## maximum subarray problem

A bit of a background: Kadane's algorithm is based on splitting up the set of possible solutions into mutually exclusive (disjoint) sets. We exploit the fact that any solution (i.e., any member of the set of solutions) will always have a last element {\displaystyle i} i (this is what is meant by "sum ending at position {\displaystyle i} i"). Thus, we simply have to examine, one by one, the set of solutions whose last element's index is {\displaystyle 1} 1, the set of solutions whose last element's index is {\displaystyle 2} 2, then {\displaystyle 3} 3, and so forth to {\displaystyle n} n. It turns out that this process can be carried out in linear time.
Kadane's algorithm begins with a simple inductive question: if we know the maximum subarray sum ending at position {\displaystyle i} i (call this {\displaystyle B{i}} B{i}), what is the maximum subarray sum ending at position {\displaystyle i+1} i+1 (equivalently, what is {\displaystyle B{i+1}} B{{i+1}})? The answer turns out to be relatively straightforward: either the maximum subarray sum ending at position {\displaystyle i+1} i+1 includes the maximum subarray sum ending at position {\displaystyle i} i as a prefix, or it doesn't (equivalently, {\displaystyle B{i+1}=max(A{i+1},A{i+1}+B{i})} {\displaystyle B{i+1}=max(A{i+1},A{i+1}+B{i})}, where {\displaystyle A{i+1}} A{i+1} is the element at index {\displaystyle i+1} i+1).
Thus, we can compute the maximum subarray sum ending at position {\displaystyle i} i for all positions {\displaystyle i} i by iterating once over the array. As we go, we simply keep track of the maximum sum we've ever seen. Thus, the problem can be solved with the following code, expressed here in Python:
``````1 def max_subarray(A):
2    max_ending_here = max_so_far = A[0]
3    for x in A[1:]:
4        max_ending_here = max(x, max_ending_here + x)
5        max_so_far = max(max_so_far, max_ending_here)
6    return max_so_far``````

## 121. Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

## Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5
max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

## Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.

## version 1:

``````class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size() == 0)return 0;
int maxCur = 0, maxSoFar = 0;
for(int i = 1; i<prices.size();++i){
maxCur = max(0, maxCur += prices[i] - prices[i - 1]);
maxSoFar = max(maxCur, maxSoFar);
}
return maxSoFar;
}
};``````

version 1 plus:

``````static const auto _____ = []()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
return nullptr;
}();

class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size() == 0)return 0;
int maxCur = 0, maxSoFar = 0;
for(int i = 1; i<prices.size();++i){
maxCur = max(0, maxCur += prices[i] - prices[i - 1]);
maxSoFar = max(maxCur, maxSoFar);
}
return maxSoFar;
}
};``````

1. 首先sync_with_stdio(false)是为了打断iostream输入输出到缓存，可以节约很多时间，使之与scanf相差无几。
2. tie是将两个stream绑定的函数，空参数的话返回当前的输出指针，即tie(0)与tie(nullptr)来解决cin与cout的绑定。