Answer 2
What is the computational complexity of your answer in Question 1? Can you explain why?
The computational complexity of Question 1 is O(n2).
Explanation:
for i in range(len(arr2)):
{% for key, value in test_codes_dict.items %} | |
<option value="{{ key }}">{{ value }}</option> | |
{% endfor %} |
<!DOCTYPE html> | |
<html lang="en"> | |
<head> | |
<meta charset="utf-8" /> | |
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" /> | |
<meta content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=0" name="viewport" /> | |
<meta name="viewport" content="width=device-width" /> | |
<title>ATE-Dashboard</title> |
function createArrayOfFunctions(y){ | |
var arr = []; | |
x = 1 | |
for (let i = 0; i < y ; i++){ | |
arr[i] = function(x){ | |
return x + i; | |
} | |
} | |
return arr; |
import time | |
def recursiveFibonacci(number, fn1, fn2): | |
fnNext = fn1 + fn2 | |
fn1 = fn2 | |
if number < fnNext: | |
print(fnNext) | |
else: | |
recursiveFibonacci(number, fn1,fnNext) |
Answer 2
What is the computational complexity of your answer in Question 1? Can you explain why?
The computational complexity of Question 1 is O(n2).
Explanation:
for i in range(len(arr2)):
def isSubset(arr1, arr2): | |
count = 0 | |
flagSubset = False | |
for i in range(len(arr2)): | |
for j in range(len(arr1)): | |
if(arr2[i].strip().upper() == arr1[j].strip().upper()): | |
count+=1 | |
break | |
curl -H 'Content-Type: application/json' -X POST http://localhost:5984/_replicate -d ' {"source": "http://localhost:5984/bec-ember-three/", "target": "http://128.199.160.40:5984/bec-ember-recover"}' | |
curl -X POST http://128.199.160.40:5984/_replicate -H "Content-Type: application/json" -d '{"source": "test_database", "target":"bec-practice", "doc_ids": ["practices"]}' |