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@ -76,27 +76,31 @@ for a given function. Say f(n) is your algorithm runtime, and g(n) is an arbitra
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you are trying to relate to your algorithm. f(n) is O(g(n)), if for any real constant c (c>0),
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f(n) <= c g(n) for every input size n (n>0).
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Example 1
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f(n) = 3log n + 100
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*Example 1*
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```
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f(n) = 3log n + 100
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g(n) = log n
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```
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is f(n) O(g(n))?
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is 3 log n + 100 O(log n)?
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Let's look to the definition of Big-Oh.
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3log n + 100 <= c * log n
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Is there some constant c that satisfies this for all n?
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3log n + 100 <= 150 * log n, n > 2 (undefined at n = 1)
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is f(n) O(g(n))?
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is 3 log n + 100 O(log n)?
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Let's look to the definition of Big-Oh.
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3log n + 100 <= c * log n
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Is there some constant c that satisfies this for all n?
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3log n + 100 <= 150 * log n, n > 2 (undefined at n = 1)
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Yes! The definition of Big-Oh has been met therefore f(n) is O(g(n)).
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Example 2
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f(n) = 3*n^2
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*Example 2*
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```
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f(n) = 3*n^2
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g(n) = n
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```
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is f(n) O(g(n))?
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is 3*n^2 O(n)?
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Let's look at the definition of Big-Oh.
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3*n^2 <= c * n
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Is there some constant c that satisfies this for all n?
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is f(n) O(g(n))?
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is 3*n^2 O(n)?
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Let's look at the definition of Big-Oh.
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3*n^2 <= c * n
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Is there some constant c that satisfies this for all n?
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No there isn't, f(n) is NOT O(g(n)).
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### Big-Omega
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