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Wiggle Subsequence

Finding the length of a longest wiggle subsequence is an interesting problem. We’ll solve it first using dynamic programming and then using a greedy algorithm. We’ll compare and contrast the two different approaches that we used to solve the problem towards the end of the article. Problem Statement A wiggle sequence is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences. For example, [1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences (6, -3, 5, -7, 3) alternate between positive and negative. In contrast, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero. A subsequence is obtained by deleting some elements (possi
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SELECT with groups of 3

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Exercise 9.3-1 asks you to show that the $SELECT$ algorithm still runs in linear time if the elements are divided into groups of 7. This problem asks about dividing into groups of 3. Show that $SELECT$ runs in linear time if you divide the elements into groups whose size is any odd constant greater than 3. Show that $SELECT$ runs in $O(n\; lg\; n)$ time if you divide the elements into groups of size 3 Because the bound in part (b) is just an upper bound, we do not know whether the groups-of-3 strategy actually runs in $O(n)$ time. But by repeating the groups-of-3 idea on the middle groups of medians, we can pick a pivot that guarantees $O(n)$ time. The $SELECT3$ algorithm on the next page determines the $i^{\;th}$ smallest of an input array of $n \gt 1$ distinct elements. Describe in English how the $SELECT3$ algorithm works.Include in your description one or more suitable diagrams. Show that $SELECT3$ runs in $O(n)$ time in the worst case. [1]   The worst-case linear-time median-findi
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kth quantiles of an $n$-element set

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Problem statement Exercise 9.3-8 The $k$th quantiles of an $n$-element set are the $k$ - 1 order statistics that divide the sorted set into $k$ equal sized sets (to within 1). Give an $O(n \cdot lg k)$-time algorithm to list the $k$th quantiles of a set. [1] Definitions In statistics and probability, quantiles are cut points dividing the range of a probability distribution into continuous intervals with equal probabilities, or dividing the observations in a sample in the same way. There is one fewer quantile than the number of groups created. [2] Approach Let's write a divide and conquer algorithm using selection to solve this problem. What if we find the $l = \lfloor\frac{n \cdot \lfloor\frac{k}{2}\rfloor}{k}\rfloor$ th smallest number  and partition the array around it. Then, we can find the first $\lfloor\frac{k}{2}\rfloor$ cut points in the first half $l$ and the remaining $\lceil\frac{k}{2}\rceil$ cut points in the second half $n - l$. At each step, we keep on dividing $k$ by
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Solving recurrences and proving bounds

Exercise 7.4-1 Show that the recurrence $T(n) = max\{T(q) + T(n - q - 1): 0 \le q \le n - 1\} + \Theta(n)$ has a lower bound of $T(n) = \Omega(n^2)$. [1] Inductive hypothesis: $T(n) \ge n^2$ $T(n) = max\{q^2 + (n - q - 1)^2: 0 \le q \le n - 1\} + \Theta(n)$ $f(q) = q^2 + (n - q - 1)^2: 0 \le q \le n - 1$ By expanding the expression we obtain $f(q) = q^2 + (n - q)^2 - 2 \cdot (n - q) + 1$ $f(q) = 2 \cdot q^2 + n^2 - 2nq - 2n + 2q + 1$ Now let’s find the local maxima of the function using its first derivative. Taking the first derivative with respect to $q$ we obtain: $f'(q) = \frac {d}{d q}2q^2 + \frac {d}{d q} n^2 + \frac {d}{d q} -2nq + \frac {d}{d q}-2n + \frac {d}{d q} 2q + \frac {d}{d q} 1$ Since we are looking for a critical point in the graph, $f'(q) = 4q - 2n + 2 = 0$ $q = \frac{n - 1}{2}$ This can be either a local minima or a maxima of our function. Let’s determine it next. Let $q = 1$ for sufficiently large $n$, plugging in the values in the above equation we get: $4
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Establishing a lower bound for searching

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Today, we are going to establish a lower bound for searching and prove that binary search is optimal, and no comparison based search in this universe would be better than that by more than a constant factor. Let’s define our problem more formally first. Theorem Let us assume that we have $n$ pre-processed items in the array. Finding a given item among them in a comparison model requires $\Omega(log_2 n)$ in the worst case. Computation Model Let’s try to model the situation using a decision tree. In this decision tree, an internal node represents a decision that our algorithm makes and a leaf represents an answer. Here’s our representation of the problem using a decision tree computation model for $n = 3$ element array for the sake of simplicity. I mentioned here that the Items are preprocessed to mean you could do whatever you want with items ahead of time. For instance, I can sort them, build them into an AVL tree et.al. Proof Let $n$ be the number of items in the input array. Then, w
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Depth First Search

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 We return today to graph search. Last time we saw breadth-first search, today we're going to do depth-first search. It's a simple algorithm, but you can do lots of cool things with it. DFS is often used to explore the whole graph, and so we are going to see how to do that today. So high-level description is we're going to just recursively explore the graph, backtracking as necessary, kind of like how you solve a maze. Let’s now consider a real world application of the DFS algorithm. Number of Closed Islands - Problem Definition Given a 2D grid consists of $0s$ (land) and $1s$ (water).  An island is a maximal 4-directionally connected group of $0s$ and a closed island is an island totally (all left, top, right, bottom) surrounded by $1s$. Return the number of closed islands. Output: 5 Output: 4 Note that closed islands are colored in the tables above. Output: 2 Output: 1 Number of Closed Islands - Approach We use the table $d$ to keep track of the discovery state of each bl
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Minimum Insertion Steps to Make a String Palindrome

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A word, phrase, or sequence that reads the same backwards as forwards is a palindrome. For example the sequence "Step on no pets" is a palindrome. Today, let's focus on converting a sequence into a palindrome by inserting characters at arbitrary positions in the original input sequence. Minimum Insertion Steps to Make a String Palindrome - Problem Statement Given an input sequence $s$, in one step you can insert any character at any position of the input sequence. Return the minimum number of steps to make $s$ palindrome. For example, given the input sequence "leetcode", inserting five characters the string becomes "leetcodocteel". This is the minimum number of insertions required to make the input sequence a palindrome. Minimum Insertion Steps to Make a String Palindrome - Definition Let $s$ be an input sequence, $i$ and $j$ denote the start and end positions of a subsequence. Let $p[i][j]$ be the minimum number of insertions required to make subseque
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