Given an array of length N and an integer x, you need to find all the indexes where x is present in t indexes in an array (in increasing order). Do this recursively. Indexing in the array starts from 0. Input Format: Line 1: An Integer Ni.e. size of array Line 2: N integers which are elements of the array, separated by spaces Line 3: Integer x Output Format : indexes where x is present in the array (separated by space) Constraints
Given an array of length N and an integer x, you need to find all the indexes where x is present in t indexes in an array (in increasing order). Do this recursively. Indexing in the array starts from 0. Input Format: Line 1: An Integer Ni.e. size of array Line 2: N integers which are elements of the array, separated by spaces Line 3: Integer x Output Format : indexes where x is present in the array (separated by space) Constraints
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![Given an array of length N and an integer x, you need to find all the indexes where x is present in the input array. Save all the
indexes in an array (in increasing order).
Do this recursively. Indexing in the array starts from 0.
Input Format :
Line 1: An Integer Ni.e. size of array
Line 2: N integers which are elements of the array, separated by spaces
Line 3: Integer x
Output Format :
indexes where x is present in the array (separated by space)
Constraints:
1 <= N <= 10^3
Sample Input:
5
981088
8
Sample Output :
134
Solution://///
public class Solution {
public static int[] allIndexes (int input[], int x, int startIndex) {
if(startIndex==input.length) {
int output[] = new int[0];
return output;
}
int smallOutput[] = allIndexes (input,x, startIndex+1);
if(input[startIndex]==x){
int output[] = new int[smallOutput.length +1];
}
}
output[0] = startIndex;
for(int i=0;i<smallOutput.length;i++){
output[i+1] =smallOutput[i];
}
return output;
}
else{
}
return smallOutput;
public static int[] allIndexes (int input[], int x) {
return allIndexes(input,x,0);](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9e5eb73-c433-419a-ab5d-bbffff55a5d5%2F05191acd-12fc-4a5b-96d7-c6e89add48f0%2Fgg3ny1e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given an array of length N and an integer x, you need to find all the indexes where x is present in the input array. Save all the
indexes in an array (in increasing order).
Do this recursively. Indexing in the array starts from 0.
Input Format :
Line 1: An Integer Ni.e. size of array
Line 2: N integers which are elements of the array, separated by spaces
Line 3: Integer x
Output Format :
indexes where x is present in the array (separated by space)
Constraints:
1 <= N <= 10^3
Sample Input:
5
981088
8
Sample Output :
134
Solution://///
public class Solution {
public static int[] allIndexes (int input[], int x, int startIndex) {
if(startIndex==input.length) {
int output[] = new int[0];
return output;
}
int smallOutput[] = allIndexes (input,x, startIndex+1);
if(input[startIndex]==x){
int output[] = new int[smallOutput.length +1];
}
}
output[0] = startIndex;
for(int i=0;i<smallOutput.length;i++){
output[i+1] =smallOutput[i];
}
return output;
}
else{
}
return smallOutput;
public static int[] allIndexes (int input[], int x) {
return allIndexes(input,x,0);
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