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An online coding test problem

An online coding test problem# JobHunting - 待字闺中

g*y2014-09-02 07:09

1 楼

Consider N coins aligned in a row. Each coin is showing either heads or
tails. The adjacency of these coins is the number of adjacent pairs of coins
with the same side facing up.
It must return the maximum possible adjacency that can be obtained by
reversing exactly one coin (that is, one of the coins must be reversed).
Consecutive elements of array A represent consecutive coins in the row.
Array A contains only 0s and/or 1s: 0 represents a coin with heads facing up
; 1 represents a coin with tails facing up. For example, given array A
consisting of six numbers, such that:
A[0] = 1
A[1] = 1
A[2] = 0
A[3] = 1
A[4] = 0
A[5] = 0
the function should return 4. The initial adjacency is 2, as there are two
pairs of adjacent coins with the same side facing up, namely (0, 1) and (4,
5). After reversing the coin represented by A[2], the adjacency equals 4, as
there are four pairs of adjacent coins with the same side facing up, namely
higher adjacency. The same adjacency can be obtained by reversing the coin
represented by A[3].
I should mention here that number of elements can be from 1....10'000. And
it has to be done as O(N)

y*a2014-09-02 07:09

2 楼

int longestConsecutiveCoins(boolean[]A) {
int n=A.length;
if (n<=1) return n;
int[] f=new int[n], b=new int[n]; f[0]=1; b[n-1]=1;
for (int i=1;if[i]=A[i]==A[i-1]?f[i-1]+1:1;
b[n-i-1]=A[i]==A[i+1]?b[n-i]+1:1;
}
int rel=0;
for (int i=1;irel=Math.max(rel, 1+(!A[i]==A[i-1]?f[i-1]:0)+
(!A[i]==A[i+1]?b[i+1]:0));
if (!A[0]==A[1]) rel=Math.max(rel, 1+b[1]);
if (!A[n-1]==A[n-2]) rel=Math.max(rel, f[n-2]+1);
return rel;
}

g*y2014-09-02 07:09

3 楼

The additional space is not allowed.

【在 y**********a 的大作中提到】

: int longestConsecutiveCoins(boolean[]A) {
: int n=A.length;
: if (n<=1) return n;
: int[] f=new int[n], b=new int[n]; f[0]=1; b[n-1]=1;
: for (int i=1;i: f[i]=A[i]==A[i-1]?f[i-1]+1:1;
: b[n-i-1]=A[i]==A[i+1]?b[n-i]+1:1;
: }
: int rel=0;
: for (int i=1;i

s*i2014-09-02 07:09

4 楼

Java，space 是O(1) time O(n)

coins
up

【在 g******y 的大作中提到】

: Consider N coins aligned in a row. Each coin is showing either heads or
: tails. The adjacency of these coins is the number of adjacent pairs of coins
: with the same side facing up.
: It must return the maximum possible adjacency that can be obtained by
: reversing exactly one coin (that is, one of the coins must be reversed).
: Consecutive elements of array A represent consecutive coins in the row.
: Array A contains only 0s and/or 1s: 0 represents a coin with heads facing up
: ; 1 represents a coin with tails facing up. For example, given array A
: consisting of six numbers, such that:
: A[0] = 1