The following algorithm determines if all the numbers in the vector are different from each other, i.e. they are all distinct values. Explain when the best-case running time happens. Determine the function for this algorithm. bool all Distinct(const vector& a) { for (int i = 0; i < a.size()-1; i++) { for (int j =i+1; j < a.size(); j++){ if (a[i] == a[j]) } return false; # of primitive operations
The following algorithm determines if all the numbers in the vector are different from each other, i.e. they are all distinct values. Explain when the best-case running time happens. Determine the function for this algorithm. bool all Distinct(const vector& a) { for (int i = 0; i < a.size()-1; i++) { for (int j =i+1; j < a.size(); j++){ if (a[i] == a[j]) } return false; # of primitive operations
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![The following algorithm determines if all the numbers in the vector are different from each other, i.e., they are all distinct values. Explain when the best-case running time happens. Determine the Ω function for this algorithm.
```cpp
bool allDistinct(const vector<int>& a) {
for (int i = 0; i < a.size() - 1; i++) {
for (int j = i+1; j < a.size(); j++) {
if (a[i] == a[j])
return false;
}
}
return true;
}
```
| | # of primitive operations |
|------------------|---------------------------|
| (Content missing) | |
**Explanation:**
- **Outer Loop:** Iterates from 0 to `a.size() - 1`.
- **Inner Loop:** Starts from `i+1` up to `a.size()`.
- **Comparison:** Checks if any two elements are equal (`a[i] == a[j]`).
- **Return:** If a duplicate is found, the function returns `false`. If the loop completes, it returns `true`.
**Best-Case Scenario:**
The best-case running time occurs when the first pair `(a[i], a[j])` has an equal value, causing the algorithm to return `false` immediately. This scenario gives a best-case complexity of Ω(1) because the loop exits after the first comparison.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d8d9faf-b9df-42a1-8480-6ec7c9a1f4ad%2F1dff68da-dc77-4ce3-884f-3c71bcd80469%2Fdu3es9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The following algorithm determines if all the numbers in the vector are different from each other, i.e., they are all distinct values. Explain when the best-case running time happens. Determine the Ω function for this algorithm.
```cpp
bool allDistinct(const vector<int>& a) {
for (int i = 0; i < a.size() - 1; i++) {
for (int j = i+1; j < a.size(); j++) {
if (a[i] == a[j])
return false;
}
}
return true;
}
```
| | # of primitive operations |
|------------------|---------------------------|
| (Content missing) | |
**Explanation:**
- **Outer Loop:** Iterates from 0 to `a.size() - 1`.
- **Inner Loop:** Starts from `i+1` up to `a.size()`.
- **Comparison:** Checks if any two elements are equal (`a[i] == a[j]`).
- **Return:** If a duplicate is found, the function returns `false`. If the loop completes, it returns `true`.
**Best-Case Scenario:**
The best-case running time occurs when the first pair `(a[i], a[j])` has an equal value, causing the algorithm to return `false` immediately. This scenario gives a best-case complexity of Ω(1) because the loop exits after the first comparison.
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