📅 Day 2 Plan – Core Concepts & Application SSC CGL

 ⏱️ Total Study Time: 6–8 hours

Subject	Topic
🧠 Reasoning	Alphabet Series + Coding-Decoding
📐 Quant	HCF, LCM Tricks + Application
📖 English	Synonyms/Antonyms + Sentence Correction
🌏 GK	Fundamental Rights + Static GK

🧠 1. Reasoning – Alphabet Series + Coding-Decoding

✅ Alphabet Series

• A=1, B=2, ..., Z=26

• Understand forward & backward movement

• Practice patterns like:

  • A, C, E, G, ?, ?

  • Z, X, V, T, ?

✅ Coding-Decoding Basics

• Example:
  If CAT = DBU, then code logic is +1 for each letter.
  So, DOG = ? → EPH

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🎯 Practice Now:

1. A, C, F, J, O, ?

2. If TABLE = UBCMF, then CHAIR = ?

3. If MANGO = NZOHQ, what is the code for GRAPE?

📌 Try and send your answers — I’ll explain them.

---

📐 2. Quant – HCF & LCM

✅ Concepts:

• HCF: Highest number that divides both

• LCM: Smallest common multiple

• Formula:
  HCF × LCM = Product of two numbers

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🎯 Practice:

1. HCF of 60 and 48?

2. LCM of 15, 20?

3. If HCF = 6 and LCM = 120, what are possible numbers?

4. Two numbers have HCF = 12 and LCM = 180. If one number is 36, find the other.

📌 Solve & reply — we’ll do step-by-step correction.

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📖 3. English – Synonyms & Error Spotting

✅ 10 Important Synonyms (Learn these):

1. Rapid – Fast

2. Generous – Kind

3. Ancient – Old

4. Brave – Courageous

5. Enormous – Huge

6. Genuine – Real

7. Polite – Courteous

8. Silent – Mute

9. Answer – Reply

10. Famous – Renowned

📝 Task: Write sentences using 5 of them!

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✅ Grammar Practice (Fix errors):

1. He don’t goes to gym daily.

2. Neither of the boy are present.

3. She is knowing the truth.

📌 Reply with corrected versions — I’ll guide.

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🌏 4. GK – Indian Constitution: Fundamental Rights

🇮🇳 6 Fundamental Rights:

1. Right to Equality

2. Right to Freedom

3. Right against Exploitation

4. Right to Freedom of Religion

5. Cultural and Educational Rights

6. Right to Constitutional Remedies

Q&A:

• Who is called the “Father of Indian Constitution”?

• How many articles are there in the Indian Constitution?

• Which Article gives Right to Education?

---

📰 Current Affairs Task:

👉 Read top 5 headlines of June 27 & 28 from GKToday
📌 You can ask me for a summary too!

---

✅ Homework Summary for Day 2

Subject	Task
Reasoning	3 coding/series questions
Quant	4 HCF/LCM questions
English	Learn synonyms + fix 3 sentences
GK	Learn 6 rights + 3 polity questions
CA	Read today’s current affairs

📘 DAY 1 (1-50): FOUNDATION - SSC CGL

 

📘 DAY 1: FOUNDATION

We’ll cover today:

1. Reasoning – Number Series

2. Quant – Number System Basics

3. English – Vocabulary & Error Spotting

4. GK – Indian Polity + Current Affairs

---

🧠 1. Reasoning – Number Series

🔍 Concept:

A number series is a sequence that follows a logical rule. Your task is to find the next number or find the wrong one.

🧩 Example Patterns:

• Addition: 2, 4, 6, 8, __ (Answer: 10)

• Multiplication: 3, 6, 12, 24, __ (Answer: 48)

• Squares: 1, 4, 9, 16, __ (Answer: 25)

• Fibonacci: 1, 1, 2, 3, 5, 8, __ (Answer: 13)

---

🎯 Practice Now (Try these):

1. 2, 6, 12, 20, 30, ?

2. 5, 10, 20, 40, ?

3. 100, 81, 64, 49, ?

4. 1, 4, 9, 16, 25, ?

📌 Write answers and reply — I’ll check and explain.

---

📐 2. Quant – Number System Basics

✍️ Concepts:

• Types of Numbers:

  • Natural: 1, 2, 3...

  • Whole: 0, 1, 2...

  • Even: 2, 4, 6...

  • Prime: 2, 3, 5, 7...

• Divisibility Rules (Learn today for 2 to 6)

  • Div by 2: Ends in 0, 2, 4, 6, 8

  • Div by 3: Sum of digits divisible by 3

  • Div by 5: Ends in 0 or 5

  • Div by 6: Divisible by 2 and 3 both

---

🎯 Practice Now:

1. Find LCM of 18 and 24

2. Is 527 divisible by 3?

3. Which is the smallest number divisible by 6, 8 and 12?

4. What type of number is 1? (Prime/Composite?)

📌 Write your answers — I’ll review and teach more if needed.

---

📖 3. English – Vocab + Grammar

📝 Task:

• Learn these 10 words:

  • Abandon

  • Benevolent

  • Candid

  • Deter

  • Elusive

  • Frugal

  • Hinder

  • Jubilant

  • Keen

  • Lucid

Write 2 sentences for each to learn better!

---

🧐 Grammar Practice:

Find the error:

1. She go to school everyday.

2. He don’t like tea.

3. They has gone to the market.

📌 Try fixing these. I’ll explain each.

---

🌏 4. GK – Indian Polity (Basics)

🇮🇳 Learn Today:

• First President of India – Dr. Rajendra Prasad

• Current President (2025) – You tell me?

• Rajya Sabha – Upper House

• Lok Sabha – Lower House

• Members in Lok Sabha – 543

• Members in Rajya Sabha – 245

---

📰 Current Affairs (June 2025):

👉 Visit AffairsCloud or GKToday and read:

• National events

• Sports

• Government schemes launched in June

📌 I can send today’s curated current affairs summary too, just say yes.

---

✅ Homework Summary

Subject	Task
Reasoning	Solve 5 number series
Quant	Practice 4 number system questions
English	Learn 10 words + correct 3 sentences
GK	Revise polity + read June 2025 current affairs

✅ Topics for Day 6 DSA with Java

Focus: Two Pointer Technique

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🔁 1. Learn & Implement Two Pointer Technique

• Understand how the two-pointer technique works on sorted arrays.

• Implement the following problems:

// Java Example: Two Sum (Sorted Array)
public int[] twoSum(int[] numbers, int target) {
    int left = 0, right = numbers.length - 1;
    while (left < right) {
        int sum = numbers[left] + numbers[right];
        if (sum == target) {
            return new int[] { left + 1, right + 1 }; // 1-indexed
        } else if (sum < target) {
            left++;
        } else {
            right--;
        }
    }
    return new int[] {}; // no solution
}

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🧠 2. Solve 2 LeetCode Problems Using Two Pointers

1. LeetCode 167. Two Sum II – Input Array Is Sorted

2. LeetCode 283. Move Zeroes

   • Optimize with minimal swaps using the two-pointer pattern.

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✨ Bonus Challenges (Optional but Recommended)

• LeetCode 26. Remove Duplicates from Sorted Array

• LeetCode 844. Backspace String Compare

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📊 Explore Visualization

Try two-pointer visualizations at:
👉 https://visualgo.net/en/list

---

🧩 Concept Summary

• Two pointers are most useful for:

  • Sorted arrays

  • Reverse traversal

  • Sliding window (upcoming)

  • Partitioning logic

✅ Day 5 Solution - DSA with Java

 

✅ 1. Implement Merge Sort and Quick Sort

🔹 Merge Sort

public class MergeSort {
    public static void mergeSort(int[] arr, int left, int right) {
        if (left < right) {
            int mid = (left + right) / 2;

            mergeSort(arr, left, mid);
            mergeSort(arr, mid + 1, right);

            merge(arr, left, mid, right);
        }
    }

    public static void merge(int[] arr, int left, int mid, int right) {
        int n1 = mid - left + 1, n2 = right - mid;
        int[] L = new int[n1];
        int[] R = new int[n2];

        for (int i = 0; i < n1; i++) L[i] = arr[left + i];
        for (int j = 0; j < n2; j++) R[j] = arr[mid + 1 + j];

        int i = 0, j = 0, k = left;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) arr[k++] = L[i++];
            else arr[k++] = R[j++];
        }

        while (i < n1) arr[k++] = L[i++];
        while (j < n2) arr[k++] = R[j++];
    }
}

---

🔹 Quick Sort

public class QuickSort {
    public static void quickSort(int[] arr, int low, int high) {
        if (low < high) {
            int pi = partition(arr, low, high);

            quickSort(arr, low, pi - 1);
            quickSort(arr, pi + 1, high);
        }
    }

    public static int partition(int[] arr, int low, int high) {
        int pivot = arr[high];
        int i = low - 1;

        for (int j = low; j < high; j++) {
            if (arr[j] < pivot) {
                i++;
                // Swap
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }

        // Final pivot swap
        int temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;

        return i + 1;
    }
}

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✅ 2. LeetCode Problems

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🔸 Merge Sorted Array

Merge nums2 into nums1 sorted in-place.

public class MergeSortedArray {
    public static void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m - 1, j = n - 1, k = m + n - 1;

        while (i >= 0 && j >= 0) {
            nums1[k--] = (nums1[i] > nums2[j]) ? nums1[i--] : nums2[j--];
        }

        while (j >= 0) {
            nums1[k--] = nums2[j--];
        }
    }
}

---

🔸 Sort an Array

Use any sorting method. Here's Merge Sort used:

public class SortAnArray {
    public static int[] sortArray(int[] nums) {
        mergeSort(nums, 0, nums.length - 1);
        return nums;
    }

    public static void mergeSort(int[] arr, int l, int r) {
        if (l < r) {
            int m = (l + r) / 2;
            mergeSort(arr, l, m);
            mergeSort(arr, m + 1, r);
            merge(arr, l, m, r);
        }
    }

    public static void merge(int[] arr, int l, int m, int r) {
        int n1 = m - l + 1, n2 = r - m;
        int[] L = new int[n1];
        int[] R = new int[n2];
        for (int i = 0; i < n1; ++i) L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j) R[j] = arr[m + 1 + j];

        int i = 0, j = 0, k = l;
        while (i < n1 && j < n2)
            arr[k++] = (L[i] <= R[j]) ? L[i++] : R[j++];
        while (i < n1) arr[k++] = L[i++];
        while (j < n2) arr[k++] = R[j++];
    }
}

---

✅ 3. Non-Recursive Quick Sort (Iterative)

import java.util.Stack;

public class QuickSortIterative {
    public static void quickSort(int[] arr) {
        Stack<int[]> stack = new Stack<>();
        stack.push(new int[]{0, arr.length - 1});

        while (!stack.isEmpty()) {
            int[] range = stack.pop();
            int low = range[0], high = range[1];

            if (low < high) {
                int pi = partition(arr, low, high);
                stack.push(new int[]{low, pi - 1});
                stack.push(new int[]{pi + 1, high});
            }
        }
    }

    private static int partition(int[] arr, int low, int high) {
        int pivot = arr[high], i = low - 1;
        for (int j = low; j < high; j++) {
            if (arr[j] < pivot) {
                i++;
                // Swap
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
        // Final pivot placement
        int temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;
        return i + 1;
    }
}

---

✅ 4. Visualize Merge vs Quick Sort

You can interactively visualize how Merge and Quick Sort work at:

🔗 Visualgo.net – Sorting Visualizations

Click:

• Merge Sort → observe divide → merge steps

• Quick Sort → watch pivot partitioning → recursive reduction

💡 Use small array sizes (e.g. 10 elements) to step-through for learning.

🚀 Day 5: Merge Sort & Quick Sort in Java 🎯 Goals for Today

• Understand the Divide and Conquer strategy

• Learn and implement:

  • Merge Sort

  • Quick Sort

• Analyze Time & Space Complexities

• Solve intermediate-level sorting problems

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📘 1. Merge Sort

A stable, divide-and-conquer sorting algorithm.
Time Complexity: O(n log n) in all cases
Space Complexity: O(n) (extra array)

✅ Logic:

• Divide the array into two halves

• Sort each half recursively

• Merge the two sorted halves

🔹 Java Implementation

public class MergeSort {
    public static void mergeSort(int[] arr, int left, int right) {
        if (left < right) {
            int mid = (left + right) / 2;

            // Sort the left and right halves
            mergeSort(arr, left, mid);
            mergeSort(arr, mid + 1, right);

            // Merge them
            merge(arr, left, mid, right);
        }
    }

    public static void merge(int[] arr, int left, int mid, int right) {
        int n1 = mid - left + 1;
        int n2 = right - mid;

        int[] L = new int[n1];
        int[] R = new int[n2];

        for (int i = 0; i < n1; i++) L[i] = arr[left + i];
        for (int j = 0; j < n2; j++) R[j] = arr[mid + 1 + j];

        int i = 0, j = 0, k = left;

        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) arr[k++] = L[i++];
            else arr[k++] = R[j++];
        }

        while (i < n1) arr[k++] = L[i++];
        while (j < n2) arr[k++] = R[j++];
    }
}

---

⚡ 2. Quick Sort

Quick and efficient, especially on average cases.
Time Complexity:

• Best/Average: O(n log n)

• Worst: O(n²) (when pivot is worst chosen)
  Space: O(log n) (due to recursive calls)

✅ Logic:

• Choose a pivot

• Partition the array: elements < pivot go left, > pivot go right

• Recursively sort left and right

🔹 Java Implementation

public class QuickSort {
    public static void quickSort(int[] arr, int low, int high) {
        if (low < high) {
            int pi = partition(arr, low, high);

            quickSort(arr, low, pi - 1);
            quickSort(arr, pi + 1, high);
        }
    }

    public static int partition(int[] arr, int low, int high) {
        int pivot = arr[high];
        int i = low - 1; // Index of smaller element

        for (int j = low; j < high; j++) {
            if (arr[j] < pivot) {
                i++;
                // Swap
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }

        // Swap pivot to the right place
        int temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;

        return i + 1;
    }
}

---

📊 3. Merge Sort vs Quick Sort

Feature	Merge Sort	Quick Sort
Time (Best)	O(n log n)	O(n log n)
Time (Worst)	O(n log n)	O(n²)
Space	O(n)	O(log n)
Stable	✅ Yes	❌ No
In-place	❌ No	✅ Yes
Use Case	Linked lists	Arrays

🧠 4. Practice Problems

Try these problems:

1. ✅ Merge Sorted Array

2. ✅ Sort an Array

3. 🔁 Kth Largest Element in Array – Use Quick Select

4. 🧠 Count Inversions in Array – Use Merge Sort logic

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📚 Homework for Day 5

• Implement Merge Sort and Quick Sort

• Solve 2 LeetCode problems involving sorting or partitioning

• Try to write non-recursive quick sort (advanced)

• Visualize Merge vs Quick: Try https://visualgo.net/en/sorting

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🔜 Coming Up on Day 6:

• Recursion Mastery

• Backtracking intro (Subset, Permutations)

• Understanding call stack and dry runs

✅ Day 4 Solution - DSA with Java

 

✅ 1. Implement All Three Sorting Algorithms

🔹 Bubble Sort

public class BubbleSort {
    public static void bubbleSort(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n - 1; i++) {
            boolean swapped = false;
            for (int j = 0; j < n - i - 1; j++) {
                if (arr[j] > arr[j + 1]) {
                    // Swap
                    int temp = arr[j];
                    arr[j] = arr[j + 1];
                    arr[j + 1] = temp;
                    swapped = true;
                }
            }
            if (!swapped) break; // Optimization
        }
    }
}

---

🔹 Selection Sort

public class SelectionSort {
    public static void selectionSort(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n - 1; i++) {
            int minIdx = i;
            for (int j = i + 1; j < n; j++) {
                if (arr[j] < arr[minIdx]) {
                    minIdx = j;
                }
            }
            // Swap
            int temp = arr[i];
            arr[i] = arr[minIdx];
            arr[minIdx] = temp;
        }
    }
}

---

🔹 Insertion Sort

public class InsertionSort {
    public static void insertionSort(int[] arr) {
        for (int i = 1; i < arr.length; i++) {
            int key = arr[i];
            int j = i - 1;

            while (j >= 0 && arr[j] > key) {
                arr[j + 1] = arr[j];
                j--;
            }
            arr[j + 1] = key;
        }
    }
}

---

✅ 2. LeetCode Problems

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🔸 Problem 1: Sort Colors

Sort an array containing only 0s, 1s, and 2s.
Use Dutch National Flag Algorithm (O(n) time, O(1) space)

public class SortColors {
    public static void sortColors(int[] nums) {
        int low = 0, mid = 0, high = nums.length - 1;

        while (mid <= high) {
            if (nums[mid] == 0) {
                // Swap with low
                int temp = nums[low];
                nums[low] = nums[mid];
                nums[mid] = temp;
                low++;
                mid++;
            } else if (nums[mid] == 1) {
                mid++;
            } else {
                // nums[mid] == 2
                int temp = nums[mid];
                nums[mid] = nums[high];
                nums[high] = temp;
                high--;
            }
        }
    }
}

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🔸 Problem 2: Check if Array Is Sorted and Rotated

Return true if array is sorted and rotated at most once.

🔍 Logic:

Count how many times arr[i] > arr[i+1]. If it happens more than once, it's not valid.

public class CheckSortedRotated {
    public static boolean check(int[] nums) {
        int count = 0;
        int n = nums.length;

        for (int i = 0; i < n; i++) {
            if (nums[i] > nums[(i + 1) % n]) {
                count++;
            }
        }

        return count <= 1;
    }

    public static void main(String[] args) {
        int[] arr = {3, 4, 5, 1, 2};
        System.out.println("Is Sorted and Rotated? " + check(arr)); // true
    }
}

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🧠 Summary of Day 4

• ✅ Bubble, Selection, Insertion Sort implemented

• ✅ LeetCode problems solved

  • Dutch National Flag (0s, 1s, 2s)

  • Check Sorted & Rotated

✅ Day 5 Solution - DSA with C

 

✅ 1. Implement your own strlen()

#include <stdio.h>

int my_strlen(char str[]) {
    int i = 0;
    while (str[i] != '\0') {
        i++;
    }
    return i;
}

int main() {
    char str[100];
    printf("Enter a string: ");
    scanf("%s", str);
    printf("Length of string = %d\n", my_strlen(str));
    return 0;
}

---

✅ 2. Implement your own strcpy()

#include <stdio.h>

void my_strcpy(char dest[], char src[]) {
    int i = 0;
    while (src[i] != '\0') {
        dest[i] = src[i];
        i++;
    }
    dest[i] = '\0'; // Null-terminate
}

int main() {
    char src[100], dest[100];
    printf("Enter a string to copy: ");
    scanf("%s", src);

    my_strcpy(dest, src);
    printf("Copied string: %s\n", dest);
    return 0;
}

---

✅ 3. Implement your own strcmp()

#include <stdio.h>

int my_strcmp(char str1[], char str2[]) {
    int i = 0;
    while (str1[i] != '\0' && str2[i] != '\0') {
        if (str1[i] != str2[i])
            return str1[i] - str2[i];
        i++;
    }
    return str1[i] - str2[i]; // Also handles different lengths
}

int main() {
    char str1[100], str2[100];
    printf("Enter two strings:\n");
    scanf("%s %s", str1, str2);

    int result = my_strcmp(str1, str2);
    if (result == 0)
        printf("Strings are equal.\n");
    else if (result < 0)
        printf("First string is smaller.\n");
    else
        printf("First string is greater.\n");

    return 0;
}

---

✅ 4. LeetCode Problem: Valid Anagram

Two strings are anagrams if they contain the same characters in any order.

✨ Custom C Solution:

#include <stdio.h>
#include <string.h>

int isAnagram(char s[], char t[]) {
    int count[26] = {0};

    if (strlen(s) != strlen(t))
        return 0;

    for (int i = 0; s[i] != '\0'; i++) {
        count[s[i] - 'a']++;
        count[t[i] - 'a']--;
    }

    for (int i = 0; i < 26; i++) {
        if (count[i] != 0)
            return 0;
    }

    return 1;
}

int main() {
    char s[100], t[100];
    printf("Enter first string: ");
    scanf("%s", s);
    printf("Enter second string: ");
    scanf("%s", t);

    if (isAnagram(s, t))
        printf("Valid Anagram\n");
    else
        printf("Not an Anagram\n");

    return 0;
}

✅ This C code solves the Valid Anagram LeetCode problem without using libraries like sort.