# Use the algorithm in the book intro to algorithm has Memory Limit Exceeded problem

• Very weird...
on P72 of the MIT book intro to algorithm

• Could you please post the code here, with some explanation of the algorithm idea?

• This post is deleted!

• @ceresgu's previous comment. Please leave long code post as an answer, or it will be messy.

struct sumIndex{
int low;
int high;
int sumValue;
};

``````class Solution {

public:
sumIndex findMaxCross(vector<int> changes, int low, int mid, int high){
int leftSum= -std::numeric_limits<int>::max();
int sum=0;
int maxLeft;
for(int i=mid; i>=low; i--){
sum += changes[i];
if (sum>leftSum){
leftSum=sum;
int maxLeft=i;
}
}
int rightSum= -std::numeric_limits<int>::max();
sum=0;
int maxRight;
for (int j=mid+1;j<=high;j++){
sum +=changes[j];
if (sum>rightSum){
rightSum=sum;
int maxRight=j;

}
}
sumIndex cross;
cross.low=maxLeft;
cross.high=maxRight;
cross.sumValue=leftSum+rightSum;
return cross;
}

sumIndex findMax(vector<int> changes, int low, int high){
sumIndex result;
sumIndex left, right, cross;
if(low==high){
result.low=low;
result.high=high;
result.sumValue=changes[low];
}
else{
int mid =(low+high)/2;
left= findMax(changes, low, mid);
right=findMax(changes, mid+1, high);
cross=findMaxCross(changes,low, mid, high);

if (left.sumValue>right.sumValue && left.sumValue>cross.sumValue)
result=left;
else if(right.sumValue>left.sumValue && right.sumValue>cross.sumValue)
result=right;
else
result=cross;
}

return result;
}

int maxProfit(vector<int> &prices) {

vector<int> changes;

changes.push_back(0);
for (int i=1; i<prices.size(); i++)
changes.push_back(prices[i]-prices[i-1]);

return findMax(changes, 0, changes.size()-1).sumValue;
}
};
``````

it uses divide and conquer.
there are three situations.
the max is in the left half or right half or cross the mid.
and then recursively find the max

• ``````struct sumIndex{
int low;
int high;
int sumValue;
};

class Solution {

public:
sumIndex findMaxCross(vector<int> changes, int low, int mid, int high){
int leftSum= -std::numeric_limits<int>::max();
int sum=0;
int maxLeft;
for(int i=mid; i>=low; i--){
sum += changes[i];
if (sum>leftSum){
leftSum=sum;
int maxLeft=i;
}
}
int rightSum= -std::numeric_limits<int>::max();
sum=0;
int maxRight;
for (int j=mid+1;j<=high;j++){
sum +=changes[j];
if (sum>rightSum){
rightSum=sum;
int maxRight=j;

}
}
sumIndex cross;
cross.low=maxLeft;
cross.high=maxRight;
cross.sumValue=leftSum+rightSum;
return cross;
}

sumIndex findMax(vector<int> changes, int low, int high){
sumIndex result;
sumIndex left, right, cross;
if(low==high){
result.low=low;
result.high=high;
result.sumValue=changes[low];
}
else{
int mid =(low+high)/2;
left= findMax(changes, low, mid);
right=findMax(changes, mid+1, high);
cross=findMaxCross(changes,low, mid, high);

if (left.sumValue>right.sumValue && left.sumValue>cross.sumValue)
result=left;
else if(right.sumValue>left.sumValue && right.sumValue>cross.sumValue)
result=right;
else
result=cross;
}

return result;
}

int maxProfit(vector<int> &prices) {

vector<int> changes;

changes.push_back(0);
for (int i=1; i<prices.size(); i++)
changes.push_back(prices[i]-prices[i-1]);

return findMax(changes, 0, changes.size()-1).sumValue;
}
};``````

• I'm sure your book doesn't say this is the most efficient way to find the answer

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