal.

This is how I did it:
public int maxSpan(int[] nums) {
for (int span=nums.length; span>0; span--) {
for (int i=0; i<nums.length-span+1; i++) {
if (nums[i] == nums[i+span-1]) return span;
}
}
return 0;
}

I am not sure, if I have to use 2 for-loops... or any loop at all?
If not, this version functions without any loop.
At first you check, if the length of the array is > 0. If not, you simply return the length of the array, which will correspond to the correct answer.
If it is longer than 0, you check if the first and last position in the array have the same value.
If yes, you return the length of the array as the maxSpan.
If not, you substract a 1, since the value appears twice in the array.
Done.
public int maxSpan(int[] nums) {
if(nums.length > 0){
if(nums[0] == nums[nums.length - 1]){
return nums.length;
}
else{
return nums.length - 1;
}
}
return nums.length;
}

public int maxSpan(int[] nums) {
Stack stack = new Stack();
int count = 1;
int value = 0;
int temp = 0;
if(nums.length < 1) {
return value;
}
for(int i = 0; i < nums.length; i++) {
for(int j = nums.length - 1; j >= i; j--) {
if(nums[i] == nums[j]) {
count = (j - i) + 1;
stack.push(count);
count = 1;
break;
}
}
}
if(stack.peek() != null) {
while(stack.size() != 0) {
temp = (Integer) stack.pop();
if(value <= temp) {
value = temp;
} else {
value = value;
}
}
}
return value;
}

public int maxSpan(int[] nums) {
int totalspan=0;
int span=0;
for(int i=0;i<nums.length;i++)
{
for (int j=nums.length-1;j>i-1;j--)
{
if(nums[i]==nums[j])
{
span=j-i+1;
if (span>totalspan)
totalspan=span;
break;
}
}
}
return totalspan;
}

public int maxSpan(int[] nums) {
int max_span=0, j;
for (int i=0; i<nums.length; i++){
j=nums.length-1;
while(nums[i]!=nums[j]) j--;
if (j-i+1>max_span) max_span=j-i+1;
}
return max_span;
}

Linear solution with a Map storing the first occurrence, and calculating the distance from it for the next occurrences:
public int maxSpan(int[] nums) {
int span = 0;
Map<Integer, Integer> first = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; i++) {
if (!first.containsKey(nums[i]))
first.put(nums[i], i);
span = Math.max(span, (i - first.get(nums[i])) + 1);
}
return span;
}