如何计算受±1或±2步幅约束的简单路径? [英] How to count simple paths constrained by ±1 or ±2 steps?
问题描述
我发现了这个有趣的动态编程问题,并且想知道这种方法。
我们给了一个大小为'n'的数组'a'。 / p>
数组中的每个元素均为'1'或'2'。
我们从索引 0开始。如果a [i] = 1,我们可以转到i + 1或i-1。
相反,如果a [i] = 2,我们可以转到i + 1或i + 2或i-1或i-2。
我们必须找到所有可能路径的数量。
**主要约束**:-1)我们只能一次访问数组中的特定索引。
2)我们总是从索引'0'开始。
3)路径可以在我们想要的任何时间结束:-)
示例数组:-> [1 ,1,1,1]
答案:-4
第1条可能的路径:[0]
第二种可能的路径:[0,1]
第三种可能的路径:[0,1,2]
第四个可能路径:[0,1,2,3]
另一个示例:-
[2,2,2]
答案:-5
路径:-[0],[0,1],[0,1,2],[0,2,1],[0,2]。
(此问题分为3部分!)
n的值在范围内:-1)[1,100000]
2)[1,10]
3)[1,1000]
考虑使用的空格。
0 1 2 3 4 5 6
^
为了从右边获得一个数字,必须已使用它之前的单元格。因此,以 x
结尾的所有方式都不能包含来自左侧的数字。以 x
结尾的所有方式都来自右侧,使用了 x-1
和向右移动的一组动作 x
从左侧不相交。
让 f(A,x)= l(A,x)+ r(A,x)
,其中 l(A,x)
表示以x结尾的所有方式左边; r(A,x)
,从右边开始。
要获得 l( A,x)
,我们需要:
(1)所有到达途径(x-1 )
= l(A,x-1)
(x右边没有数字用于
,并且由于
x最后使用,所以我们可以不是
从右边到达x-1。)
(2)所有到达(x-2)的方式:
显然我们需要l(A,x-2 )。现在
将从右边到达(x-2),
唯一有效的路径是
...(x-3)->(x-1)-> (x-2)
等于
从左侧到达(x-3)的路数。
= l(A,x-2)+ l(A,x-3)
要获得 r(A,x)
,我们需要:
(1)所有到达(x + 1)的方式,因此
直接从那里到达x
= l(A,x-1)
(我们只能从(x-1)到达(x + 1)。)
(2)从(x + 1)开始在
之后到达(x + 2)的所有方式
= l(A,x-1)* f(A [x + 1 ...],1)
(到达
A [ x + 1 ...],我们必须先到达
(x-1)。)
所以看来
f(A,x)= l(A,x)+ r(A,x )
l(A,x)=
l(A,x-1)+ l(A,x-2)+ l(A,x-3)
r(A,x)=
l(A,x-1)+ l(A,x-1)* f(A [x + 1 ...],1)
下面的JavaScript代码每次运行时都会尝试使用不同的7元素数组。我将备忘录和优化留给读者使用(为了有效地制表 f(_,1)
,请注意 l(_,1)= 1
)。
function f(A,x){ if(x< 0 || x> A.length-1)返回0返回l(A,x)+ r(A,x)函数l(A,x){if(x< 0 || x > A.length-1)如果(x == 0)则返回0,如果(A [x-2]&& A [x-2] ==,则返回1 let结果= 1(A,x-1) 2){结果+ = l(A,x-2)如果(A [x-3]&& A [x-3] == 2)结果+ = l(A,x-3)}返回结果}函数r(A,x){如果(x< 0 || x> = A.length-1 ||!(A [x-1]&& A [x-1] == 2) )如果((A [x + 2]&& A [x + 2] == 2)结果+ = l(A,x-1)* f( A.slice(x + 1),1)返回结果}}函数validate(A){令n = A.length函数g(i,s){如果(调试)console.log(s)let result = 1 let [a,b] = [i + 1,i-1]如果(< n&& !s.includes(a))结果+ = g(a,s.slice()。concat(a))如果(b> = 0&&!s.includes(b))结果+ = g( b,s.slice()。concat(b))如果(A [i] == 2){[a,b] = [i + 2,i-2]如果(a< n&! s.includes(a))结果+ = g(a,s.slice()。concat(a))如果(b> = 0&&!s.includes(b))结果+ = g(b ,s.slice()。concat(b))}返回结果}返回g(0,[0])}让debug = falselet arr = []让n = 7for(让i = 0; i
I have found this interesting dynamic-programming problem and want to know the approach .
We are given an array 'a' of size-'n'.
Each element of the array is either '1' or '2'.
We start at index '0' . If a[i]=1 , we can go to i+1 or i-1.
On the contrary, If a[i]=2 , we can go to i+1 or i+2 or i-1 or i-2.
We have to find the number of all possible paths .
**Main Constraint ** : - 1) We can go to a particular index in an array only once .
2) We always start at the index-'0' .
3) A path can end anytime we want :- )
Example array : --> [1,1,1,1]
Answer : - 4
1ST possible path : [0]
2ND possible path : [0,1]
3rd possible path : [0,1,2]
4th possible path : [0,1,2,3]
Another example : -
[2,2,2]
Answer:- 5
Paths : - [0],[0,1],[0,1,2] , [0,2,1] , [0,2] .
(This question is divided into-3-parts!)
Value(s) of n are in range : - 1) [1,100000]
2) [1,10]
3)[1,1000]
Consider used spaces.
0 1 2 3 4 5 6
^
In order to reach a number from the right, the cell just before it must have been used. Therefore, all the ways to end with x
coming from the left cannot include numbers from the right. And all the ways to end with x
coming from the right used x-1
and a set of moves to the right of x
disjoint from the left side.
Let f(A, x) = l(A, x) + r(A, x)
, where l(A, x)
represents all ways to end at x coming from the left; r(A, x)
, coming from the right.
To obtain l(A, x)
, we need:
(1) all ways to reach (x-1)
= l(A, x-1)
(there are no numbers used to
the right of x, and since
x is used last, we could not
have reached x-1 from the right.)
(2) all ways to reach (x-2):
cleary we need l(A, x-2). Now
to reach (x-2) from the right,
the only valid path would have
been ...(x-3)->(x-1)->(x-2)
which equals the number of ways
to reach (x-3) from the left.
= l(A, x-2) + l(A, x-3)
To obtain r(A, x)
, we need:
(1) all ways to reach (x+1) so as
to directly go from there to x
= l(A, x-1)
(We can only reach (x+1) from (x-1).)
(2) all ways to reach (x+2) after
starting at (x+1)
= l(A, x-1) * f(A[x+1...], 1)
(To get to the starting point in
A[x+1...], we must first get to
(x-1).)
So it seems that
f(A, x) = l(A, x) + r(A, x)
l(A, x) =
l(A, x-1) + l(A, x-2) + l(A, x-3)
r(A, x) =
l(A, x-1) + l(A, x-1) * f(A[x+1...], 1)
The JavaScript code below tries a different 7-element array each time we run it. I leave memoisation and optimisation to the reader (for efficiently tabling f(_, 1)
, notice that l(_, 1) = 1
).
function f(A, x){
if (x < 0 || x > A.length - 1)
return 0
return l(A, x) + r(A, x)
function l(A, x){
if (x < 0 || x > A.length - 1)
return 0
if (x == 0)
return 1
let result = l(A, x-1)
if (A[x-2] && A[x-2] == 2){
result += l(A, x-2)
if (A[x-3] && A[x-3] == 2)
result += l(A, x-3)
}
return result
}
function r(A, x){
if (x < 0 || x >= A.length - 1 || !(A[x-1] && A[x-1] == 2))
return 0
let result = l(A, x-1)
if (A[x+2] && A[x+2] == 2)
result += l(A, x-1) * f(A.slice(x+1), 1)
return result
}
}
function validate(A){
let n = A.length
function g(i, s){
if (debug)
console.log(s)
let result = 1
let [a, b] = [i+1, i-1]
if (a < n && !s.includes(a))
result += g(a, s.slice().concat(a))
if (b >= 0 && !s.includes(b))
result += g(b, s.slice().concat(b))
if (A[i] == 2){
[a, b] = [i+2, i-2]
if (a < n && !s.includes(a))
result += g(a, s.slice().concat(a))
if (b >= 0 && !s.includes(b))
result += g(b, s.slice().concat(b))
}
return result
}
return g(0, [0])
}
let debug = false
let arr = []
let n = 7
for (let i=0; i<n; i++)
arr[i] = Math.ceil(Math.random() * 2)
console.log(JSON.stringify(arr))
console.log('')
let res = 0
for (let x=0; x<arr.length; x++){
let c = f(arr, x)
if (debug)
console.log([x, c])
res += c
}
if (debug)
console.log('')
let v = validate(arr)
if (debug)
console.log('')
console.log(v)
console.log(res)
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