Advanced Python Interview Questions for Experienced Developers

This section presents challenging Python programming problems and coding scenarios designed to assess the problem-solving and coding skills of experienced Python developers. It emphasizes efficient solutions and an understanding of Python's nuances.

Shuffling a List

Python's random.shuffle() method shuffles the elements of a list in place (modifies the original list).

import random
myList = [1, 2, 3, 4, 5]
random.shuffle(myList)
print(myList)  # Output: A shuffled version of myList

Splitting Strings

The split() method splits a string into a list of substrings based on a delimiter.

myString = "apple,banana,orange"
myList = myString.split(',')
print(myList)  # Output: ['apple', 'banana', 'orange']

Converting Strings to Lists

Use the list() constructor to convert a string to a list of characters.

myString = "hello"
myList = list(myString)
print(myList)  # Output: ['h', 'e', 'l', 'l', 'o']

Exception Handling in Python

Python uses the try...except...finally block for exception handling. The try block contains the code that might raise an exception. The except block handles specific exceptions. The finally block (optional) executes regardless of whether an exception occurred.

List and Dictionary Comprehensions

List and dictionary comprehensions provide concise ways to create lists and dictionaries.

myList = [x * 2 for x in range(5)]  # Creates [0, 2, 4, 6, 8]

myDict = {x: x * 2 for x in range(5)}  # Creates {0: 0, 1: 2, 2: 4, 3: 6, 4: 8}

Copying Objects in Python

Methods for copying objects:

• copy.copy() (shallow copy).
• copy.deepcopy() (deep copy).
• Using the copy() method (for dictionaries).

Getting Object Names

Python objects don't intrinsically have names. You can only get a reference to the object and its assigned name within the specific scope.

Checking Object Type

Use isinstance(obj, MyClass) to check if an object is an instance of a specific class or its subclasses. Avoid relying on type(obj) == MyClass as this only checks for an exact type match.

Late Binding in Python

In the example provided, the lambda function captures the variable index by reference, not by value. Therefore, it will return the final value of `index` at the time the lambda function is actually called.

Mutable Default Arguments

Default arguments in Python are evaluated only once when the function is defined, not each time the function is called. This is important because if you use a mutable default argument (like a list) that is modified within the function, that modification persists across function calls. In this example, each call appends a '1' to the list, accumulating the value for each invocation.

String Slicing in Python

Python's string slicing allows extracting parts of a string (e.g., myString[start:end]).

Getting Unique Elements from a List

Use list(set(myList)) to get a list of unique elements, and then sort them.

myList = ['a', 'b', 'c', 'a', 'b']
uniqueList = sorted(list(set(myList)))
print(uniqueList)  # Output: ['a', 'b', 'c']

Counting Word Frequencies

[Include Python code to count word frequencies in a list of words using a dictionary.]

Immutability of Object Attributes

In the provided example, the attribute x1 of class Test is not modified. The multiple print statements would display the initial value.

Mutable Default Arguments (Problem)

Using mutable objects (like lists) as default arguments can lead to unexpected results because the default argument is only created once, when the function is defined, not each time the function is called.

Standard Input/Output in Python

sys.stdin, sys.stdout, and sys.stderr represent the standard input, standard output, and standard error streams.

PyTables

PyTables is a Python library for managing hierarchical datasets, useful for handling large amounts of data efficiently.

Order of Operations in Python

In Python, the order of operations follows the PEMDAS/BODMAS rule. This is a standard order of precedence that determines the order in which operations are performed in an expression. Understanding this rule ensures that your calculations are done correctly.

1. Parentheses/Brackets

Operations enclosed in parentheses or brackets are performed first. This allows you to control the order of execution explicitly.

result = (2 + 3) * 4  # Parentheses first

result = 20  # (2 + 3) = 5, and then 5 * 4 = 20

2. Exponents

After parentheses, exponentiation is performed. Exponentiation is denoted by the ** operator in Python.

result = 2 ** 3  # Exponentiation (2 raised to the power of 3)

result = 8  # 2 ** 3 = 8

3. Multiplication and Division

Multiplication and division come next, and they are performed from left to right. In Python, multiplication is denoted by * and division by /.

result = 4 * 3 / 2  # Multiplication and division from left to right

result = 6.0  # 4 * 3 = 12, then 12 / 2 = 6

4. Addition and Subtraction

Finally, addition and subtraction are performed, and like multiplication and division, they are evaluated from left to right. In Python, addition is denoted by + and subtraction by -.

result = 5 + 3 - 2  # Addition and subtraction from left to right

result = 6  # 5 + 3 = 8, then 8 - 2 = 6

Summary of Order of Operations (PEMDAS)

• Parentheses first
• Exponents (powers and square roots)
• Multiplication and Division (left to right)
• Addition and Subtraction (left to right)

By following this order, Python ensures that calculations are performed in a predictable and correct sequence.

Finding the Next Permutation (Continued)

import java.util.Arrays;
public class NextPermutation {
    // ... (reverse method as before) ...
    public int[] nextPermutation(int arr[], int s) {
        if (s == 1) return arr;
        int i;
        for (i = s - 2; i >= 0; i--) {
            if (arr[i] < arr[i + 1]) break;
        }
        if (i < 0) {
            reverse(arr, 0);
            return arr;
        }
        int j;
        for (j = s - 1; j > i; j--) {
            if (arr[j] > arr[i]) break;
        }
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        reverse(arr, i + 1);
        return arr;
    }
    // ... (findNo method as before) ...
    public static void main(String argvs[]) {
        NextPermutation obj = new NextPermutation();
        int arr[] = {1, 3, 7, 9, 4};
        int size = arr.length;
        int ans[] = obj.nextPermutation(arr, size);
        System.out.println(Arrays.toString(ans)); // Output: [1, 3, 9, 4, 7]
    }
}

Time Complexity: O(n), Space Complexity: O(1)

Finding Four Elements with Equal Sum

Given an array of integers, find four distinct indices (i, j, k, l) such that arr[i] + arr[j] = arr[k] + arr[l]. If multiple solutions exist, return any one.

import java.util.*;
public class EqualSum {
    public ArrayList<Integer> findInidices(ArrayList<Integer> A) {
        Map<Integer, ArrayList<Integer>> map = new HashMap<>();
        ArrayList<Integer> result = new ArrayList<>();
        for (int i = 0; i < A.size() - 1; i++) {
            for (int j = i + 1; j < A.size(); j++) {
                int sum = A.get(i) + A.get(j);
                if (map.containsKey(sum)) {
                    int k = map.get(sum).get(0);
                    int l = map.get(sum).get(1);
                    if (k == i || k == j || l == i || l == j) continue;
                    result.addAll(Arrays.asList(i, j, k, l));
                    return result;
                } else map.put(sum, new ArrayList<>(Arrays.asList(i, j)));
            }
        }
        return new ArrayList<>();
    }
    public static void main(String args[]) {
        EqualSum obj = new EqualSum();
        ArrayList<Integer> al = new ArrayList<>(Arrays.asList(-2, 0, 2, 4, -2, -4));
        System.out.println(obj.findInidices(al)); //Output: [0, 1, 2, 5]
    }
}

Time Complexity: O(n²), Space Complexity: O(n)

Finding the Smallest Window Containing All Characters

Given two strings, str (the text) and X (the characters to find), find the smallest window within str that contains all characters from X. The time complexity must be O(n).

import java.util.*;
public class SmallestWindow {
    int[] a;
    int[] b;
    public String minWindow(String A, String B) {
        a = new int[256];
        b = new int[256];
        for (char c : B.toCharArray()) b[(int) c]++;
        int n = A.length();
        int l = 0;
        int x = -1;
        int y = n + 1;
        for (int r = 0; r < n; r++) {
            char c = A.charAt(r);
            a[(int) c]++;
            while (l <= r && good(A.charAt(l))) {
                a[(int) A.charAt(l)]--;
                l++;
            }
            if (good() && (r - l) < (y - x)) {
                y = r;
                x = l;
            }
        }
        if (x == -1 && y == n + 1) return "";
        return A.substring(x, y + 1);
    }
    public boolean good(char c) {
        for (int i = 0; i < 256; i++) if (a[i] < b[i]) return false;
        return (a[(int) c] - b[(int) c] >= 1);
    }
    public boolean good() {
        for (int i = 0; i < 256; i++) if (a[i] < b[i]) return false;
        return true;
    }
    public static void main(String args[]) {
        SmallestWindow obj = new SmallestWindow();
        String str = "APPLBEXODEBLBNC";
        String t = "LNC";
        System.out.println(obj.minWindow(str, t)); // Output: LBNC
    }
}

Time Complexity: O(n), Space Complexity: O(1)

Longest Increasing-Decreasing Subsequence

Given an array of integers, find the length of the longest subsequence that's first increasing and then decreasing. This problem involves dynamic programming and finding the longest increasing subsequence (LIS) and longest decreasing subsequence (LDS).

import java.util.*;
public class LongestSubsequence {
    public int longestSubsequenceLength(final List<Integer> A) {
        List<Integer> leftLIS = new ArrayList<>();
        List<Integer> rightLIS = new ArrayList<>();
        if (A.isEmpty()) return 0;
        int size = A.size();
        for (int i = 0; i < size; i++) {
            leftLIS.add(1);
            rightLIS.add(1);
        }
        int max = 0;
        for (int i = 0; i < size; i++) {
            for (int j = 0; j < i; j++) {
                if (A.get(i) > A.get(j)) leftLIS.set(i, Math.max(leftLIS.get(i), leftLIS.get(j) + 1));
            }
        }
        for (int i = size - 2; i >= 0; i--) {
            for (int j = size - 1; j > i; j--) {
                if (A.get(i) > A.get(j)) rightLIS.set(i, Math.max(rightLIS.get(i), rightLIS.get(j) + 1));
            }
        }
        for (int i = 0; i < size; i++) {
            max = Math.max(max, leftLIS.get(i) + rightLIS.get(i) - 1);
        }
        return max;
    }
    public static void main(String argvs[]) {
        ArrayList<Integer> al = new ArrayList<>(Arrays.asList(2, 11, 3, 10, 5, 4, 3, 2, 1));
        System.out.println(new LongestSubsequence().longestSubsequenceLength(al)); //Output: 8
    }
}

Time Complexity: O(n²), Space Complexity: O(n)

Counting Ways to Get a Sum with Unlimited Coins

Given a set of coin denominations and a target sum, find the number of ways to make that sum using the coins. You can use each coin denomination as many times as needed. The order of coins does not matter.

import java.util.*;
public class CoinSum {
    public int countWays(ArrayList<Integer> A, int B) {
        int dp[] = new int[B + 1];
        Arrays.fill(dp, 0);
        dp[0] = 1;
        for (int i = 0; i < A.size(); i++) {
            for (int j = A.get(i); j <= B; j++) {
                dp[j] = dp[j] + dp[j - A.get(i)];
            }
        }
        return dp[B];
    }
    public static void main(String argvs[]) {
        CoinSum obj = new CoinSum();
        ArrayList<Integer> X = new ArrayList<>(Arrays.asList(2, 3, 5, 7));
        int sum = 15;
        System.out.println(obj.countWays(X, sum)); //Output: 10
    }
}

Time Complexity: O(n*sum), Space Complexity: O(sum)

Locust for Load Testing

Locust is a flexible, scalable, and easy-to-use load testing tool written in Python. Unlike some other load testing tools, Locust is designed for distributed testing, which means you can easily simulate hundreds or even thousands of concurrent users. This lets you stress test the performance limits of your servers effectively.

Key Features and Benefits of Locust

• User-friendly interface
• Scalability (simulate a large number of users)
• Distributed testing (run tests across multiple machines)
• Python-based (leveraging Python's ease of use and libraries)
• Flexibility (easily customize test scenarios)

Using Locust

[Include a brief explanation of how to use Locust for load testing. This might include steps like defining user behavior, running tests, and analyzing results. You could also include links to Locust documentation or tutorials.]

Conclusion

Locust offers a valuable approach to performance testing. Its usability and flexibility are key advantages.

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