Isotonic Regression | Algorithms Notes

Isotonic Regression | Algorithms Notes
Given a sequence A of n integers, design an algorithm to find an non-decreasing sequence B of n integers, such that is minimized.
http://stackoverflow.com/questions/28149331/given-a-sequence-a1-a2-an-find-an-array-b1-b2-bn-s

Given a sequence a[1], a[2], ..., a[n], find an array b[1] < b[2] < ... < b[n] such that |a[1]-b[1]| + |a[2]-b[2]| + ... + |a[n]-b[n]| is minimum.

I have already come up with some observations, but I still couldn't figure out any feasible solution.

it is given that n <= 10^6.

The problem is from BOI 2003, and is called Sequence.

Input

9 4 8 20 14 15 18

Output

6 7 8 13 14 15 18

Since the difference is 13.

Solution:
This problem can be solved by minimum cost flow. Here we use dynamic programming.
Let F(k, z) be the optimal objective value for the first k elements with B[k] = z. Then, we have a recurrence relation: F(k, z) = min F(k-1, x) + |z – A[k]|. Thus, the problem becomes how to efficiently find the z minimizing F(k, z) for all fixed k. It can be showed that F(k, z) is a piecewise linear convex function in z and the minimizing value must be in A. Thus, the dynamic programming can be speed up by using slope optimization and find the solution in O(n lg n) time.

令 F(k, z) 為前 k 個元素且  B[k] = z 時的最佳解之值。當給定 k ，我們可得一遞回關係式：F(k, z) = min F(k-1, x) + |z – A[k]|。因此問題就變成如何快速的找出最小化 F(k, z) 的 z 值。我們可以證明 F(k, z) 對於 z 是一個 piecewise linear convex 函數 ，而且可以最小化 F 的 z 值必定在 A 中。因此使用動態規劃加上斜率最佳化可以在O(n lg n)時間內找到答案。

http://stackoverflow.com/questions/10460861/finding-an-increasing-sequence-a-which-minimizes-sigmaabsaici
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algorithm - How to tell if an array is a permutation in O(n)? - Stack Overflow

algorithm - How to tell if an array is a permutation in O(n)? - Stack Overflow
Input: A read-only array of N elements containing integer values from 1 to N (some integer values can appear more than once!). And a memory zone of a fixed size (10, 100, 1000 etc - not depending on N).
How to tell in O(n) if the array represents a permutation?

Check out the following solution. It uses O(1) additional space. It alters the array during the checking process, but returns it back to its initial state at the end.

The idea is:

1. Check if any of the elements is out of the range [1, n] => O(n).
2. Go over the numbers in order (all of them are now assured to be in the range [1, n]), and for each number x (e.g. 3):
   • go to the x'th cell (e.g. a[3]), if it's negative, then someone already visited it before you => Not permutation. Otherwise (a[3] is positive), multiply it by -1. => O(n).
3. Go over the array and negate all negative numbers.

This way, we know for sure that all elements are in the range [1, n], and that there are no duplicates => The array is a permutation.

int is_permutation_linear(int a[], int n) {
    int i, is_permutation = 1;

    // Step 1.
    for (i = 0; i < n; ++i) {
        if (a[i] < 1 || a[i] > n) {
            return 0;
        }
    }

    // Step 2.
    for (i = 0; i < n; ++i) {
        if (a[abs(a[i]) - 1] < 0) {
            is_permutation = 0;
            break;
        }
        a[i] *= -1;
    }

    // Step 3.
    for (i = 0; i < n; ++i) {
        if (a[i] < 0) {
            a[i] *= -1;
        }
    }

    return is_permutation;
}

I'm very slightly skeptical that there is a solution. Your problem seems to be very close to one posed several years ago in the mathematical literature, with a summary given here ("The Duplicate Detection Problem", S. Kamal Abdali, 2003) that uses cycle-detection -- the idea being the following:

If there is a duplicate, there exists a number j between 1 and N such that the following would lead to an infinite loop:

x := j;
do
{
   x := a[x];
}
while (x != j);

because a permutation consists of one or more subsets S of distinct elements s₀, s₁, ... s_k-1 where s_j = a[s_j-1] for all j between 1 and k-1, and s₀ = a[s_k-1], so all elements are involved in cycles -- one of the duplicates would not be part of such a subset.

e.g. if the array = [2, 1, 4, 6, 8, 7, 9, 3, 8]

then the element in bold at position 5 is a duplicate because all the other elements form cycles: { 2 -> 1, 4 -> 6 -> 7 -> 9 -> 8 -> 3}. Whereas the arrays [2, 1, 4, 6, 5, 7, 9, 3, 8] and [2, 1, 4, 6, 3, 7, 9, 5, 8] are valid permutations (with cycles { 2 -> 1, 4 -> 6 -> 7 -> 9 -> 8 -> 3, 5 } and { 2 -> 1, 4 -> 6 -> 7 -> 9 -> 8 -> 5 -> 3 } respectively).

Read full article from algorithm - How to tell if an array is a permutation in O(n)? - Stack Overflow

algorithm - Maximize sum-distance for integer array - Stack Overflow

algorithm - Maximize sum-distance for integer array - Stack Overflow
Given an integer array A, return the maximum possible sum-distance between two elements. The sum-distance is defined as A[i] + A[j] + (i - j) for i > j
For example with A = [8, 2, 4, 9, 5, 8, 0, 3, 8, 2] the max sum-distance is 24 achieved with i=0 and j=8
An O(n²) solution is trivial. Is there any O(n) solution (where n is the length of the array)?

For each index i, we only need to know one index that maximize the sum A[i] + A[index] + (i - index) = A[i] + i + (A[index] - index). Which mean, we only need to maintain an index, which A[index] - index is maximum.

int index = 0;
int result = 0;

for(int i = 1; i < n; i++){
    int total = A[i] + i + A[index] - index;
    result = max(result, total);
    if(A[i] - i > A[index] - index){
        index = i;
    }
}
return result;

Read full article from algorithm - Maximize sum-distance for integer array - Stack Overflow

algorithm - Calculating number of moves from top left corner to bottom right with move in any direction - Stack Overflow

algorithm - Calculating number of moves from top left corner to bottom right with move in any direction - Stack Overflow
There is a robot situated at (1,1) in a N X N grid, the robot can move in any direction left, right ,up and down. Also I have been given an integer k, which denotes the maximum steps in the path. I had to calculate the number of possible ways to move from (1,1) to (N,N) in k or less steps.
Read full article from algorithm - Calculating number of moves from top left corner to bottom right with move in any direction - Stack Overflow

Second largest element in BST

http://quiz.geeksforgeeks.org/second-largest-element-in-binary-search-tree-bst/

The second largest element is second last element in inorder traversal and second element in reverse inorder traversal. We traverse given Binary Search Tree in reverse inorder and keep track of counts of nodes visited. Once the count becomes 2, we print the node.

// A function to find 2nd largest element in a given tree.

void secondLargestUtil(Node *root, int &c)

{

    // Base cases, the second condition is important to

    // avoid unnecessary recursive calls

    if (root == NULL || c >= 2)

        return;

 

    // Follow reverse inorder traversal so that the

    // largest element is visited first

    secondLargestUtil(root->right, c);

 

    // Increment count of visited nodes

    c++;

 

    // If c becomes k now, then this is the 2nd largest

    if (c == 2)

    {

        cout << "2nd largest element is "

             << root->key << endl;

        return;

    }

 

    // Recur for left subtree

    secondLargestUtil(root->left, c);

}

algorithm - Second max in BST - Stack Overflow
The max element is the rightmost leaf in the BST. The second max is either its parent or its left child. So the solution is to traverse the BST to find the rightmost leaf and check its parent and left child.

public static int findSecondLargestValueInBST(Node root)
    {
        int secondMax;
        Node pre = root;
        Node cur = root;
        while (cur.Right != null)
        {
            pre = cur;
            cur = cur.Right;
        }
        if (cur.Left != null)
        {
            cur = cur.Left;
            while (cur.Right != null)
                cur = cur.Right;
            secondMax = cur.Value;
        }
        else
        {
            if (cur == root && pre == root)
                //Only one node in BST
                secondMax = int.MinValue;
            else
                secondMax = pre.Value;
        }
        return secondMax;
    }

http://blog.gainlo.co/index.php/2016/06/03/second-largest-element-of-a-binary-search-tree/
Read full article from algorithm - Second max in BST - Stack Overflow