These are chat archives for **Codewars/codewars.com/kata-solving-help**

@itamargal hey just looking at this now!

so in Java ArrayList class is jst like an array except it has methods such as add so you dont have to worry about arraysize

@jk16 Gotcha.

not sure if I should post in kata-solving-help as I'm fairly confident my code is correct as I have written my own tests. Although it's not even taking a second to run locally, it times out on codewars. Any suggestions how to tackle this? It passes the basic test ftiw.

Kata in question: https://www.codewars.com/kata/biggest-sum/

After trying a few things, I ended up implementing bellman-ford where I convert the input matrix into a graph

Kata in question: https://www.codewars.com/kata/biggest-sum/

After trying a few things, I ended up implementing bellman-ford where I convert the input matrix into a graph

```
def initialize(graph, source):
"""Set up each node within the graph where we assume that rest of the nodes
are very far away."""
d = {}
p = {}
for node in graph:
d[node] = float('Inf')
p[node] = None
d[source] = 0 # For the source we know how to reach
return d, p
def relax(node, neighbour, graph, d, p):
"""Check if the distance between node and neighbour is lower than the one
we know of and record if it is."""
if d[neighbour] > d[node] + graph[node][neighbour]:
d[neighbour] = d[node] + graph[node][neighbour]
p[neighbour] = node
def bellman_ford(graph, source):
"""Returns two dictionaries, d and p where d holds all nodes and the cost
to reach each one and p holds the predecessors which show which path
to take each node with the lost cost."""
d, p = initialize(graph, source)
for i in range(len(graph)-1): #Run this until it converges
for u in graph:
for v in graph[u]:
relax(u, v, graph, d, p)
#check for negative-weight cycles
for u in graph:
for v in graph[u]:
assert d[v] <= d[u] + graph[u][v]
return d, p
def make_node_matrix(matrix):
"""Convert input matrix to matrix of tuples with node name and value to be
input into function to convert to graph. Values are inverted for input
into shortest path algorithm."""
node_names = ['n{0}'.format(p) for p in range(1, len(matrix)**2 + 1)]
nodes = zip(node_names, [ -n for row in matrix for n in row])
node_matrix = []
for i in range(len(matrix)):
sub_list = []
for idx, t in enumerate(nodes):
sub_list.append(t)
if idx == len(matrix) - 1:
break
node_matrix.append(sub_list)
return node_matrix
def make_graph_from_matrix(grid):
"""Convert an input matrix of tuples into a graph where we assume the
direction of traversal in the matrix only allowed to be one to the right
or down."""
graph= {}
for row_count, row in enumerate(grid):
for col, node in enumerate(row):
vertex = {}
ref, value = node
if col == len(row) - 1:
if row_count == len(grid) - 1:
key, value = grid[len(grid) - 1][len(grid) -1]
graph[key] = vertex
break
key, value = grid[row_count + 1][col]
vertex[key] = value
elif row_count == len(grid) - 1:
key, value = grid[row_count][col + 1]
vertex[key] = value
else:
vertex.update([grid[row_count][col + 1], grid[row_count + 1][col]])
graph[ref] = vertex
return graph
def find_sum(m):
"""Find the highest sum in a matrix from top left to bottom right."""
node_matrix = make_node_matrix(m)
graph = make_graph_from_matrix(node_matrix)
last_node = 'n' + str(len(m) ** 2)
d, p = bellman_ford(graph, 'n1')
return abs(d[last_node]) + m[0][0]
```

@MSKW76_twitter Do you know on what input your solution times out?