Zscaler Coding Questions 2025: Recruitment Process, Eligibility Criteria

Coding problems are a significant part of technical interviews, especially for top tech firms such as Zscaler. They examine the problem-solving ability of the candidate together with educating them about algorithms and data structures and their capacity to code clean and efficient code under stress. Practice in solving coding problems will help the candidates prepare for the technical phases of Zscaler and enhance their general coding abilities. This article will discuss Zscaler coding questions asked in 2025, tips to crack Zscaler coding rounds, and valuable approaches towards an interview.

Table of Contents

About Zscaler

Zscaler is the most significant security cloud, enabling organizations worldwide to transform digitally as fast and securely as possible. Founded in 2007, Zscaler serves over 7,500 customers, including 30% of the Forbes Global 2000, and secures over 500 billion daily transactions. The cloud-native Zero Trust Exchange platform helps organizations provide secure access to users, devices, and applications to protect against cyberattacks and data loss regardless of location.

Zscaler allows organisations to accelerate their digital transformation by becoming nimbler, more efficient, more resilient, and more secure. Zero-trust principles empower organizations to modernise infrastructure, securely enable the workplace, and protect cloud workloads and SaaS data. Zscaler focuses on security transformation that protects IoT and OT devices and allows remote access to critical systems.

Regarded as a Leader in the Gartner MQ for Security Service Edge (SSE), Zscaler solutions have transformed how organisations secure their digital environments and accelerate growth, ensuring a seamless and secure information exchange.

Zscaler Recruitment Process 2025

The Zscaler recruitment process consists of multiple stages designed to assess technical skills, problem-solving abilities, and cultural fit:

1. Online Coding Assessment

A HackerRank online coding assessment that tests your programming and problem-solving skills.

2. Pre-Placement Talk

This is an introductory session in which Zscaler explains the company culture, job roles, and recruitment process.

3. Technical Rounds of Interview (2-3 Rounds)

You will be examined on data structures, algorithms, and programming, with some specifications on languages like C (for Dev roles).

4. HR Round

Final interview round that tests behavioral questions, culture fit, and salary negotiation.

Eligibility Criteria for Zscaler

Aspect Details
Position Offered Associate Software Engineer – Dev
Associate Software Engineer – DevTest
Eligible Batch 2019 – 2023
Course B.Tech (All Branches) and M.C.A
Academic Qualification Minimum 60% or equivalent CGPA in:
- 10th
- 12th
- Graduation/P.G.
Backlog Criteria No Active Backlogs
Cost to Company (CTC) - Interns: 50K per month
- Dev Roles: 12 LPA
- DevTest Role: 9 LPA
Salary for Dev Roles - 2 Lakhs Retention Bonus
- 12K USD RSUs
Specific Programming Skills - Dev Roles: C Programming
- DevTest Roles: C/C++, Java Programming

Most Asked Coding Questions in Zscaler

Below are the top 10 common coding questions asked in Zscaler interviews, with complete solutions in C++, Java, and Python.

Problem Statement 1: Find the Missing Number

You are given an array of size N-1 containing integers from 1 to N. One number is missing. Your task is to find and return the missing number.

Solution in C++

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    int sum = (n * (n + 1)) / 2;
    int arr_sum = 0;
    for (int i = 0; i < n - 1; i++) {
        int num;
        cin >> num;
        arr_sum += num;
    }
    cout << sum - arr_sum << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int sum = (n * (n + 1)) / 2;
        int arr_sum = 0;
        for (int i = 0; i < n - 1; i++) {
            arr_sum += sc.nextInt();
        }
        System.out.println(sum - arr_sum);
    }
}

Solution in Python

n = int(input())
total_sum = n * (n + 1) // 2
arr_sum = sum(int(input()) for _ in range(n - 1))
print(total_sum - arr_sum)

Explanation

The code calculates the sum of numbers from 1 to N using the formula N*(N+1)/2. Then, it computes the sum of the given array. The missing number is found by subtracting the array sum from the total sum.

Example Input

5
1 2 4

Example Output

3

Problem Statement 2: Count Devil Groups

There are some groups of devils. People are represented as a string where "@" and "$" are devils and "P" is a person. The devils make people their group, and the largest group is the one that will be killed. Determine how many groups can be formed based on this representation.

Solution in C++

#include <bits/stdc++.h>
using namespace std;

int main() {
    string s;
    cin >> s;
    int groups = 1;
    
    for (int i = 1; i < s.length(); i++) {
        if (s[i] == '@' || s[i] == '$') {
            if (s[i - 1] != '@' && s[i - 1] != '$') {
                groups++;
            }
        }
    }
    
    cout << groups << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        int groups = 1;
        
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i) == '@' || s.charAt(i) == '$') {
                if (s.charAt(i - 1) != '@' && s.charAt(i - 1) != '$') {
                    groups++;
                }
            }
        }
        
        System.out.println(groups);
    }
}

Solution in Python

s = input()
groups = 1

for i in range(1, len(s)):
    if s[i] == '@' or s[i] == '$':
        if s[i - 1] != '@' and s[i - 1] != '$':
            groups += 1

print(groups)

Explanation

It keeps a count of groups formed by consecutive @ or $ characters. It starts with the initial count of groups as 1. If the character in the string is @ or $ and the previous character is not @ or $, then a new group starts, increasing the count.

Example Input

PPPPPP@PPP@PP$PP

Example Output

7

Problem Statement 3: Minimum Starting Coins for Game

Raman plays a game where he wins or loses coins in each step. To avoid going into credit, help him find the minimum starting coins required to ensure he never runs out of money.

Solution in C++

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    int a[n], sum = 0, minAmount = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
        if (sum < 1) {
            minAmount = max(minAmount, -sum + 1);
            sum = 0;
        }
    }
    cout << minAmount << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int sum = 0, minAmount = 0;
        for (int i = 0; i < n; i++) {
            int outcome = sc.nextInt();
            sum += outcome;
            if (sum < 1) {
                minAmount = Math.max(minAmount, -sum + 1);
                sum = 0;
            }
        }
        System.out.println(minAmount);
    }
}

Solution in Python

n = int(input())
outcomes = list(map(int, input().split()))
sum, minAmount = 0, 0
for outcome in outcomes:
    sum += outcome
    if sum < 1:
        minAmount = max(minAmount, -sum + 1)
        sum = 0
print(minAmount)

Explanation

The program calculates the minimum amount required to keep the total sum positive throughout the game. As each outcome is added to the total, if the sum falls below 1, it adjusts by ensuring the starting amount is enough to avoid a negative balance, resulting in a minimum of 7.

Example Input

4
2 -9 15 2

Example Output

7

Problem Statement 4: Count Borrow Operations in Subtraction

Compute number of borrow operation to subtract number2 from number1. Return "Not possible" if that is impossible.

Solution in C++

#include <bits/stdc++.h>
using namespace std;

int main() {
    string s1, s2;
    cin >> s1 >> s2;
    if (stoi(s1) < stoi(s2)) {
        cout << "Impossible" << endl;
        return 0;
    }
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    int borrowCount = 0, carry = 0;
    for (int i = 0; i < s1.length(); i++) {
        int digit1 = s1[i] - '0', digit2 = s2[i] - '0';
        if (digit1 < digit2 + carry) {
            borrowCount++;
            carry = 1;
        } else {
            carry = 0;
        }
    }
    cout << borrowCount << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s1 = sc.nextLine();
        String s2 = sc.nextLine();
        if (Integer.parseInt(s1) < Integer.parseInt(s2)) {
            System.out.println("Impossible");
            return;
        }
        s1 = new StringBuilder(s1).reverse().toString();
        s2 = new StringBuilder(s2).reverse().toString();
        int borrowCount = 0, carry = 0;
        for (int i = 0; i < s1.length(); i++) {
            int digit1 = s1.charAt(i) - '0', digit2 = s2.charAt(i) - '0';
            if (digit1 < digit2 + carry) {
                borrowCount++;
                carry = 1;
            } else {
                carry = 0;
            }
        }
        System.out.println(borrowCount);
    }
}

Solution in Python

s1, s2 = input(), input()
if int(s1) < int(s2):
    print("Impossible")
else:
    s1, s2 = s1[::-1], s2[::-1]
    borrowCount, carry = 0, 0
    for i in range(len(s1)):
        digit1, digit2 = int(s1[i]), int(s2[i])
        if digit1 < digit2 + carry:
            borrowCount += 1
            carry = 1
        else:
            carry = 0
    print(borrowCount)

Explanation

The code calculates the borrow operations required to subtract s2 from s1. The strings are reversed to simulate the subtraction from the least significant digit. The count of borrow is incremented each time there's a requirement for a borrow operation in subtraction. The final result is 2.

Example Input

754
658

Example Output

2

Problem Statement 5: Minimum Coins to Avoid Debt (Alternate Version)

Given the actions of a game in which Raman wins or loses coins, dictate the least that he should begin with so that he will not go into debt.

Solution in C++

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    int sum = 0, minCoins = 0;
    for (int i = 0; i < n; i++) {
        int change;
        cin >> change;
        sum += change;
        if (sum < 1) {
            minCoins = max(minCoins, -sum + 1);
            sum = 0;
        }
    }
    cout << minCoins << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int sum = 0, minCoins = 0;
        for (int i = 0; i < n; i++) {
            int change = sc.nextInt();
            sum += change;
            if (sum < 1) {
                minCoins = Math.max(minCoins, -sum + 1);
                sum = 0;
            }
        }
        System.out.println(minCoins);
    }
}

Solution in Python

n = int(input())
steps = list(map(int, input().split()))
sum, minCoins = 0, 0
for change in steps:
    sum += change
    if sum < 1:
        minCoins = max(minCoins, -sum + 1)
        sum = 0
print(minCoins)

Explanation

The program calculates the minimum amount of coins to avoid running losses. It resets the total when it is processing changes. When the total becomes negative, it rebalances in order to have sufficient funds. The minimum amount of coins required here is 7.

Example Input

4
2 -9 15 2

Example Output

7

Problem Statement 6: Find Kth Good Prime Number

A prime number is called a "good prime number" if the sum of its digits is also a prime number. Given an integer N, find the Kth good prime number greater than N.

Solution in C++

#include <bits/stdc++.h>
using namespace std;

bool isPrime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}

int solve(int N, int K) {
    int count = 0;
    for (int i = N + 1; count < K; i++) {
        if (isPrime(i)) {
            int sum = 0, temp = i;
            while (temp > 0) {
                sum += temp % 10;
                temp /= 10;
            }
            if (isPrime(sum)) {
                count++;
                if (count == K) return i;
            }
        }
    }
    return -1;
}

int main() {
    int N, K;
    cin >> N >> K;
    cout << solve(N, K) << endl;
    return 0;
}

Solution in Java

import java.util.*;

public class Main {
    static boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) return false;
        }
        return true;
    }

    static int solve(int N, int K) {
        int count = 0;
        for (int i = N + 1; count < K; i++) {
            if (isPrime(i)) {
                int sum = 0, temp = i;
                while (temp > 0) {
                    sum += temp % 10;
                    temp /= 10;
                }
                if (isPrime(sum)) {
                    count++;
                    if (count == K) return i;
                }
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        System.out.println(solve(N, K));
    }
}

Solution in Python

def is_prime(n):
    if n <= 1:
        return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False
    return True

def solve(N, K):
    count = 0
    i = N + 1
    while count < K:
        if is_prime(i):
            digit_sum = sum(map(int, str(i)))
            if is_prime(digit_sum):
                count += 1
                if count == K:
                    return i
        i += 1
    return -1

N = int(input())
K = int(input())
print(solve(N, K))

Explanation

The program finds the K-th prime number greater than N such that the sum of its digits is also prime. In this case, for N = 10 and K = 3, the program checks consecutive primes after 10. The 3rd prime whose digit sum is prime is 31.

Example Input

10
3

Example Output

31

Problem Statement 7: Check if Two Strings are Anagrams

Verify if two strings are anagrams. Two strings are anagrams if they contain the same characters in any order with repetition.

Solution in C++

#include <iostream>
#include <algorithm>
using namespace std;

bool areAnagrams(string s1, string s2) {
    sort(s1.begin(), s1.end());
    sort(s2.begin(), s2.end());
    return s1 == s2;
}

int main() {
    string s1, s2;
    cin >> s1 >> s2;
    cout << (areAnagrams(s1, s2) ? "Yes" : "No") << endl;
    return 0;
}

Solution in Java

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    static boolean areAnagrams(String s1, String s2) {
        char[] arr1 = s1.toCharArray();
        char[] arr2 = s2.toCharArray();
        Arrays.sort(arr1);
        Arrays.sort(arr2);
        return Arrays.equals(arr1, arr2);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s1 = sc.next();
        String s2 = sc.next();
        System.out.println(areAnagrams(s1, s2) ? "Yes" : "No");
    }
}

Solution in Python

def are_anagrams(s1, s2):
    return sorted(s1) == sorted(s2)

s1, s2 = input(), input()
print("Yes" if are_anagrams(s1, s2) else "No")

Explanation

The program verifies whether two strings are anagrams or not. Two strings are referred to as anagrams if they contain the same characters, but in different places. In this instance, "listen" and "silent" contain the same characters, and therefore are anagrams, and the output will be "Yes".

Example Input

listen
silent

Example Output

Yes

Problem Statement 8: Check if String is Palindrome

Check if a given string is a palindrome. A palindrome is a word, phrase, or sequence that reads the same backward as forward.

Solution in C++

#include <iostream>
using namespace std;

bool isPalindrome(string s) {
    int start = 0, end = s.length() - 1;
    while (start < end) {
        if (s[start] != s[end]) return false;
        start++;
        end--;
    }
    return true;
}

int main() {
    string s;
    cin >> s;
    cout << (isPalindrome(s) ? "Yes" : "No") << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    static boolean isPalindrome(String s) {
        int start = 0, end = s.length() - 1;
        while (start < end) {
            if (s.charAt(start) != s.charAt(end)) return false;
            start++;
            end--;
        }
        return true;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        System.out.println(isPalindrome(s) ? "Yes" : "No");
    }
}

Solution in Python

def is_palindrome(s):
    return s == s[::-1]

s = input()
print("Yes" if is_palindrome(s) else "No")

Explanation

The program checks whether a given string is a palindrome. A palindrome is a word that reads the same forward and backward. The function compares characters from both ends of the string, moving towards the center. If all characters match, the string is a palindrome, and the program prints "Yes".

Example Input

madam

Example Output

Yes

Problem Statement 9: Count Vowels in String

Write a program that counts the number of vowels in a given string. The vowels are 'a', 'e', 'i', 'o', 'u' (case-insensitive).

Solution in C++

#include <iostream>
using namespace std;

int countVowels(string s) {
    int count = 0;
    for (char c : s) {
        if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') count++;
    }
    return count;
}

int main() {
    string s;
    cin >> s;
    cout << countVowels(s) << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    static int countVowels(String s) {
        int count = 0;
        for (char c : s.toCharArray()) {
            if ("aeiou".indexOf(c) != -1) count++;
        }
        return count;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        System.out.println(countVowels(s));
    }
}

Solution in Python

def count_vowels(s):
    return sum(1 for c in s if c in 'aeiou')

s = input()
print(count_vowels(s))

Explanation

The program counts the number of vowels in the given string. It iterates through each character in the string, checking if it is one of the vowels ('a', 'e', 'i', 'o', 'u'). If it is, the counter is incremented. Finally, the result (vowel count) is printed.

Example Input

hello

Example Output

2

Problem Statement 10: Calculate Power Using Loop

Write a program to calculate the power of a number using a loop (without using the built-in pow function).

Solution in C++

#include <iostream>
using namespace std;

int main() {
    int base, exponent;
    cin >> base >> exponent;

    int result = 1;

    for (int i = 1; i <= exponent; i++) {
        result *= base;
    }

    cout << result << endl;
    return 0;
}

Solution in Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int base = sc.nextInt();
        int exponent = sc.nextInt();

        int result = 1;

        for (int i = 1; i <= exponent; i++) {
            result *= base;
        }

        System.out.println(result);
    }
}

Solution in Python

base, exponent = map(int, input().split())

result = 1

for _ in range(exponent):
    result *= base

print(result)

Explanation

The program calculates the power of a number by multiplying the base by itself for the number of times specified by the exponent using a loop. It initializes the result to 1 and multiplies it by the base in each loop iteration. Finally, it outputs the calculated result.

Example Input

2 3

Example Output

8

Tips for Cracking Zscaler Coding Rounds

Here are the essential tips to clear Zscaler coding rounds:

1. DSA Fundamentals

Cover all Data Structures & Algorithms, concentrating mainly on arrays, linked lists, trees, graphs, and dynamic programming.

2. Problem-Solving Practice

Review problems regularly on LeetCode, HackerRank, and Codeforces platforms.

3. Time Management

Learn to solve problems within the time limit. Practice under pressure.

4. Language Skills

Dev roles require a good command of C, and DevTest requires some comfort with C/C++ and Java.

5. Understand Problem Patterns

Identify common algorithmic patterns like sliding windows, backtracking, etc.

6. Explain Your Approach

Before coding, you must clearly share your thought process.

7. Mock Interviews

Practice interviews to become faster, more accurate, and more confident in conducting actual interviews.

8. Review Complexity

Consider your solutions' time and space complexity (Big O notation).

Conclusion

In conclusion, getting through Zscaler's coding rounds requires a good grasp of problem-solving's demand in programming with languages like C and Java (for Dev roles). A good understanding of concepts and zeal dedicated to regular practice can get one to the desired place with Zscaler through effective preparation. Calm down, think logically, and communicate clearly during the interview process.

Frequently Asked Questions

What Questions Are Asked in a Coding Test?

Questions in Zscaler coding tests focus on data structures and algorithms. Expect some questions on arrays (find missing numbers), linked lists (detect cycles), trees (DFS/BFS), and graphs (shortest path). These questions test for the efficiency of problem-solving.

Is a Zscaler Interview Hard?

Zscaler interviews are rated as tough because of a strong focus on data structures, algorithms, and system design. They test your problem-solving abilities, coding speed, and a bit of logical reasoning pressure; this makes preparation important. One has to practice repeatedly tackling complex questions with some ease.

How Many Coding Questions Are in a 1-Hour Interview?

Generally, the Zscaler interview includes two to three coding questions related to the effectiveness of algorithms, data structures, and problem-solving. Hence, the candidate must focus on writing quick, efficient code to be tested within one hour.

What Is the Passing CGPA in Zscaler?

Zscaler generally considers candidates with a minimum CGPA of 7.5 to 8.0 (or equivalent). While CGPA is important, Zscaler prioritises problem-solving skills, coding proficiency, and technical knowledge. A high CGPA goes a long way in your direction, but it is not the only criterion.

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