Assignment of Advanced data structure and algorithms ..pdf

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Assignment of ADSA.
Answer 1:
To find the largest element in a given array, you can follow these general steps:
Initialize a variable to hold the largest value. You can set this variable to the first element of
the array initially.
Iterate through the array, comparing each element to the current largest value.
Update the largest value if the current element is greater than the current largest value.
Return the largest value after completing the iteration.
Here's an example in Python:
python
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def find_largest_element(array):
if not array:
raise ValueError("The array is empty")
largest = array[0]
for num in array:
if num > largest:
largest = num
return largest
In this function:
largest is initially set to the first element of the array.
We loop through each element (num) in the array.
If num is greater than largest, we update largest.
Finally, we return the largest value found.
You can adapt this approach to other programming languages with similar logic.
Answer 2:
To reverse a given array, you can follow these steps:
Create a new array to store the reversed elements, or
Modify the array in place if you don't need to keep the original order.
Here are methods to achieve this in different programming languages:
Python
Using slicing (creates a new reversed array):
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def reverse_array(arr):
return arr[::-1]
In-place reversal:

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def reverse_array_in_place(arr):
left, right = 0, len(arr) - 1
while left < right:
arr[left], arr[right] = arr[right], arr[left]
left += 1
right -= 1
return arr
JavaScript
Using the reverse method (modifies the original array):
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function reverseArray(arr) {
return arr.reverse();
}
Java
Using a loop to reverse in place:
java
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public class Main {
public static void reverseArray(int[] arr) {
int left = 0;
int right = arr.length - 1;
while (left < right) {
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left++;
right--;
}
}
}
C++
Using a loop to reverse in place:
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#include <algorithm> // For std::swap

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void reverseArray(int arr[], int size) {
int left = 0;
int right = size - 1;
while (left < right) {
std::swap(arr[left], arr[right]);
left++;
right--;
}
}
Each method effectively reverses the order of the elements in the array. Choose the one that best
fits your needs based on whether you want to modify the array in place or create a new reversed
array.
Answer 3:
To find the second largest element in a given array, you can follow these steps:
Initialize two variables: one for the largest element and one for the second largest element.
Set both initially to a very low value or the smallest possible value for your data type.
Iterate through the array. For each element:
If the element is greater than the current largest element, update the second largest to
be the current largest and then update the largest to this new element.
If the element is not the largest but is greater than the second largest, update the
second largest.
Here is how you can implement this in various programming languages:
Python
python
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def find_second_largest(arr):
if len(arr) < 2:
raise ValueError("Array must contain at least two elements")
largest = second_largest = float('-inf')
for num in arr:
if num > largest:
second_largest = largest
largest = num
elif num > second_largest and num != largest:
second_largest = num

if second_largest == float('-inf'):
raise ValueError("There is no second largest element")
return second_largest
JavaScript
javascript
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function findSecondLargest(arr) {
if (arr.length < 2) {
throw new Error("Array must contain at least two elements");
}
let largest = -Infinity;
let secondLargest = -Infinity;
for (let num of arr) {
if (num > largest) {
secondLargest = largest;
largest = num;
} else if (num > secondLargest && num < largest) {
secondLargest = num;
}
}
if (secondLargest === -Infinity) {
throw new Error("There is no second largest element");
}
return secondLargest;
}
Java
java
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public class Main {
public static int findSecondLargest(int[] arr) {
if (arr.length < 2) {
throw new IllegalArgumentException("Array must contain at least two elements");

}
int largest = Integer.MIN_VALUE;
int secondLargest = Integer.MIN_VALUE;
for (int num : arr) {
if (num > largest) {
largest = num;
} else if (num > secondLargest && num < largest) {
secondLargest = num;
}
}
if (secondLargest == Integer.MIN_VALUE) {
throw new IllegalArgumentException("There is no second largest element");
}
}
}
C++
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#include <iostream>
#include <limits.h>
int findSecondLargest(int arr[], int size) {
if (size < 2) {
throw std::invalid_argument("Array must contain at least two elements");
}
int largest = INT_MIN;
int secondLargest = INT_MIN;
for (int i = 0; i < size; ++i) {
if (arr[i] > largest) {

largest = arr[i];
} else if (arr[i] > secondLargest && arr[i] < largest) {
secondLargest = arr[i];
}
}
if (secondLargest == INT_MIN) {
throw std::invalid_argument("There is no second largest element");
}
}
Each of these implementations assumes that there are at least two distinct elements in the array. If
there are fewer than two distinct values, the function will indicate that a second largest element
cannot be determined.
Answer 4:
To check if a given array is sorted, you need to determine if the array is sorted in either ascending
or descending order. Here’s how you can implement this check in various programming languages:
Python
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def is_sorted(arr):
if len(arr) < 2:
return True
ascending = all(arr[i] <= arr[i + 1] for i in range(len(arr) - 1))
descending = all(arr[i] >= arr[i + 1] for i in range(len(arr) - 1))
return ascending or descending
JavaScript
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function isSorted(arr) {
if (arr.length < 2) return true;
let ascending = true;
let descending = true;

for (let i = 0; i < arr.length - 1; i++) {
if (arr[i] > arr[i + 1]) ascending = false;
if (arr[i] < arr[i + 1]) descending = false;
}
return ascending || descending;
}
Java
java
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public class Main {
public static boolean isSorted(int[] arr) {
if (arr.length < 2) return true;
boolean ascending = true;
boolean descending = true;
for (int i = 0; i < arr.length - 1; i++) {
}
}
}
C++
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#include <iostream>
#include <vector>
bool isSorted(const std::vector<int>& arr) {
if (arr.size() < 2) return true;
bool ascending = true;
bool descending = true;

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for (size_t i = 0; i < arr.size() - 1; ++i) {
}
}
Explanation
Ascending Check: Iterate through the array and ensure that every element is less than or
equal to the next element.
Descending Check: Iterate through the array and ensure that every element is greater than
or equal to the next element.
Return Result: If the array meets either the ascending or descending condition, it is
considered sorted.
These implementations will return true if the array is sorted in either ascending or descending order,
and falseotherwise.
Answer 5:
To remove duplicates from a given array, you can use various methods depending on the
programming language and requirements. Here are some common approaches:
Python
Using a Set:
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def remove_duplicates(arr):
return list(set(arr))
This method leverages Python's set data structure to automatically remove duplicates. The list
function converts the set back into a list.
Using a List Comprehension (preserving order):
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def remove_duplicates(arr):
seen = set()
result = []
for item in arr:
if item not in seen:
result.append(item)
seen.add(item)

return result
This method maintains the original order of elements while removing duplicates.
JavaScript
Using a Set:
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function removeDuplicates(arr) {
return [...new Set(arr)];
}
This method uses JavaScript's Set to automatically filter out duplicates and then spreads the set
back into an array.
Using a Loop (preserving order):
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function removeDuplicates(arr) {
let seen = new Set();
let result = [];
for (let item of arr) {
if (!seen.has(item)) {
result.push(item);
seen.add(item);
}
}
return result;
}
This method preserves the order of elements while removing duplicates.
Java
Using a Set:
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import java.util.*;
public class Main {
public static List<Integer> removeDuplicates(List<Integer> list) {
Set<Integer> set = new HashSet<>(list);
return new ArrayList<>(set);
}
}

This method uses a HashSet to filter out duplicates and then converts it back to a List.
Using a Loop (preserving order):
java
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import java.util.*;
public class Main {
public static List<Integer> removeDuplicates(List<Integer> list) {
Set<Integer> seen = new HashSet<>();
List<Integer> result = new ArrayList<>();
for (Integer item : list) {
if (seen.add(item)) {
result.add(item);
}
}
return result;
}
}
This method preserves the order of elements while removing duplicates.
C++
Using a Set:
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#include <vector>
#include <set>
#include <algorithm>
std::vector<int> removeDuplicates(const std::vector<int>& vec) {
std::set<int> unique_set(vec.begin(), vec.end());
return std::vector<int>(unique_set.begin(), unique_set.end());
}
This method uses a std::set to remove duplicates and then converts it back to a std::vector.
Using Sorting and Unique (preserving order):
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#include <vector>

std::vector<int> removeDuplicates(std::vector<int> vec) {
std::sort(vec.begin(), vec.end());
auto last = std::unique(vec.begin(), vec.end());
vec.erase(last, vec.end());
return vec;
}
This method first sorts the array, then uses std::unique to remove duplicates, and finally erases the
duplicates.
Each method has its own advantages depending on whether you need to maintain the order of
elements or are okay with any order.
Answer 6:
To rotate a given array, you can shift the elements of the array by a certain number of positions.
Depending on the direction (left or right) and the number of positions to rotate, the approach may
differ. Here’s how you can do it in various programming languages:
Python
Rotate Array to the Right by k Positions:
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def rotate_right(arr, k):
n = len(arr)
k = k % n # Handle cases where k is greater than the length of the array
return arr[-k:] + arr[:-k]
Rotate Array to the Left by k Positions:
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def rotate_left(arr, k):
n = len(arr)
k = k % n # Handle cases where k is greater than the length of the array
return arr[k:] + arr[:k]
JavaScript
javascript
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function rotateRight(arr, k) {
const n = arr.length;
k = k % n; // Handle cases where k is greater than the length of the array

return arr.slice(-k).concat(arr.slice(0, -k));
}
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function rotateLeft(arr, k) {
const n = arr.length;
return arr.slice(k).concat(arr.slice(0, k));
}
Java
java
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import java.util.Arrays;
public class Main {
public static void rotateRight(int[] arr, int k) {
int n = arr.length;
reverse(arr, 0, n - 1);
reverse(arr, 0, k - 1);
reverse(arr, k, n - 1);
}
private static void reverse(int[] arr, int start, int end) {
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
}
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import java.util.Arrays;
public class Main {
public static void rotateLeft(int[] arr, int k) {
int n = arr.length;
reverse(arr, 0, k - 1);
reverse(arr, k, n - 1);
reverse(arr, 0, n - 1);
}
private static void reverse(int[] arr, int start, int end) {
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
}
C++
cpp
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#include <vector>
void rotateRight(std::vector<int>& arr, int k) {
int n = arr.size();
std::reverse(arr.begin(), arr.end());
std::reverse(arr.begin(), arr.begin() + k);
std::reverse(arr.begin() + k, arr.end());
}
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#include <vector>
void rotateLeft(std::vector<int>& arr, int k) {
int n = arr.size();
std::reverse(arr.begin(), arr.begin() + k);
std::reverse(arr.begin() + k, arr.end());
std::reverse(arr.begin(), arr.end());
}
Explanation
Modular Arithmetic: k % n ensures that the rotation amount k is within the bounds of the
array length.
Reversal Method:
Right Rotation: Reverse the whole array, then reverse the first k elements and the
remaining n - k elements.
Left Rotation: Reverse the first k elements, reverse the remaining n - k elements, then
reverse the whole array.
These methods efficiently rotate the array in-place or return a new rotated array depending on the
approach used.
Answer 7:
To find the frequency of elements in a given array, you can use a hash map (or dictionary) to count
occurrences of each element. This approach works efficiently in most programming languages.
Below are implementations for various languages:
Python
Using a dictionary:
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from collections import Counter
def count_frequencies(arr):
return dict(Counter(arr))
Using a manual approach:
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def count_frequencies(arr):
frequency = {}

for item in arr:
if item in frequency:
frequency[item] += 1
else:
frequency[item] = 1
return frequency
JavaScript
Using an object:
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function countFrequencies(arr) {
const frequency = {};
arr.forEach(item => {
frequency[item] = (frequency[item] || 0) + 1;
});
return frequency;
}
Java
Using a HashMap:
java
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import java.util.HashMap;
import java.util.Map;
public class Main {
public static Map<Integer, Integer> countFrequencies(int[] arr) {
Map<Integer, Integer> frequencyMap = new HashMap<>();
frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1);
}
return frequencyMap;
}
}
C++
Using an unordered_map:
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#include <vector>

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#include <unordered_map>
std::unordered_map<int, int> countFrequencies(const std::vector<int>& arr) {
std::unordered_map<int, int> frequencyMap;
frequencyMap[num]++;
}
return frequencyMap;
}
Explanation
Hash Map (or Dictionary): This data structure allows you to store key-value pairs, where
keys are array elements, and values are their counts.
Iteration: Traverse through the array and update the count of each element in the hash map.
Get Frequency: After processing the array, the hash map contains each unique element and
its frequency.
These methods provide a simple and efficient way to count the occurrences of each element in an
array.
Answer 8:
#include <vector>
std::vector<int> mergeSortedArrays(const std::vector<int>& arr1, const std::vector<int>& arr2) {
std::vector<int> merged;
size_t i = 0, j = 0;
while (i < arr1.size() && j < arr2.size()) {
if (arr1[i] < arr2[j]) {
merged.push_back(arr1[i]);
i++;
} else {
merged.push_back(arr2[j]);
j++;
}
}
// Add remaining elements
merged.insert(merged.end(), arr1.begin() + i, arr1.end());
merged.insert(merged.end(), arr2.begin() + j, arr2.end());

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return merged;
}
Explanation
Two Pointers: Initialize two pointers (i and j) to traverse both arrays.
Comparison: Compare elements from both arrays and append the smaller element to the
merged array.
Remaining Elements: Once one array is exhausted, append the remaining elements from the
other array.
Efficiency: This approach runs in O(n + m) time, where n and m are the lengths of the two
arrays, making it efficient for merging sorted arrays.
This method ensures that the merged array is also sorted, leveraging the fact that both input arrays
are already sorted.

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