The Painter's Partition Problem Part I
(http://leetcode.com/2011/04/the-painters-partition-problem.html)
You have to paint N boards of lenght {A0, A1, A2 ... AN-1}. There are K painters available and you are also given how much time a painter takes to paint 1 unit of board. You have to get this job done as soon as possible under the constraints that any painter will only paint continues sections of board, say board {2, 3, 4} or only board {1} or nothing but not board {2, 4, 5}.
We define M[n, k] as the optimum cost of a partition arrangement with n total blocks from the first block and k patitions, so
n n-1
M[n, k] = min { max { M[j, k-], ∑ Ai } }
j=1 i=j
The base cases are:
M[, k] = A0
n-1
M[n, ] = Σ Ai
i=0
Therefore, the brute force solution is:
int sum(int A[], int from, int to)
{
int total = ;
for (int i = from; i <= to; i++)
total += A[i];
return total;
} int partition(int A[], int n, int k)
{
if (n <= || k <= )
return -;
if (n == )
return A[];
if (k == )
return sum(A, , n-); int best = INT_MAX;
for (int j = ; j <= n; j++)
best = min(best, max(partition(A, j, k-), sum(A, j, n-))); return best;
}
It is exponential in run time complexity due to re-computation of the same values over and over again.
The DP solution:
int findMax(int A[], int n, int k)
{
int M[n+][k+];
int sum[n+];
for (int i = ; i <= n; i++)
sum[i] = sum[i-] + A[i-]; for (int i = ; i <= n; i++)
M[i][] = sum[i];
for (int i = ; i <= k; i++)
M[][k] = A[]; for (int i = ; i <= k; i++)
{
for (int j = ; j <= n; j++)
{
int best = INT_MAX;
for (int p = ; p <= j; p++)
{
best = min(best, max(M[p][i-], sum[j]-sum[p]));
}
M[j][i] = best;
}
}
return M[n][k];
}
Run time: O(kN*N), space complexity: O(kN).
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