07 Maximum Slice Problem

题目 1：MaxDoubleSliceSum

A non-empty zero-indexed array A consisting of N integers is given.

A triplet (X, Y, Z), such that 0 ≤ X < Y < Z < N, is called a double slice.

The sum of double slice (X, Y, Z) is the total of A[X + 1] + A[X + 2] + ... + A[Y − 1] + A[Y + 1] + A[Y + 2] + ... + A[Z − 1].

For example, array A such that:
A[0] = 3 A[1] = 2 A[2] = 6 A[3] = -1 A[4] = 4 A[5] = 5 A[6] = -1 A[7] = 2

contains the following example double slices:

double slice (0, 3, 6), sum is 2 + 6 + 4 + 5 = 17,
double slice (0, 3, 7), sum is 2 + 6 + 4 + 5 − 1 = 16,
double slice (3, 4, 5), sum is 0.

The goal is to find the maximal sum of any double slice.

Write a function:

int solution(vector<int> &A);

that, given a non-empty zero-indexed array A consisting of N integers, returns the maximal sum of any double slice.

For example, given:
A[0] = 3 A[1] = 2 A[2] = 6 A[3] = -1 A[4] = 4 A[5] = 5 A[6] = -1 A[7] = 2

the function should return 17, because no double slice of array A has a sum of greater than 17.

Assume that:

N is an integer within the range [3..100,000];
each element of array A is an integer within the range [−10,000..10,000].

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

#include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
// write your code in C++11
int n=A.size();
vector<int> left(n,0);
vector<int> right(n,0);
for(int i=1;i<n;i++)
{
left[i]=max((A[i]+left[i-1]),0);
right[n-i-1]=max((A[n-i-1]+right[n-i]),0);
}
int max_sum=-10002;
for(int i=1;i<n-1;i++)
{
max_sum=max(max_sum,left[i-1]+right[i+1]);
}
return max_sum;
}

题目 2：MaxProfit

A zero-indexed array A consisting of N integers is given. It contains daily prices of a stock share for a period of N consecutive days. If a single share was bought on day P and sold on day Q, where 0 ≤ P ≤ Q < N, then the profit of such transaction
is equal to A[Q] − A[P], provided that A[Q] ≥ A[P]. Otherwise, the transaction brings loss of A[P] − A[Q].

For example, consider the following array A consisting of six elements such that:
A[0] = 23171 A[1] = 21011 A[2] = 21123 A[3] = 21366 A[4] = 21013 A[5] = 21367

If a share was bought on day 0 and sold on day 2, a loss of 2048 would occur because A[2] − A[0] = 21123 − 23171 = −2048. If a share was bought on day 4 and sold on day 5, a profit of 354 would occur because A[5] − A[4] = 21367 − 21013 = 354. Maximum possible
profit was 356. It would occur if a share was bought on day 1 and sold on day 5.

Write a function,

int solution(vector<int> &A);

that, given a zero-indexed array A consisting of N integers containing daily prices of a stock share for a period of N consecutive days, returns the maximum possible profit from one transaction during this period. The function should return 0 if it was impossible
to gain any profit.

For example, given array A consisting of six elements such that:
A[0] = 23171 A[1] = 21011 A[2] = 21123 A[3] = 21366 A[4] = 21013 A[5] = 21367

the function should return 356, as explained above.

Assume that:

N is an integer within the range [0..400,000];
each element of array A is an integer within the range [0..200,000].

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

#include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
// write your code in C++11
//find the min number in the left side!
int min=2e6,maxprofit=0;
int n=A.size();
for(int i=0;i<n;i++)
{

if(A[i]<min)
min=A[i];
maxprofit=max(maxprofit,A[i]-min);
}
return maxprofit;

}

题目 3:MaxSliceSum

A non-empty zero-indexed array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The sum of a slice (P, Q) is the total of A[P] + A[P+1] + ... + A[Q].

Write a function:

int solution(vector<int> &A);

that, given an array A consisting of N integers, returns the maximum sum of any slice of A.

For example, given array A such that:
A[0] = 3 A[1] = 2 A[2] = -6A[3] = 4 A[4] = 0

the function should return 5 because:

(3, 4) is a slice of A that has sum 4,
(2, 2) is a slice of A that has sum −6,
(0, 1) is a slice of A that has sum 5,
no other slice of A has sum greater than (0, 1).

Assume that:

N is an integer within the range [1..1,000,000];
each element of array A is an integer within the range [−1,000,000..1,000,000];
the result will be an integer within the range [−2,147,483,648..2,147,483,647].

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

#include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
// write your code in C++11
int maxslicesum=-1e6-1,slicesum=-1e6-1;
int n=A.size();
for(int i=0;i<n;i++)
{
slicesum=max(slicesum+A[i],A[i]);
maxslicesum=max(maxslicesum,slicesum);
}
return maxslicesum;
}