LintCode "Copy Books"

Classic DP. The initial intuitive O(k*n^2) solution is like this:

class Solution {
public:
/**
* @param pages: a vector of integers
* @param k: an integer
* @return: an integer
*/
int copyBooks(vector<int> &pages, int k) {
size_t n = pages.size();
if(k > n)
{
return *max_element(pages.begin(), pages.end());
}

//    Prefix Sums
vector<long long> psum(n);
for(int i = 0; i < n; i ++)
psum[i] = i == 0? pages[i] : (psum[i - 1] + pages[i]);

//  DP
vector<vector<long long>> dp(n + 1, vector<long long>(k + 1, INT_MAX));
for(int i = 1; i <= n; i ++)
dp[i][1] = psum[i - 1];

for(int i = 2; i <= k; i ++) // person
for(int b = i; b <= n; b ++) // book
for(int c = i-1; c < b; c ++) // prev book
{
long long last = dp[c][i - 1];
long long cur  = psum[b-1] - psum[c - 1];
dp[b][i] = min(dp[b][i], max(cur, last));
}

return dp
[k];
}
};

O(nk): http://sidbai.github.io/2015/07/25/Copy-Books/
Point above:

long long last = dp[c][i - 1];
long long cur  = psum[b-1] - psum[c - 1];
min(dp[b][i], max(cur, last));

dp[c][i-1] is mono-inc by c, cur is mono-dec. min(.., max(cur,last)) is V-like in 2D plane. So we can use 2-pointers to find the bottom of the V!

Or, binary search with O(nlg(sum/k)): https://github.com/kamyu104/LintCode/blob/master/C++/copy-books.cpp