pavlos-io

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

typedef struct list List;
typedef struct node Node;

struct node {
  int data;
  Node* next;

pavlos-io

class Solution {
public:
  int findCircleNum(vector<vector<int>>& M) {
    int n = M.size();
    int partitions = 0;
    if(!n) return partitions;

    for(int i = 0; i < n; i++){
      if(M[i][i]==1){
        M[i][i] = 0; //instead of using extra mem. to store visited, I alter the matrix.

pavlos-io

import java.awt.Point;
import java.util.ArrayList;
import java.util.List;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

pavlos-io

    /// Gets the value of a tile.
    /**
     * Given an X and Y coordinate, returns the value of the tile at that position.
     * This function also marks that location as visited.  Part of your grade on this
     * assignment will be determined by how many tiles your solution visits.
     * 
     * @param x The X coordinate.
     * @param y The Y coordinate.
     * @return The value of the tile at that location.
     */

pavlos-io

ostream& operator<<(ostream& os, const vector<int>& v){
    os << "[ ";
    for(int i = 0; i<=v.size()-1; i++){
        os << "'" << v.at(i) << "'";
        if(i != v.size()-1) os << ',';
    }
    os << " ]";
    return os;
}

pavlos-io

function fn = intcompare(x, f, a, b)

    hp = build_hermite(x, hermitecalc(x, f));
    sp = build_spline( x,splinecalc(x,f) );

    i_f = integral(f,a,b);
    sp_i = spline_integral(x, a, b, sp);
    hp_i = integral( matlabFunction(hp),a,b );
    
    first_row = [i_f,sp_i,hp_i];

pavlos-io

# **DP SOLUTION**
def longest_inc_subseq_dp a
  n = a.length
  r = [1]
  s = Array.new(n) {[]}
  s[0] << a[0]

  for i in 1..(n-1)
    res = get_max(a, r, i)
    r[i] = res[0] + 1

pavlos-io

# **PROBLEM STATEMENT**
# Assembly line scheduling w/ 2 lines.

# **OPTIMAL SUBSTRUCTURE**
# Claim: A candidate solution is a finite sequence of entries of A
# and the last element in S is either A[0][n-1] or A[1][n-1].
# If S is the opt. solution for A, then S' = S[:-1] is optimal to
# the subproblem with force S[-1].

# **RECURRENCE RELATION**

pavlos-io

# **PROBLEM STATEMENT**
#Given an mxn matrix A of non-negative integers,
# find the finite sequence S of adjacent entries of
# A (starting from top left and moving right, or bottom)
# s.t Σ(S) is minimum.

# **OPTIMAL SUBSTRUCTURE**
# Claim: If S is the opt. solution for A, then S' = S[:-1] is optimal to
# the subproblem with force S[-1].

pavlos-io

# **PROBLEM STATEMENT**
#Given a finite sequence of integers A,
# find the finite contiguous subsequence S s.t. the Σ(S) := "sum of the elements of S", is maximum.

# **OPTIMAL SUBSTRUCTURE**
# Claim: Consider the problem of obtaining the contig. subsequence w/ max sum
# up to and including the ith element of A. Suppose S = S(k) is the optimal solution
# of the sequence A, where k is the index in A of the last element in the solution.
# That is, among all candidate solutions, Σ(S) is maximum and A[k] is the last element in S.
# Let F = S[-1] be a focring factor. Then, S' = S[:-1] is the contig. subsequence