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Minimum Hops - Medium

Problem: Given an array of positive non-zero integers. Find out the minimum hops required to traverse the array, if the value at an index denotes the maximum length you can hop from that index.


To perform a brute force solution, we can use a depth first search with pruning. This usually explodes with arrays of size greater than 50. But never the less, it's important to understand how it works.

def minhops_dfs(lst):
sz = len(lst)
min_hop = [0] * sz
def dfs(index,s,hops,min_hop):
if index >= sz:
if len(min_hop) > s:
min_hop[:] = hops
#min_hop.extend(hops)
elif s > len(min_hop):
pass
else:
hops.append(lst[index])
for i in xrange(1,lst[index] + 1):
dfs(index + i, s + 1,hops,min_hop)
hops.pop()
dfs(0,0,[],min_hop)
return len(min_hop)
view raw minhop dfs hosted with ❤ by GitHub
Following our dismal performance using DFS, we move on to the big guns, Dynamic Programming. We define minHop[i]  as the minimum hops you need from index i to reach the end of the array.

Therefore,

minHop[i] = 1 { if arr[i] + i > len(arr) } needs just one hop, trivial.
minHop[i] = 1 + min ( minHop[i+1 : i+arr[i]] ) { the minimum hop from my range which will reach the end of the array }

We start with i = len(arr) - 1 and count down to i  = 0 . At which point minHop[0] will be the minimum hops required to reach the end of the array from index = 0.
import random
def random_array(sz,low=10,high = 100):
return map(lambda i:random.randint(low,high),xrange(sz))
R = random_array(50,1,10)
print R
def minhops_dfs(lst):
sz = len(lst)
min_hop = [0] * sz
def dfs(index,s,hops,min_hop):
if index >= sz:
if len(min_hop) > s:
min_hop[:] = hops
#min_hop.extend(hops)
elif s > len(min_hop):
pass
else:
hops.append(lst[index])
for i in xrange(1,lst[index] + 1):
dfs(index + i, s + 1,hops,min_hop)
hops.pop()
dfs(0,0,[],min_hop)
return len(min_hop)
def min_hops_dp(lst):
sz = len(lst)
min_hops = [0] * sz
for l in xrange(sz - 1,-1,-1):
if lst[l] + l >= sz:
min_hops[l] = 1
else:
end = min(lst[l] + l,sz)
min_hops[l] = 1 + min(min_hops[l+1:end+1])
return min_hops[0]
print minhops_dfs(R)
print min_hops_dp(R)
view raw Minimum Hops hosted with ❤ by GitHub
So here we have an O(n^2) solution, at the expense of O(n) space. Until Next time.

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