{
"translations" :{
"en": {
"sc1": "A mistake that people often make is to confuse the upper bound and the worst case.",
"sc2": "
The upper bound of an algorithm indicates the upper or highest growth rate that the algorithm can have for a problem of size $n$.",
"sc3": "The sequential search algorithm accepts as its input an array of $n$ keys and the target key $K$ to search for.",
"sc4": "
So what is the upper bound of the sequential search algorithm?",
"sc5": "
Do you think this is the correct way to ask this question?",
"sc6": "Remember that there are three input cases that affect the running time of sequential search.",
"sc7": "
1- When the target key $K$ is located at the first position in the input array.",
"sc8": "
2- When the target key $K$ is located at the last position in the input array.",
"sc9": "
3- The average cost over all possible positions for $K$, which comes out to about $n/2$.",
"sc10": "In the best case for the algorithm, only a single element is visited. $\\mathbf{T}(n) = 1$",
"sc11": "
Accordingly, an upper bound for the algorithm in the best case is $O(1)$. Even when $n$ increases, the cost for the best case does not grow.",
"sc12": "In the worst case for the algorithm, $n$ elements must be visited. $\\mathbf{T}(n) = n$",
"sc13": "
So an upper bound for the algorithm in the worst case is $O(n)$. No matter the value of $n$, for some constant $c$, $cn$ is bigger than $n$.",
"sc14": "In the average case for the algorithm, about $n/2$ elements are visited. $T(n) = \\frac{n}{2}$",
"sc15": "
So an upper bound for the algorithm in the average case is also $O(n)$. No matter the value of $n$, for some constant $c$, $cn$ is bigger than $n/2$.",
"sc16": "Accordingly, the correct way to ask the question is:
What is the upper bound of sequential search in the best/average/worst case?",
"sc17": "
And the answer should be...",
"lab1": "$O(1)$ in the Best Case.",
"lab2": "$O(n)$ in the Worst Case.",
"lab3": "$O(n)$ in the Average Case."
}
}
}