Related
what is the name of this sort? its just like bubble sort but its easy to write but harder in term of complexity.codes are in python language.
def sort(arr):
n = len(arr)
for i in range(n):
for j in range(n):
if arr[i] < arr[j] :
arr[j], arr[i] = arr[i], arr[j]
but bubble sort is like this :
def bubbleSort(arr):
n = len(arr)
# Traverse through all array elements
for i in range(n):
# Last i elements are already in place
for j in range(0, n-i-1):
# traverse the array from 0 to n-i-1
# Swap if the element found is greater
# than the next element
if arr[j] > arr[j+1] :
arr[j], arr[j+1] = arr[j+1], arr[j]
It is neither selection sort nor bubble sort, but an unnecessary bad sort. All of them, your sort, selection sort and bubble sort are of the same complexity, O(n^2).
Your sort sweeps over the array in two loops without consideration for if an element has reached its proper place (as in selection sort) or if an element has gained a place relative to the next (as in bubble sort). The extra code in bubble and selection sort make them far better than yours.
Compare, yourself:
def sort(arr):
print(arr)
n = len(arr)
for i in range(n):
for j in range(n):
if arr[i] < arr[j] :
arr[j], arr[i] = arr[i], arr[j]
print(arr)
def selectionsort(arr):
print(arr)
n = len(arr)
for i in range(n-1):
min_index = i
for j in range(i+1, n):
if arr[j] < arr[min_index]:
min_index = j
arr[i], arr[min_index] = arr[min_index], arr[i]
print(arr)
def bubblesort(arr):
print(arr)
n = len(arr)
for i in range(n):
for j in range(0, n-i-1):
if arr[j] > arr[j+1] :
arr[j], arr[j+1] = arr[j+1], arr[j]
print(arr)
print("sort:")
sort([4,6,3,2,7,1,8,5])
print("selectionsort:")
selectionsort([4,6,3,2,7,1,8,5])
print("bubblesort:")
bubblesort([4,6,3,2,7,1,8,5])
with the results:
sort:
[4, 6, 3, 2, 7, 1, 8, 5]
[6, 4, 3, 2, 7, 1, 8, 5]
[7, 4, 3, 2, 6, 1, 8, 5]
[8, 4, 3, 2, 6, 1, 7, 5]
[4, 8, 3, 2, 6, 1, 7, 5]
[3, 8, 4, 2, 6, 1, 7, 5]
[3, 4, 8, 2, 6, 1, 7, 5]
[2, 4, 8, 3, 6, 1, 7, 5]
[2, 3, 8, 4, 6, 1, 7, 5]
[2, 3, 4, 8, 6, 1, 7, 5]
[2, 3, 4, 6, 8, 1, 7, 5]
[1, 3, 4, 6, 8, 2, 7, 5]
[1, 2, 4, 6, 8, 3, 7, 5]
[1, 2, 3, 6, 8, 4, 7, 5]
[1, 2, 3, 4, 8, 6, 7, 5]
[1, 2, 3, 4, 6, 8, 7, 5]
[1, 2, 3, 4, 6, 7, 8, 5]
[1, 2, 3, 4, 5, 7, 8, 6]
[1, 2, 3, 4, 5, 6, 8, 7]
[1, 2, 3, 4, 5, 6, 7, 8]
selectionsort:
[4, 6, 3, 2, 7, 1, 8, 5]
[1, 6, 3, 2, 7, 4, 8, 5]
[1, 2, 3, 6, 7, 4, 8, 5]
[1, 2, 3, 6, 7, 4, 8, 5]
[1, 2, 3, 4, 7, 6, 8, 5]
[1, 2, 3, 4, 5, 6, 8, 7]
[1, 2, 3, 4, 5, 6, 8, 7]
[1, 2, 3, 4, 5, 6, 7, 8]
bubblesort:
[4, 6, 3, 2, 7, 1, 8, 5]
[4, 3, 6, 2, 7, 1, 8, 5]
[4, 3, 2, 6, 7, 1, 8, 5]
[4, 3, 2, 6, 1, 7, 8, 5]
[4, 3, 2, 6, 1, 7, 5, 8]
[3, 4, 2, 6, 1, 7, 5, 8]
[3, 2, 4, 6, 1, 7, 5, 8]
[3, 2, 4, 1, 6, 7, 5, 8]
[3, 2, 4, 1, 6, 5, 7, 8]
[2, 3, 4, 1, 6, 5, 7, 8]
[2, 3, 1, 4, 6, 5, 7, 8]
[2, 3, 1, 4, 5, 6, 7, 8]
[2, 1, 3, 4, 5, 6, 7, 8]
[1, 2, 3, 4, 5, 6, 7, 8]
It doesn't make a big difference on small arrays but on large arrays it make a lot of difference (but then there are other methods).
I am trying to solve this problem but I can't manage to figure out how.
Let's suppose I have a list of positive and negative numbers whose sum is guaranteed to be 0.
[-10, 1, 2, 20, 5, -100, -80, 10, 15, 15, 60, 100, -20, -18]
I want to obtain a list with the largest number of sub-sets, using all the elements of the initial list only once. And each subset must have the sum 0.
So in the case of this simple input:
[-5, -4, 5, 2, 3, -1]
The best results that can be obtained are:
1. [[-5, 5], [[-4, -1, 2, 3]]] #2 subsets
2. [[-5, 2, 3], [-4, -1, 5]] #2 subsets
These, for example, would be totally wrong answers:
1. [[-5, -4, -1, 2, 3, 5]] #1 subsets that is the initial list, NO
2. [[-5,5]] #1 subset, and not all elements are used, NO
Even if it's NP-Complete, how can I manage to solve it even with a brute-force approach? I just need a solution for small list of numbers.
def get_subsets(lst):
N = len(lst)
cand = []
dp = [0 for x in range(1<<N)] # maximum number of subsets using elements represented by bitset
back = [0 for x in range(1<<N)]
# Section 1
for i in range(1,1<<N):
cur = 0
for j in range(N):
if i&(1<<j):
cur += lst[j]
if not cur:
cand.append(i) # if subset sums to 0, it's viable
dp[0] = 1
# Section 2
for i in range(1<<N):
while cand and cand[0] <= i:
cand.pop(0)
if not dp[i]:
continue
for j in cand:
if i&j: # if subsets intersect, it cannot be added
continue
if dp[i]+1 > dp[i|j]:
back[i|j] = j
dp[i|j] = dp[i]+1
# Section 3
ind = dp.index(max(dp))
res = []
while back[ind]:
cur = []
for i in range(N):
if back[ind]&(1<<i):
cur.append(lst[i])
res.append(cur)
ind = ind^back[ind]
return res
print (get_subsets([-5, -4, 5, 2, 3, -1]))
Basically, this solution collects all subsets of the original list that can sum to zero, then attempts to merge as many of them together as possible without colliding. It runs in worst-case O(2^{2N}) time, where N is the length of the list, but it should hit an average case of around O(2^N), since there typically shouldn't be too many subsets summing to 0.
EDIT: I added sections to facilitate explanation of the algorithm
Section 1: I iterate through all possible 2^N-1 nonempty subsets of the original list, and check which of these subsets sum to 0; any viable zero-sum subsets are added to the list cand (represented as an integer in the range [1,2^N-1] with bits set at the indices making up the subset).
Section 2: dp is a dynamic programming table storing the maximum number of subsets summing to 0 that can be formed using the subset represented by the integer i at dp[i]. Initially, all entries of dp are set to 0 except dp[0] = 1, since the empty set has a sum of 0. Then I iterate through each subset from 0 to 2^N-1, and I run through the list of candidate subsets and attempt to merge the two subsets.
Section 3: This is just backtracking to find the answer: while filling in dp, I also kept an array back that stores the most recent subset added to achieve the subset i at back[i]. So I find the subset that maximizes the number of sub-subsets summing to 0 with ind = dp.index(max(dp)), and then I backtrack from there, shrinking the subset by removing the most recently added subset until I finally arrive back to the empty set.
This problem is NP-complete, since it is a combination of two NP-complete problems:
finding a single subset whose sum is 0 is known as the subset sum problem
when you find all the subsets whose sum is 0, you have to solve an exact cover problem with a special condition: you want to maximize the number of subsets.
The following steps will provide a solution:
use dynamic programming to find the subsets whose sum is 0 ('https://en.wikipedia.org/wiki/Subset_sum_problem#Pseudo-polynomial_time_dynamic_programming_solution)
to maximize the number of subsets, one would use D. Knuth's Algorithm X to find the exact cover.
A few remarks:
First, we know that there is an exact cover because the list of numbers has a sum of 0.
Second, we can use only the subsets that are not supersets of any other subset. Because, if A is a superset of X (both sum to 0), A can't be in the cover that has the largest number of subsets. Let A, B, C, ... be the cover with the maximum number of subsets, then we can replace A by X and A\X (it is trivial to see that the sum of A\X elements is 0) and we get the cover X, A\X, B, C, ... that is better.
Third, when we use Algorithm X, all paths in the search tree will lead to a success. Let A, B, C, ... be a path composed of non overlapping subsets having each a sum of 0. Then the complent has also a sum of 0 (which may be a superset of another subset, and then we'll use 2.).
As you see, nothing new here, and I will use only well known techniques/algorithms.
Find the subsets having a sum of 0.
The algorithm is well known. Here's a Python implementation based on Wikipedia explanations
class Q:
def __init__(self, values):
self.len = len(values)
self.min = sum(e for e in values if e <= 0)
self.max = sum(e for e in values if e >= 0)
self._arr = [False] * self.len * (self.max - self.min + 1)
def __getitem__(self, item):
index, v = item
return self._arr[v * self.len + index]
def __setitem__(self, item, value):
index, v = item
self._arr[v * self.len + index] = value
class SubsetSum:
def __init__(self, values):
self._values = values
self._q = Q(values)
def prepare(self):
for s in range(self._q.min, self._q.max + 1):
self._q[0, s] = (self._values[0] == s)
for i in range(self._q.len):
self._q[i, 0] = True
for i in range(1, self._q.len):
v = self._values[i]
for s in range(self._q.min, self._q.max + 1):
self._q[i, s] = (v == s) or self._q[i - 1, s] or self._q[
i - 1, s - v]
def subsets(self, target=0):
yield from self._subsets(self._q.len - 1, target, [])
def _subsets(self, i, target, p):
assert i >= 0
v = self._values[i]
c = self._q[i - 1, target]
b = self._q[i - 1, target - v]
if i == 0:
if target == 0:
if p:
yield p
elif self._q[0, target]:
yield p + [i]
else:
if self._q.min <= target - v <= self._q.max and self._q[
i - 1, target - v]:
yield from self._subsets(i - 1, target - v, p + [i])
if self._q[i - 1, target]:
yield from self._subsets(i - 1, target, p)
Here's how it works:
arr = [-10, 1, 2, 20, 5, -100, -80, 10, 15, 15, 60, 100, -20, -18]
arr = sorted(arr)
s = SubsetSum(arr)
s.prepare()
subsets0 = list(s.subsets())
print(subsets0)
Output:
[[13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0], [13, 12, 11, 10, 9, 7, 6, 5, 3, 2, 1, 0], [13, 12, 11, 10, 9, 4, 2, 1, 0], [13, 12, 11, 10, 8, 7, 4, 2, 1, 0], [13, 12, 11, 10, 8, 6, 5, 4, 3, 1, 0], [13, 12, 11, 10, 7, 2, 1, 0], [13, 12, 11, 10, 6, 5, 3, 1, 0], [13, 12, 11, 9, 8, 7, 4, 2, 1, 0], [13, 12, 11, 9, 8, 6, 5, 4, 3, 1, 0], [13, 12, 11, 9, 7, 2, 1, 0], [13, 12, 11, 9, 6, 5, 3, 1, 0], [13, 12, 11, 8, 7, 6, 5, 3, 1, 0], [13, 12, 11, 8, 4, 1, 0], [13, 12, 11, 1, 0], [13, 12, 10, 9, 8, 7, 6, 5, 4, 3, 1, 0], [13, 12, 10, 9, 8, 2, 1, 0], [13, 12, 10, 9, 7, 6, 5, 3, 1, 0], [13, 12, 10, 9, 4, 1, 0], [13, 12, 10, 8, 7, 4, 1, 0], [13, 12, 10, 7, 1, 0], [13, 12, 9, 8, 7, 4, 1, 0], [13, 12, 9, 7, 1, 0], [13, 11, 10, 8, 6, 5, 4, 3, 2, 0], [13, 11, 10, 6, 5, 3, 2, 0], [13, 11, 9, 8, 6, 5, 4, 3, 2, 0], [13, 11, 9, 6, 5, 3, 2, 0], [13, 11, 8, 7, 6, 5, 3, 2, 0], [13, 11, 8, 4, 2, 0], [13, 11, 7, 6, 5, 4, 3, 2, 1], [13, 11, 7, 6, 5, 4, 3, 0], [13, 11, 2, 0], [13, 10, 9, 8, 7, 6, 5, 4, 3, 2, 0], [13, 10, 9, 7, 6, 5, 3, 2, 0], [13, 10, 9, 4, 2, 0], [13, 10, 8, 7, 4, 2, 0], [13, 10, 8, 6, 5, 4, 3, 2, 1], [13, 10, 8, 6, 5, 4, 3, 0], [13, 10, 7, 2, 0], [13, 10, 6, 5, 3, 2, 1], [13, 10, 6, 5, 3, 0], [13, 9, 8, 7, 4, 2, 0], [13, 9, 8, 6, 5, 4, 3, 2, 1], [13, 9, 8, 6, 5, 4, 3, 0], [13, 9, 7, 2, 0], [13, 9, 6, 5, 3, 2, 1], [13, 9, 6, 5, 3, 0], [13, 8, 7, 6, 5, 3, 2, 1], [13, 8, 7, 6, 5, 3, 0], [13, 8, 4, 2, 1], [13, 8, 4, 0], [13, 7, 6, 5, 4, 3, 1], [13, 2, 1], [13, 0], [12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1], [12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 0], [12, 11, 10, 9, 8, 2, 0], [12, 11, 10, 9, 7, 6, 5, 3, 2, 1], [12, 11, 10, 9, 7, 6, 5, 3, 0], [12, 11, 10, 9, 4, 2, 1], [12, 11, 10, 9, 4, 0], [12, 11, 10, 8, 7, 4, 2, 1], [12, 11, 10, 8, 7, 4, 0], [12, 11, 10, 8, 6, 5, 4, 3, 1], [12, 11, 10, 7, 2, 1], [12, 11, 10, 7, 0], [12, 11, 10, 6, 5, 3, 1], [12, 11, 9, 8, 7, 4, 2, 1], [12, 11, 9, 8, 7, 4, 0], [12, 11, 9, 8, 6, 5, 4, 3, 1], [12, 11, 9, 7, 2, 1], [12, 11, 9, 7, 0], [12, 11, 9, 6, 5, 3, 1], [12, 11, 8, 7, 6, 5, 3, 1], [12, 11, 8, 4, 1], [12, 11, 1], [12, 10, 9, 8, 7, 6, 5, 4, 3, 1], [12, 10, 9, 8, 2, 1], [12, 10, 9, 8, 0], [12, 10, 9, 7, 6, 5, 3, 1], [12, 10, 9, 4, 1], [12, 10, 8, 7, 4, 1], [12, 10, 7, 1], [12, 9, 8, 7, 4, 1], [12, 9, 7, 1], [11, 10, 8, 6, 5, 4, 3, 2], [11, 10, 6, 5, 3, 2], [11, 9, 8, 6, 5, 4, 3, 2], [11, 9, 6, 5, 3, 2], [11, 8, 7, 6, 5, 3, 2], [11, 8, 4, 2], [11, 7, 6, 5, 4, 3], [11, 2], [10, 9, 8, 7, 6, 5, 4, 3, 2], [10, 9, 7, 6, 5, 3, 2], [10, 9, 4, 2], [10, 8, 7, 4, 2], [10, 8, 6, 5, 4, 3], [10, 7, 2], [10, 6, 5, 3], [9, 8, 7, 4, 2], [9, 8, 6, 5, 4, 3], [9, 7, 2], [9, 6, 5, 3], [8, 7, 6, 5, 3], [8, 4]]
Reduce the number of subsets
We have 105 subsets that sum to 0, but we can remove the subsets that are superset of other subsets. We need a function to find if a list of elements contains all elements in another list. In Python:
import collections
def contains(l1, l2):
"""
Does l1 contain all elements of l2?
"""
c = collections.Counter(l1)
for e in l2:
c[e] -= 1
return all(n >= 0 for n in c.values())
Now, we can remove the subsets that are supersets of another subset.
def remove_supersets(subsets):
subsets = sorted(subsets, key=len)
new_subsets = []
for i, s1 in enumerate(subsets):
for s2 in subsets[:i]: # smaller subsets
if contains(s1, s2):
break
else: # not a superset
new_subsets.append(s1)
return new_subsets
In our situation:
subsets0 = remove_supersets(subsets0)
print(len(subsets0))
Output:
[[13, 0], [11, 2], [8, 4], [13, 2, 1], [12, 11, 1], [10, 7, 2], [9, 7, 2], [12, 10, 7, 1], [12, 9, 7, 1], [10, 9, 4, 2], [10, 6, 5, 3], [9, 6, 5, 3], [12, 11, 10, 7, 0], [12, 11, 9, 7, 0], [12, 10, 9, 8, 0], [12, 10, 9, 4, 1], [8, 7, 6, 5, 3], [12, 11, 10, 9, 4, 0], [12, 10, 9, 8, 2, 1], [11, 7, 6, 5, 4, 3], [13, 7, 6, 5, 4, 3, 1]]
[[0, 2, 10, 6, 4], [0, 2, 10, 8, 1], [0, 2, 11, 5, 4], [0, 2, 11, 7, 1], [0, 16, 9, 4], [0, 16, 15, 1], [0, 18, 19], [3, 2, 12, 11], [3, 2, 13, 10], [3, 17, 16], [3, 19, 14], [20, 14, 1]]
We managed to reduce the number of subsets to 21, that is a good improvement since we need to explore all possibilities to find an exact cover.
Algorithm X
I do not use the dancing links here (I think that technique is we'll designed for low level languages like C, but you can implement them in Python if you want). We just need to keep track of the remaing subsets:
class Matrix:
def __init__(self, subsets, ignore_indices=set()):
self._subsets = subsets
self._ignore_indices = ignore_indices
def subset_values(self, i):
assert i not in self._ignore_indices
return self._subsets[i]
def value_subsets_indices(self, j):
return [i for i, s in self._subsets_generator() if j in s]
def _subsets_generator(self):
return ((i, s) for i, s in enumerate(self._subsets) if
i not in self._ignore_indices)
def rarest_value(self):
c = collections.Counter(
j for _, s in self._subsets_generator() for j in s)
return c.most_common()[-1][0]
def take_subset(self, i):
s = self._subsets[i]
to_ignore = {i2 for i2, s2 in self._subsets_generator() if
set(s2) & set(s)}
return Matrix(self._subsets,
self._ignore_indices | to_ignore)
def __bool__(self):
return bool(list(self._subsets_generator()))
And finally the cover function:
def cover(m, t=[]):
if m: # m is not empty
j = m.rarest_value()
for i in m.value_subsets_indices(j):
m2 = m.take_subset(i)
yield from cover(m2, t + [i])
else:
yield t
Finally, we have:
m = Matrix(subsets0)
ts = list(cover(m))
t = max(ts, key=len)
print([[arr[j] for j in subsets0[i]] for i in t])
Output:
[[100, -100], [10, -10], [15, 2, 1, -18], [15, 5, -20], [60, 20, -80]]
Below essentially the same idea as Michael Huang, with 30 more lines...
A solution with cliques
We can prebuild all subsets whose sum is 0.
Build subsets of 1 elem;
then of size 2 by reusing the previous ones
and keep those whose sum is zero along the way
Now say such subset is a node of a graph.
Then a node is in relation with another one iff their associated subset has no number in common.
We thus want to build the maximum clique of the graph:
In a clique, all nodes are in relation idem their subsets are disjoints
The maximum clique gives us the maximal number of subsets
function forall (v, reduce) {
const nexts = v.map((el, i) => ({ v: [el], i, s: el })).reverse()
while (nexts.length) {
const next = nexts.pop()
for (let i = next.i + 1; i < v.length; ++i) {
const { s, skip } = reduce(next, v[i])
if (!skip) {
nexts.push({ v: next.v.concat(v[i]), s: s, i })
}
}
}
}
function buildSubsets (numbers) {
const sums = []
forall(numbers, (next, el) => {
const s = next.s + el
if (s === 0) {
sums.push({ s, v: next.v.concat(el) })
return { s, skip: true }
}
return { s }
})
return sums
}
const bin2decs = bin => {
const v = []
const s = bin.toString(2)
for (let i = 0; i < s.length; ++i) {
if (intersects(dec2bin(i), bin)) {
v.push(i)
}
}
return v
}
const dec2bin = dec => Math.pow(2, dec)
const decs2bin = decs => decs.reduce((bin, dec) => union(dec2bin(dec), bin), 0)
// Set methods on int
const isIn = (a, b) => (a & b) === a
const intersects = (a, b) => a & b
const union = (a, b) => a | b
// if a subset contains another one, discard it
// e.g [1,2,4] should be discarded if [1,2] is present
const cleanSubsets = bins => bins.filter(big => bins.every(b => big === b || !isIn(b, big)))
function bestClique (decs) {
const cliques = []
forall(decs, (next, el) => {
if (intersects(next.s, el)) { return { skip: true } }
const s = union(next.s, el)
cliques.push({ s, v: next.v.concat(el) })
return { s }
})
return cliques.sort((a, b) => b.v.length - a.v.length)[0]
}
// in case we have duplicated numbers in the list,
// they are still uniq thanks to their id: i (idem position in the list)
const buildNumbers = v => v.map((n, i) => {
const u = new Number(n)
u.i = i
return u
})
function run (v) {
const numbers = buildNumbers(v)
const subs = buildSubsets(numbers)
const bins = subs.map(s => decs2bin(s.v.map(n => n.i)))
const clique = bestClique(cleanSubsets(bins))
const indexedSubs = clique.v.map(bin2decs)
const subsets = indexedSubs.map(sub => sub.map(i => numbers[i].valueOf()))
console.log('subsets', JSON.stringify(subsets))
}
run([1, -1, 2, -2])
run([-10, 1, 2, 20, 5, -100, -80, 10, 15, 15, 60, 100, -20, -18, 10, -10])
run([-5, -4, 5, 2, 3, -1])
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
https://projecteuler.net/problem=8
data = '''73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450'''
The code is:
def largest_product_series
mx = 0
arr = data.split("")
arr.map!{ |x| x.to_i}
(arr.size - 13).times do |i|
0.upto(13) do |c|
result = arr.inject() {|sum,c | sum * arr[i + c]}
puts result
if result >= mx
mx = result
end
end
end
#puts "#{mx}"
mx
end
It is actually from Euler project 8 which I am working on. please help me to correct my code and give me some advise, I keep getting a hundred digit output.
A Rubyish way of doing that is as follows.
data = '73167176531330624919225119674426574742355349194934' +
'96983520312774506326239578318016984801869478851843' +
'85861560789112949495459501737958331952853208805511' +
'12540698747158523863050715693290963295227443043557' +
'66896648950445244523161731856403098711121722383113' +
'62229893423380308135336276614282806444486645238749' +
'30358907296290491560440772390713810515859307960866' +
'70172427121883998797908792274921901699720888093776' +
'65727333001053367881220235421809751254540594752243' +
'52584907711670556013604839586446706324415722155397' +
'53697817977846174064955149290862569321978468622482' +
'83972241375657056057490261407972968652414535100474' +
'82166370484403199890008895243450658541227588666881' +
'16427171479924442928230863465674813919123162824586' +
'17866458359124566529476545682848912883142607690042' +
'24219022671055626321111109370544217506941658960408' +
'07198403850962455444362981230987879927244284909188' +
'84580156166097919133875499200524063689912560717606' +
'05886116467109405077541002256983155200055935729725' +
'71636269561882670428252483600823257530420752963450'
def prod(arr)
arr.reduce(1, :*)
end
arr = data.each_char.map(&:to_i).each_cons(13).max_by { |a| prod(a) }
#=> [5, 5, 7, 6, 6, 8, 9, 6, 6, 4, 8, 9, 5]
[arr.join, prod(arr)]
#=> ["5576689664895", 23514624000]
The steps are as follows. Suppose
data = '731671765313326'
(data.size #=> 15). Then
b = data.each_char.map(&:to_i)
#=> [7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 2, 6]
c = b.each_cons(13)
#=> #<Enumerator: [7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 2, 6]:each_cons(13)>
We can see the elements that will be generated by this enumerator by converting it to an array:
c.to_a
#=> [[7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3],
# [3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 2],
# [1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 2, 6]]
arr = c.max_by { |a| prod(a) }
#=> [7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3]
This was obtained by computing:
[prod [7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3],
prod [3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 2],
prod [1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3, 2, 6]].max
#=> [5_000_940, 1_428_840, 2_857_680].max
#=> 5000940
The last step is to return the array
[arr.join, prod(arr)]
#=> [[7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3].join,
#=> prod([7, 3, 1, 6, 7, 1, 7, 6, 5, 3, 1, 3, 3])]
#=> ["7316717653133", 5000940]
Why does the following piece of code result in an infinite loop of 3's?
a = [1,2,3,4,5,6,7,8,9,10]
a.each {|value| puts a.insert(value,3)}
The problem is that insert changes the original array:
a = [1,2,3,4,5,6,7,8,9,10]
a.each do |value|
a.insert(value, 3)
p a
end
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] # original, ^ marks current value
# ^
# [1, 3, 2, 3, 4, 5, 6, 7, 8, 9, 10] # inserted 3 at position 1
# ^
# [1, 3, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10] # inserted 3 at position 3
# ^
# [1, 3, 3, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10] # inserted 3 at position 2
# ^
# [1, 3, 3, 3, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10] # inserted 3 at position 2
# ^
# [1, 3, 3, 3, 3, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10] # inserted 3 at position 2
# ^
# [1, 3, 3, 3, 3, 3, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10] # inserted 3 at position 2
# ^
# ... # continues forever ...
What you probably want instead is something like this:
a = [1,2,3,4,5,6,7,8,9,10]
a.each_index {|index| p a.dup.insert(index, 3) }
# [3, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
# [1, 3, 2, 3, 4, 5, 6, 7, 8, 9, 10]
# [1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10]
# [1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10]
# [1, 2, 3, 4, 3, 5, 6, 7, 8, 9, 10]
# [1, 2, 3, 4, 5, 3, 6, 7, 8, 9, 10]
# [1, 2, 3, 4, 5, 6, 3, 7, 8, 9, 10]
# [1, 2, 3, 4, 5, 6, 7, 3, 8, 9, 10]
# [1, 2, 3, 4, 5, 6, 7, 8, 3, 9, 10]
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 3, 10]
each_index iterates over the indices, not the values. This is likely the correct thing to do here, because insert takes an index as first argument.
dup duplicates the array on every iteration so a remains unchanged.
Given an array of n elements, remove any adjacent pair of elements which are equal. Repeat this operation until there are no more adjacent pairs to remove; that will be the final array.
For e.g 1 2 2 3 4 should return the array 1 3 4.
please note array need not to be sorted.
check this test case also: 1,2,2,3,4,4,3,5 o/p should be 1,5.
(2,2) and (4,4) gets removed, then (3,3) which became adjacent after the removal of (4,4)
Any time you remove a pair of elements, you also need to see if you generated another pair that you want to remove.
The algorithm should follow naturally from that observation.
In Python:
>>> l=[1,2,2,3,4,4,3,5]
>>> [x for x in l if not l.count(x) > 1]
[1, 5]
This removes all integers that occur more than once in the list. This is a correct result for your example but I think that you are really trying to state something different. I think you are saying:
list:=(an unsorted list of integers)
while adjacent_pairs(list) is True:
remove_adjacent_pairs(list)
Once again, in Python:
#!/usr/bin/env python
def dedupe_adjacent(l):
for i in xrange(len(l) - 1, 0, -1):
if l[i] == l[i-1]:
del l[i-1:i+1]
return True
return False
def process_list(l):
print "input list: ",l
i=1
while(dedupe_adjacent(l)):
print " loop ",i,":",l
i+=1
print "processed list=",l
print
process_list([1,2,2,3,4,4,3,5])
process_list([1,2,2,3,4,4,6,3,5])
Output:
input list: [1, 2, 2, 3, 4, 4, 3, 5]
loop 1 : [1, 2, 2, 3, 3, 5]
loop 2 : [1, 2, 2, 5]
loop 3 : [1, 5]
processed list= [1, 5]
input list: [1, 2, 2, 3, 4, 4, 6, 3, 5]
loop 1 : [1, 2, 2, 3, 6, 3, 5]
loop 2 : [1, 3, 6, 3, 5]
processed list= [1, 3, 6, 3, 5]
The following:
function compress(arr) {
var prev, res = [];
for (var i in arr) {
if (i == 0 || (arr[i] != arr[i - 1]))
res.push(arr[i]);
}
return res;
}
compress([1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]);
Returns:
[1, 2, 3, 4, 3, 5, 6, 7, 8]
Also (JavaScript 1.6 solution):
[1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8].filter(function(el, i, arr) {
return i == 0 || (el != arr[i - 1]);
})
Edit: Removing any item that appears in the array more than once requires a different solution:
function dedup(arr) {
var res = [], seen = {};
for (var i in arr)
seen[arr[i]] = seen[arr[i]] ? ++seen[arr[i]] : 1;
for (var j in arr) {
if (seen[arr[j]] == 1)
res.push(arr[j]);
}
return res;
}
The following:
dedup([1, 2, 2, 3, 4, 4, 3, 5]);
Produces:
[1, 5]
I have a solution to this in Java. You need to use replaceAll method in String class in Java. You can use regular expession to remove such adjacent redundant characters:
public class MyString {
public static void main(String[] args) {
String str = "12234435";
while(!str.replaceAll("(\\w)\\1+", "").equalsIgnoreCase(str))
str = str.replaceAll("(\\w)\\1+", "");
System.out.println(str);
}
}
You can find how to give a regular expression here
I would:
Sort the array.
From the start of the array, until you are at the last element of the array do:
`count` = count the number of array[i] elements.
remove the first `count` elements of the array if `count` > 1.
The following Python 3 code will remove duplicates from a list (array). It does this by scanning the array from start towards end and compares the target element with the element one larger. If they are the same they are removed. If the element pointer is not pointing at 0, then it is reduced by 1 in order to catch nested pairs. If the two compared elements are different then the pointer is incremented.
I'm sure there's a more pythonic way to remove two adjacent elements from a list, but I'm new to Python and haven't figured that out yet. Also, you'll want to get rid of the print(indx, SampleArray) statement--I left it in there to let you follow the progress in the output listing below.
# Algorithm to remove duplicates in a semi-sorted list
def CompressArray(SampleArray):
indx=0
while(indx < len(SampleArray)-1):
print(indx, SampleArray)
if(SampleArray[indx]==SampleArray[indx+1]):
del(SampleArray[indx])
del(SampleArray[indx])
if(indx>0):
indx-=1
else:
indx+=1
return SampleArray
Here are sample runs for:
[1, 2, 2, 3, 4]
[1, 2, 2, 3, 4, 4, 3, 5]
[1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
[1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
[1, 1, 2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
================================
0 [1, 2, 2, 3, 4]
1 [1, 2, 2, 3, 4]
0 [1, 3, 4]
1 [1, 3, 4]
[1, 3, 4]
================================
0 [1, 2, 2, 3, 4, 4, 3, 5]
1 [1, 2, 2, 3, 4, 4, 3, 5]
0 [1, 3, 4, 4, 3, 5]
1 [1, 3, 4, 4, 3, 5]
2 [1, 3, 4, 4, 3, 5]
1 [1, 3, 3, 5]
0 [1, 5]
[1, 5]
================================
0 [1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 4, 3, 3, 5, 6, 7, 8, 8]
2 [1, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 4, 5, 6, 7, 8, 8]
2 [1, 4, 5, 6, 7, 8, 8]
3 [1, 4, 5, 6, 7, 8, 8]
4 [1, 4, 5, 6, 7, 8, 8]
5 [1, 4, 5, 6, 7, 8, 8]
[1, 4, 5, 6, 7]
================================
0 [1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
2 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
3 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
4 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
3 [1, 3, 4, 6, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
2 [1, 3, 4, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 3, 3, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 5, 9, 10, 10, 9, 11]
1 [1, 5, 9, 10, 10, 9, 11]
2 [1, 5, 9, 10, 10, 9, 11]
3 [1, 5, 9, 10, 10, 9, 11]
2 [1, 5, 9, 9, 11]
1 [1, 5, 11]
[1, 5, 11]
================================
0 [1, 1, 2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [2, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
2 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [4, 5, 5, 7, 8, 8, 7, 4, 9]
0 [4, 7, 8, 8, 7, 4, 9]
1 [4, 7, 8, 8, 7, 4, 9]
2 [4, 7, 8, 8, 7, 4, 9]
1 [4, 7, 7, 4, 9]
0 [4, 4, 9]
[9]
================================
I love Java, but functional solutions should get more time on this site.
In Haskell, doing things the way the question asks:
compress lst = if (length lst == length b) then lst else (compress b) where
b = helper lst
helper [] = []
helper [x] = [x]
helper (x:y:xs) = if (x == y) then (helper xs) else (x:helper (y:xs))
You can solve this problem in O(n) time, although it is a bit more complicated
compress' lst = reverse (helper [] lst) where
helper xs [] = xs
helper [] (x:xs) = helper [x] xs
helper (a:as) (x:xs)
| a == x = helper as xs
| otherwise = helper (x:a:as) xs
I think we could use a stack to check adjacent duplicated elements.
Scan the array. For each new element, if it is equal to the top element in the stack, drop it and pop the top element from the stack. Otherwise, push it into the stack.
Here is the stack based algorithm based upon the edited question.
// pseudo code, not tested
void RemoveDupp(vector<int> & vin, vector<int> & vout)
{
int i = 0, int j = -1;
vout.resize(vin.size());
while (i < vin.size())
{
if (j == -1 || vout[j] != vin[i])
vout[++j] = vin[i++]; //push
else
j--, i++; //pop
}
vout.resize(j + 1);
}