Most Critical Mistakes & Tips in Competitive Programming

The moment a beginner hears this word, an image pops up in one’s mind where students are sitting in a room full of Computers and coding some out of the world stuff, We’ll to be honest it’s nothing too hard to grasp so we will be proposing tips that help a programmer to level up faster. So let’s begin with the most underrated points on each topic where even the advanced competitive programmers slip.

So the concept of overflow can be seen in competitive coding a lot, the worst problem of not understanding it is, writing a completely perfect code but Still getting the wrong answer, this is because, at higher input ranges, the compiler strips down the result as per the data size hence it’s important to remember the basic ranges of data types.

Int => [-109  To 109]
Long Int / long => [-1012,1012]
Long Long Int / long long => [-1018,10^18]

Example:

C++

// C++ Program, to Illustrate General Tips In Competitive
// Programming

// Importing all C++ libraries
#include <bits/stdc++.h>

using namespace std;

// Main method
int main()
{

    // Case: Stack Overflow
    int a = 100000, b = 100000;
    // Expected answer ? 10^10 but
    // no the answer you get is 1410065408, error in
    // precision.
    int c = a * b;
    long long int d = a * b;

    // Note: Still error will be generated because
    // calculation was done on two ints which were later
    // converted into long long ie they already
    // lost their data before converting

    // Print and display on console
    cout << c << " " << d << endl;

    // Now if we store a value more than its capacity then
    // what happens is the number line of range of value
    // becomes a number
    // Example: Circle, If min val is 1 and max is 9, so if
    // we add 1 to 9 it will result in 1, it looped back to
    // starting, this is overflow

    int p = INT_MAX;

    // An example of overflow
    cout << "An example of overflow " << p + 1 << endl;

    // Long long is way better than double,
    // double although can store more than long long but
    // in exchange it will cost you your precision.
    // We can simply use the below command as follows:
    // cout<<fixed(no scientific
    // notation)<<setprecision(0){removes decimal}<<variable
    // to give value same form as long long

    // Question where the answer came wrong coz of
    // overflow!!
    int t, n;
    cin >> t;
    while (t--) {
        cin >> n;

        // Here, before i used int and got wrong answer,
        // then made it long long
        long long pdt = 1, temp;

        for (int i = 0; i < n; i++) {
            cin >> temp;
            pdt *= temp;
        }

        if (pdt % 10 == 2 || pdt % 10 == 3 || pdt % 10 == 5)
            cout << "YES" << endl;

        else
            cout << "NO" << endl;
    }
}

Java

// Java Program, to Illustrate General Tips In Competitive
// Programming

// Importing required Java libraries
import java.util.*;

public class Main {

  // Main method
  public static void main(String[] args)
  {

    // Case: Stack Overflow
    int a = 100000, b = 100000;
    // Expected answer ? 10^10 but
    // no the answer you get is 1410065408, error in
    // precision.
    int c = a * b;
    long d = (long)a * b;

    // Note: Still error will be generated because
    // calculation was done on two ints which were later
    // converted into long long ie they already
    // lost their data before converting

    // Print and display on console
    System.out.println(c + " " + d);

    // Now if we store a value more than its capacity
    // then what happens is the number line of range of
    // value becomes a number Example: Circle, If min
    // val is 1 and max is 9, so if we add 1 to 9 it
    // will result in 1, it looped back to starting,
    // this is overflow

    int p = Integer.MAX_VALUE;

    // An example of overflow
    System.out.println("An example of overflow "
                       + (p + 1));

    // Long is way better than double,
    // double although can store more than long but
    // in exchange it will cost you your precision.
    // We can simply use the below command as follows:
    // System.out.printf("%.0f", variable)
    // to give value same form as long

    // Scanner to read input
    Scanner sc = new Scanner(System.in);

    // Question where the answer came wrong coz of
    // overflow!!
    int t = sc.nextInt();
    while (t-- > 0) {
      int n = sc.nextInt();

      // Here, before i used int and got wrong answer,
      // then made it long
      long pdt = 1, temp;

      for (int i = 0; i < n; i++) {
        temp = sc.nextLong();
        pdt *= temp;
      }

      if (pdt % 10 == 2 || pdt % 10 == 3
          || pdt % 10 == 5)
        System.out.println("YES");

      else
        System.out.println("NO");
    }
  }
}

C#

// C# Program, to Illustrate General Tips In Competitive
// Programming

// Importing required C# libraries
using System;

public class MainClass {

    // Main method
    public static void Main(string[] args)
    {
        // Case: Stack Overflow
        int a = 100000, b = 100000;
        // Expected answer ? 10^10 but
        // no the answer you get is 1410065408, error in
        // precision.
        int c = a * b;
        long d = (long)a * b;
        // Note: Still error will be generated because
        // calculation was done on two ints which were later
        // converted into long long ie they already
        // lost their data before converting

        // Print and display on console
        Console.WriteLine(c + " " + d);

        // Now if we store a value more than its capacity
        // then what happens is the number line of range of
        // value becomes a number Example: Circle, If min
        // val is 1 and max is 9, so if we add 1 to 9 it
        // will result in 1, it looped back to starting,
        // this is overflow

        int p = int.MaxValue;

        // An example of overflow
        Console.WriteLine("An example of overflow "
                          + (p + 1));

        // Long is way better than double,
        // double although can store more than long but
        // in exchange it will cost you your precision.
        // We can simply use the below command as follows:
        // Console.WriteLine("{0:0}", variable)
        // to give value same form as long

        // Reading input
        int t = int.Parse(Console.ReadLine());
        while (t-- > 0) {
            int n = int.Parse(Console.ReadLine());

            // Here, before i used int and got wrong answer,
            // then made it long
            long pdt = 1, temp;

            for (int i = 0; i < n; i++) {
                temp = long.Parse(Console.ReadLine());
                pdt *= temp;
            }

            if (pdt % 10 == 2 || pdt % 10 == 3
                || pdt % 10 == 5)
                Console.WriteLine("YES");

            else
                Console.WriteLine("NO");
        }
    }
}

Python3

# Case: Stack Overflow
a = 100000
b = 100000
# Expected answer ? 10^10 but
# no the answer you get is 1410065408, error in
# precision.
c = a * b
d = a * b
# Note: Still error will be generated because
# calculation was done on two ints which were later
# converted into long long ie they already
# lost their data before converting

# Print and display on console
print(c, d)

# Now if we store a value more than its capacity then
# what happens is the number line of range of value
# becomes a number
# Example: Circle, If min val is 1 and max is 9, so if
# we add 1 to 9 it will result in 1, it looped back to
# starting, this is overflow

p = float("inf")

# An example of overflow
print("An example of overflow ", p + 1)

# Long long is way better than double,
# double although can store more than long long but
# in exchange it will cost you your precision.
# We can simply use the below command as follows:
# print(fixed(no scientific notation) setprecision(0){removes decimal} variable)
# to give value same form as long long

# Question where the answer came wrong coz of overflow!!
t = int(input())
while t > 0:
    n = int(input())

    # Here, before i used int and got wrong answer,
    # then made it long long
    pdt = 1
    for i in range(n):
        temp = int(input())
        pdt *= temp

    if pdt % 10 == 2 or pdt % 10 == 3 or pdt % 10 == 5:
        print("YES")
    else:
        print("NO")

    t -= 1

Output

1410065408 1410065408
An example of overflow -2147483648

Tips For Data Structures

Tip 1: In competitive programming always declare the array globally when you need to have a function along with the main function, globally declared arrays can have a size of 10^7 as they are stored in Data Segments.

Tip 2: Whenever you declare a function where you need to return multiple items, or even in general where your goal is to return an array after some modifications, always go for Pass By Reference (eg void swap(int &a, int &b)), it helps you in two ways one, by reducing the time in making a copy of the data and then processing on it and second, as you’re directly working on the main data you can modify as many values as you wish without having to worry about returning the values and dealing with them.

• The limit of the size of an array declared inside the main method is 10^5, coz it’s stored in Stack Memory with a memory limit of around 8MB, if the size is more than this then it results in Stack Overflow.
• A very peculiar error occurs sometimes when you’re taking multiple strings as input after a cin input. instead of taking required inputs, it automatically takes space as input then takes the remaining two strings, to resolve this use cin.ignore() before using the getline() to take the string as input.
• while forming a new string from another string, instead of using syntax like str = str + str2[I], ie adding character to get a string, this method’s time complexity is O(size(string)), Hence listed of adding characters to a string we use push_back(character) whose time complexity is O(n).

Tips on STL

As we already know by far now that STL stands for Standard Template Library in C++, which is a boon for competitive programmers, that contains inbuilt special Data Structures and Algorithm that reduces and optimize your code like Crazy. DO remember following below listed STL data structures tips as follows:

Tip 1: Whenever we need to sort data in a question and remove duplicates, the best practice is to store it in a Set, it automatically sorts the data and as per the property of Set, It only stores unique values.

Tip 2: In questions where we need to find the frequency of all values, the best practice is to use a Map, The property of Map makes it automatically arrange the data in lexicographical order, and just with a little trick, you’ll have the most efficient solution.

The same can be done in java as well, below is the code for above tips. 

Example

C++

// C++ Program to Illustrate STL Algorithms Tips

// Importing all C++ libraries
// Standard command
#include <bits/stdc++.h>

using namespace std;

// Main method
// From here the code starts executing
int main()
{

    // Creating vector and initializing objects
    // by passing custom integer numbers
    vector<int> v
        = { 1, 4, 2, 5, 6, 3, 9, 0, 10, 3, 15, 17, 3 };

    // Operation 1
    // Finding the minimum element in the array,

    // Note: It returns the pointer/iterator
    auto min = min_element(v.begin(), v.end());

    // Print and display elements on console
    cout << "Min element: " << *min << endl;

    // Operation 2
    // // Finding the maximum  element in the array
    auto max = max_element(v.begin(), v.end());
    cout << "Max Element: " << *max << endl;

    // Operation 3
    // Sum of all elements, the third parameter tells us
    // initial sum ki value kya hai.
    auto sum = accumulate(v.begin(), v.end(), 0);
    cout << "The sum of all elements: " << sum << endl;

    // Operation 4
    // Count the frequency of an element in the array/vector
    auto cnt = count(v.begin(), v.end(), 3);
    cout << "Frequency Of 3 is: " << cnt << endl;

    // Operation 5
    // Finds an element and returns its pointer
    auto elem = find(v.begin(), v.end(), 3);

    if (elem != v.end())
        cout << "the element is at posn: " << *elem << endl;
    else
        cout << "Element not found" << endl;

    // Operation 6
    // Reversing a string or array/vector
    // using reverse method
    reverse(v.begin(), v.end());

    cout << "Reversed Vector: ";

    for (auto val : v) {
        cout << val << " ";
    }

    // New line for better readability
    cout << endl;

    // Operation 7
    // Sort array/vector using sort() method
    sort(v.begin(), v.end());
    cout << "Sorted Vector: ";
    for (auto val : v) {
        cout << val << " ";
    }
}

Java

import java.util.*;

public class Main {

    public static void main(String[] args) {
        // Creating ArrayList and initializing objects
        // by passing custom integer numbers
        ArrayList<Integer> list = new ArrayList<>(
                Arrays.asList(1, 4, 2, 5, 6, 3, 9, 0, 10, 3, 15, 17, 3));

        // Operation 1
        // Finding the minimum element in the array,

        // Note: It returns the pointer/iterator
        int min = Collections.min(list);

        // Print and display elements on console
        System.out.println("Min element: " + min);

        // Operation 2
        // Finding the maximum element in the array
        int max = Collections.max(list);
        System.out.println("Max Element: " + max);

        // Operation 3
        // Sum of all elements, the third parameter tells us
        // initial sum ki value kya hai.
        int sum = 0;
        for (int val : list) {
            sum += val;
        }
        System.out.println("The sum of all elements: " + sum);

        // Operation 4
        // Count the frequency of an element in the array/list
        int cnt = Collections.frequency(list, 3);
        System.out.println("Frequency Of 3 is: " + cnt);

        // Operation 5
        // Finds an element and returns its pointer
        int index = list.indexOf(3);

        if (index != -1)
            System.out.println("the element is at posn: " + index);
        else
            System.out.println("Element not found");

        // Operation 6
        // Reversing a list using Collections.reverse method
        Collections.reverse(list);

        System.out.print("Reversed List: ");
        for (int val : list) {
            System.out.print(val + " ");
        }

        // New line for better readability
        System.out.println();

        // Operation 7
        // Sort array/list using Collections.sort() method
        Collections.sort(list);
        System.out.print("Sorted List: ");
        for (int val : list) {
            System.out.print(val + " ");
        }
    }
}

Python3

# Creating a list and initializing objects
# by passing custom integer numbers
my_list = [1, 4, 2, 5, 6, 3, 9, 0, 10, 3, 15, 17, 3]

# Operation 1
# Finding the minimum element in the list
min_element = min(my_list)
print("Min element:", min_element)

# Operation 2
# Finding the maximum element in the list
max_element = max(my_list)
print("Max element:", max_element)

# Operation 3
# Sum of all elements
sum_of_elements = sum(my_list)
print("The sum of all elements:", sum_of_elements)

# Operation 4
# Count the frequency of an element in the list
frequency_of_3 = my_list.count(3)
print("Frequency of 3 is:", frequency_of_3)

# Operation 5
# Finds an element and returns its index
try:
    index = my_list.index(3)
    print("The element is at position:", index)
except ValueError:
    print("Element not found")

# Operation 6
# Reversing a list using reverse() method
my_list.reverse()
print("Reversed List:", my_list)

# Operation 7
# Sort list using sort() method
my_list.sort()
print("Sorted List:", my_list)

C#

using System;
using System.Collections.Generic;
using System.Linq;

public class MainClass
{
    public static void Main(string[] args)
    {
        // Creating List and initializing objects
        // by passing custom integer numbers
        List<int> list = new List<int>
        {
            1, 4, 2, 5, 6, 3, 9, 0, 10, 3, 15, 17, 3
        };

        // Operation 1
        // Finding the minimum element in the list
        int min = list.Min();

        // Print and display elements on console
        Console.WriteLine("Min element: " + min);

        // Operation 2
        // Finding the maximum element in the list
        int max = list.Max();
        Console.WriteLine("Max Element: " + max);

        // Operation 3
        // Sum of all elements
        int sum = list.Sum();
        Console.WriteLine("The sum of all elements: " + sum);

        // Operation 4
        // Count the frequency of an element in the list
        int cnt = list.Count(x => x == 3);
        Console.WriteLine("Frequency Of 3 is: " + cnt);

        // Operation 5
        // Finds an element and returns its index
        int index = list.IndexOf(3);

        if (index != -1)
            Console.WriteLine("the element is at posn: " + index);
        else
            Console.WriteLine("Element not found");

        // Operation 6
        // Reversing a list using list.Reverse method
        list.Reverse();

        Console.Write("Reversed List: ");
        foreach (int val in list)
        {
            Console.Write(val + " ");
        }

        // New line for better readability
        Console.WriteLine();

        // Operation 7
        // Sort list using list.Sort() method
        list.Sort();
        Console.Write("Sorted List: ");
        foreach (int val in list)
        {
            Console.Write(val + " ");
        }
    }
}

Javascript

// Creating an array and initializing objects
// by passing custom integer numbers
let myArray = [1, 4, 2, 5, 6, 3, 9, 0, 10, 3, 15, 17, 3];

// Operation 1
// Finding the minimum element in the array
let minElement = Math.min(...myArray);
console.log("Min element:", minElement);

// Operation 2
// Finding the maximum element in the array
let maxElement = Math.max(...myArray);
console.log("Max element:", maxElement);

// Operation 3
// Sum of all elements
let sumOfElements = myArray.reduce((a, b) => a + b, 0);
console.log("The sum of all elements:", sumOfElements);

// Operation 4
// Count the frequency of an element in the array
let frequencyOf3 = myArray.filter(x => x === 3).length;
console.log("Frequency of 3 is:", frequencyOf3);

// Operation 5
// Finds an element and returns its index
let index = myArray.indexOf(3);
if (index !== -1) {
console.log("The element is at position:", index);
} else {
console.log("Element not found");
}

// Operation 6
// Reversing an array using reverse() method
myArray.reverse();
console.log("Reversed Array:", myArray);

// Operation 7
// Sort array using sort() method
myArray.sort((a, b) => a - b);
console.log("Sorted Array:", myArray);

Output

Min element: 0
Max Element: 17
The sum of all elements: 78
Frequency Of 3 is: 3
the element is at posn: 3
Reversed Vector: 3 17 15 3 10 0 9 3 6 5 2 4 1 
Sorted Vector: 0 1 2 3 3 3 4 5 6 9 10 15 17 

Note: These are some popular algorithms that can make your life easy all have the same pattern of usage ie the name of the function

(lower lim, upper lim+1)