ArrayProblem

Q1 - Given an array sorted in increasing order of size n and an integer x, find if there exists a
pair in the array whose absolute difference is exactly x.(n>1) (Medium)
Input:
N = 5
Arr[] = [5,10,15,20,26]
x= 10
Expected Output:
Yes
Explanation
 Traverse the array and consider every element as a possible part of the pair to be foundr
 Traverse the array again for further indices and loos for the given differencer
 If two numbers have a difference equal to x, then the pair is complete, return yesr
 Else print no after the loops which will only run if you do not find a pair.
ode:
iport java.util.Scanner;
public class Test {
public static void ain(String[] args) {
Scanner scn = new Scanner(Syste.in);
Syste.out.print(“Enter the length of the array: ”);
int n = scn.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++){
arr[i] = scn.nextInt();
}
int x = scn.nextInt();
for(int i = 0; i < n; i++){
for(int j = i+1; j < n; j++){ //check for pair with all eleents 1 by 1
if(arr[j] - arr[i]== x){
Syste.out.print("Yes");
return;
}
}
}
Syste.out.print("No");
}
}
Assignment Solutions

Cracking the Coding Interview in JAVA - Foundation

Q2 - Given an array of size n, find the total number of occurrences of given number x. (Easy)
Inpt:
n = 7
arr[] = [3, p, 0, 7, 8, 3, 0]
 = 0
Expecte
Otpt:
2
Explanation
 Keep a pointer count, and traverse the array, and check for each element being equal to 
 If current element equals z, increment couno
 Print count in the end.
Co
e:
import java.util.Scanner;
public class Test{
public static void main(String[] args){
Scanner scn = new Scanner(System.in);
System.out.println("Enter the length of array");
int n = scn.nextInt();
int[] arr = new int[n];
System.out.println("Enter the elements of array");
for(int i = 0; i < n; i++){
arr[i] = scn.nextInt();
}
System.out.println("Enter the number");
int x = scn.nextInt();
int count = 0;
for(int i = 0; i < arr.length; i++){
if(arr[i] == x){
count++;
}
}
System.out.println(count);
}
}
Assignment Solutions

Cracking the Coding Interview in JAVA - Foundation
Q3 - Given an array arr[] of size N-1 with integers in the range of [1, N], the task is to find the
missing number from the first N integers. There are no duplicates in the list. (Medium)
Inpzt:
 = 7
arr[] = {1n 2n 4n 6n 3n 7n 8}
Expected Oztpzt:
5
Explanationm
y Logic is to mark the elemet at the curret elemet as idexn egativeN Son whichever elemet would be
missign the elemet at the missig elemet as idex would be positiveM
y Traverse the give arrayn if the absolute value of curret elemet is greater tha size of the arrayn the
cotiuen else multiply the (absolute value of (curret elemet) – 1)th idex with -1M
y Iitialize a variable as = size + 1M
y Traverse the array ad if the value is positive assig as = idex + 1n prit ad retur
y Prit +1 i the edn this will ru oly whe o elemet is missig from 1 to .
Assignment Solutions

Cracking the Coding Interview in JAVA - Foundation

Code:
import java.util.Scanner;
public class Test{
public static void main(String[] args){
Scanner scn = new Scanner(System.in);
System.out.println("Enter the length of array");
int n = scn.nextInt();
int[] arr = new int[n];
System.out.println("Enter the elements of array");
for(int i = 0; i < n; i++){
arr[i] = scn.nextInt();
}
for (int i = 0; i < n; i++) {
if (Math.abs(arr[i]) - 1 == n) {
continue;
}
int ind = Math.abs(arr[i]) - 1;
arr[ind] *= -1;
}
int ans = 0;
for(int i = 0; i < n; i++) {
if (arr[i] > 0){
ans = i + 1;
System.out.println(ans);
return;
}
}
System.out.println(n+1);
}
}