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Original file line number Diff line number Diff line change
Expand Up @@ -9,21 +9,18 @@
* "product": 30 // 2 * 3 * 5
* }
*
* Time Complexity:
* Space Complexity:
* Optimal Time Complexity:
* Time Complexity: O(n)
* Space Complexity:O(1)
* Optimal Time Complexity:O(n)
*
* @param {Array<number>} numbers - Numbers to process
* @returns {Object} Object containing running total and product
*/
export function calculateSumAndProduct(numbers) {
let sum = 0;
for (const num of numbers) {
sum += num;
}

let product = 1;
for (const num of numbers) {
sum += num;
product *= num;
}

Expand All @@ -32,3 +29,4 @@ export function calculateSumAndProduct(numbers) {
product: product,
};
}
console.log(calculateSumAndProduct([1, 2, 3]));
19 changes: 13 additions & 6 deletions Sprint-1/JavaScript/findCommonItems/findCommonItems.js
Original file line number Diff line number Diff line change
@@ -1,14 +1,21 @@
/**
* Finds common items between two arrays.
*
* Time Complexity:
* Space Complexity:
* Optimal Time Complexity:
* Time Complexity:O(n*m)
* Space Complexity:O(u)
* Optimal Time Complexity:O(n+m)
*
* @param {Array} firstArray - First array to compare
* @param {Array} secondArray - Second array to compare
* @returns {Array} Array containing unique common items
*/
export const findCommonItems = (firstArray, secondArray) => [
...new Set(firstArray.filter((item) => secondArray.includes(item))),
];
export const findCommonItems = (firstArray, secondArray) => {
const allowedItems = new Set(secondArray);

const uniqueValues = [
...new Set(firstArray.filter((item) => allowedItems.has(item))),
];
return uniqueValues;
};

console.log(findCommonItems([2, 5, 7], [5, 6, 7]));
20 changes: 12 additions & 8 deletions Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js
Original file line number Diff line number Diff line change
@@ -1,21 +1,25 @@
/**
* Find if there is a pair of numbers that sum to a given target value.
*
* Time Complexity:
* Space Complexity:
* Optimal Time Complexity:
* Time Complexity:O(n^2)
* Space Complexity:O(1)
* Optimal Time Complexity:O(n)
*
* @param {Array<number>} numbers - Array of numbers to search through
* @param {number} target - Target sum to find
* @returns {boolean} True if pair exists, false otherwise
*/
export function hasPairWithSum(numbers, target) {
for (let i = 0; i < numbers.length; i++) {
for (let j = i + 1; j < numbers.length; j++) {
if (numbers[i] + numbers[j] === target) {
return true;
}
const seen = new Set();

for (let num of numbers) {
const complement = target - num;
if (seen.has(complement)) {
return true;
}
seen.add(num);
}
return false;
}

console.log(hasPairWithSum([1, 2, 3], 4));
33 changes: 6 additions & 27 deletions Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs
Original file line number Diff line number Diff line change
@@ -1,36 +1,15 @@
/**
* Remove duplicate values from a sequence, preserving the order of the first occurrence of each value.
*
* Time Complexity:
* Space Complexity:
* Optimal Time Complexity:
* Time Complexity:O(n^2)
* Space Complexity:O(n)
* Optimal Time Complexity:O(n)
*
* @param {Array} inputSequence - Sequence to remove duplicates from
* @returns {Array} New sequence with duplicates removed
*/
export function removeDuplicates(inputSequence) {
const uniqueItems = [];

for (
let currentIndex = 0;
currentIndex < inputSequence.length;
currentIndex++
) {
let isDuplicate = false;
for (
let compareIndex = 0;
compareIndex < uniqueItems.length;
compareIndex++
) {
if (inputSequence[currentIndex] === uniqueItems[compareIndex]) {
isDuplicate = true;
break;
}
}
if (!isDuplicate) {
uniqueItems.push(inputSequence[currentIndex]);
}
}

return uniqueItems;
return [...new Set(inputSequence)];
}

console.log(removeDuplicates([3, 3, 5, 6, 3, 7]));
Original file line number Diff line number Diff line change
Expand Up @@ -12,20 +12,21 @@ def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
"sum": 10, // 2 + 3 + 5
"product": 30 // 2 * 3 * 5
}
Time Complexity:
Space Complexity:
Optimal time complexity:
Time Complexity:O(n)
Space Complexity:O(1)
Optimal time complexity:O(n)
"""
# Edge case: empty list
if not input_numbers:
return {"sum": 0, "product": 1}

sum = 0
for current_number in input_numbers:
sum += current_number

product = 1
for current_number in input_numbers:
sum += current_number
product *= current_number


return {"sum": sum, "product": product}

print(calculate_sum_and_product([1,2,3,4]))
21 changes: 12 additions & 9 deletions Sprint-1/Python/find_common_items/find_common_items.py
Original file line number Diff line number Diff line change
Expand Up @@ -9,13 +9,16 @@ def find_common_items(
"""
Find common items between two arrays.

Time Complexity:
Space Complexity:
Optimal time complexity:
Time Complexity: O(n^3)
Space Complexity:O(min(n,m))
Optimal time complexity: O(n+m)
Comment on lines +12 to +14

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  • Why use $m$ in the complexity of your algorithm but not in the complexity of the original algorithm?

  • How does the algorithm use $O(min(n, m))$ space?

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You are completely right about the inconsistency between old and new Time complex:

  1. let's define n as len(first_sequence) and m as len(second_sequence), if the input sizes differ, m should be used in both.
    Original Algorithm: as explained earlier the outer loop runs n times, the inner loop runs m times, and the not in list takes up to O(n) time. Therefore, time complexity of the original code is O(n^2 \cdot m). If we assume n = m, it simplifies to O(n^3).
    New Algorithm: Converting the sequences to sets takes O(n) + O(m) time. The Python set intersection '&' average-case time complexity is O(\min(n, m)) as documented in this link :(https://wiki.python.org/moin/TimeComplexity). the overall optimal time complexity of O(n + m).

  2. Space Complexity Proof for O(\min(n, m)):
    the original algorithm allocates memory for the common_items list, an item is only appended to common_items if it exists in both first_sequence and second_sequence
    Mathematically, the intersection of two sets cannot contain more elements than the smaller of the two sets

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How much space needed by first_set and second_set?

"""
common_items: List[ItemType] = []
for i in first_sequence:
for j in second_sequence:
if i == j and i not in common_items:
common_items.append(i)
return common_items

first_set = set(first_sequence)
second_set = set(second_sequence)
# we use the & operator shortcut, of The intersection() method
# to return a set that contains the similarity between two or more sets
common_items= first_set & second_set
return list(common_items)

print(find_common_items([1,3,5,4],[1,4,8,0]))
17 changes: 10 additions & 7 deletions Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py
Original file line number Diff line number Diff line change
Expand Up @@ -7,12 +7,15 @@ def has_pair_with_sum(numbers: List[Number], target_sum: Number) -> bool:
"""
Find if there is a pair of numbers that sum to a target value.

Time Complexity:
Space Complexity:
Optimal time complexity:
Time Complexity:O(n^2)
Space Complexity:O(1)
Optimal time complexity:O(n)
"""
for i in range(len(numbers)):
for j in range(i + 1, len(numbers)):
if numbers[i] + numbers[j] == target_sum:
return True
seen = set()
for num in numbers:
complement = target_sum - num
if complement in seen:
return True
seen.add(num)
return False
print(has_pair_with_sum([1,2,3,4],3))
22 changes: 8 additions & 14 deletions Sprint-1/Python/remove_duplicates/remove_duplicates.py
Original file line number Diff line number Diff line change
Expand Up @@ -7,19 +7,13 @@ def remove_duplicates(values: Sequence[ItemType]) -> List[ItemType]:
"""
Remove duplicate values from a sequence, preserving the order of the first occurrence of each value.

Time complexity:
Space complexity:
Optimal time complexity:
Time complexity:O(n^2)
Space complexity:O(n)
Optimal time complexity:O(n)
"""
unique_items = []

for value in values:
is_duplicate = False
for existing in unique_items:
if value == existing:
is_duplicate = True
break
if not is_duplicate:
unique_items.append(value)

return unique_items

# dict.fromkeys creates a dictionary with values as keys,and automatically remove duplicates
#list() to keep the original order.
return list(dict.fromkeys(values))
print(remove_duplicates([1,2,2,3,3,4,0,0]))
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