I have n vectors with unique integers. I want to find all the common part between them. Is there any specific algorithm or data struct for this problem?
example:
std::vector<int> a = {0,1,2,3,4,5,6,7,8,9,10,11,12,13};
std::vector<int> b = {1,7,8,9,2,10,11,12};
std::vector<int> c = {4,9,8,7,0,1,2,3};
result:
ignore result with only one interge
7,8,9 between a and b
10,11,12 bewteen a and b
0,1,2,3 between a and c
if you want all common subarrays with a length greater than 1, then for each element from the first array iterate over all elements in the second array if you match two elements then go to the next element in the first and second array, and so on.
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (arr1[i] == arr2[j]) {
int ii = i, jj = j, cnt = 0;
std::vector<int> res;
res.push_back(arr1[ii]);
while (++ii < n and ++jj < m and arr1[ii] == arr2[jj])res.push_back(arr1[ii]);
if (res.size() > 1) {
for (auto x: res)std::cout << x << " ";
}
}
}
}
time complexity:O(n^3)
and this another way by LCS.
memset(dp, 0, sizeof dp);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
dp[i][j] = 0;
if (arr1[i] == arr2[j]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
}
std::cout << dp[i][j] << " ";
}
std::cout << "\n";
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (dp[i][j] > 1) {
for (int ii = i, jj = j, k = dp[i][j]; k; ii--, jj--, k--) {
std::cout << arr1[ii] << " ";
}
std::cout << "\n";
}
}
}
O(n^3)
It seems to me that you are looking for Longest Common Subsequence
These images are calculated by a diff like program which compares lines (unordered), like shortest edit distance
Blue lines : Deleted lines to come from left to right
Red lines : Changed lines
Green lines: Inserted lines
Lines without color are unchanged = longest common subsequence. Diff result looks pretty much the same as the given results.
Reference:
A Fast Algorithm for computing longest common subsequences
by James W. Hunt and Thomas G. Szymanski
from Communications of the ACM May 1977 Volume 20 no. 5
The question Maximal Square in https://leetcode.com/problems/maximal-square/description/ is easy to solve by DP. But how to solve the following up question:
Similar as Maximal Square question, but allows 0's inside a square, "inside" means the border of the square must be all 1.
For example, given the following matrix:
1 0 1 0 0
1 0 1 1 1
1 1 1 0 1
1 0 1 1 1
Return 9.
Update: Because the 3*3 matrix in the right bottom corner matches the requirement, the border must be all 1, and there can be 0 inside the square.
I thought up a O(n^3) algorithm: take maze[i][j] as the right bottom corner of the square if maze[i][j] == 1, enumerate the edge length of the square. If edge length is 3, consider whether maze[i - 2][j - 2], maze[i][j - 2], maze[i - 2][j], maze[i][j] forms a square with the numbers in each edge are all 1.
Is there any better algorithm?
Your problem can be solved in O (n * m) time and space complexity, where n is total rows and m is total columns in matrix. You may look at the code below where I have commented out to make it understandable.
Please, let me know if you have any doubt.
#include <bits/stdc++.h>
using namespace std;
void precalRowSum(vector< vector<int> >& grid, vector< vector<int> >&rowSum, int n, int m) {
// contiguous sum upto jth position in ith row
for (int i = 0; i < n; ++i) {
int sum = 0;
for (int j = 0; j < m; ++j) {
if (grid[i][j] == 1) {
sum++;
} else {
sum = 0;
}
rowSum[i][j] = sum;
}
}
}
void precalColSum(vector< vector<int> >& grid, vector< vector<int> >&colSum, int n, int m) {
// contiguous sum upto ith position in jth column
for (int j = 0; j < m; ++j) {
int sum = 0;
for (int i = 0; i < n; ++i) {
if (grid[i][j] == 1) {
sum++;
} else {
sum = 0;
}
colSum[i][j] = sum;
}
}
}
int solve(vector< vector<int> >& grid, int n, int m) {
vector< vector<int> >rowSum(n, vector<int>(m, 0));
vector< vector<int> >colSum(n, vector<int>(m, 0));
// calculate rowwise sum for 1
precalRowSum(grid, rowSum, n, m);
// calculate colwise sum for 1
precalColSum(grid, colSum, n, m);
vector< vector<int> >zerosHeight(n, vector<int>(m, 0));
int ans = 0;
for (int i = 0; i < (n - 1); ++i) {
for (int j = 0; j < m; ++j) {
zerosHeight[i][j] = ( grid[i][j] == 0 );
if (grid[i][j] == 0 && i > 0) {
zerosHeight[i][j] += zerosHeight[i - 1][j];
}
}
if (i == 0) continue;
// perform calculation on ith row
for (int j = 1; j < m; ) {
int height = zerosHeight[i][j];
if (!height) {
j++;
continue;
}
int cnt = 0;
while (j < m && height == zerosHeight[i][j]) {
j++;
cnt++;
}
if ( j == m) break;
if (cnt == height && (i - cnt) >= 0 ) {
// zeros are valid, now check validity for boundries
// Check validity of upper boundray, lower boundary, left boundary, right boundary respectively
if (rowSum[i - cnt][j] >= (cnt + 2) && rowSum[i + 1][j] >= (cnt + 2) &&
colSum[i + 1][j - cnt - 1] >= (cnt + 2) && colSum[i + 1][j] >= (cnt + 2) ){
ans = max(ans, (cnt + 2) * (cnt + 2) );
}
}
}
}
return ans;
}
int main() {
int n, m;
cin>>n>>m;
vector< vector<int> >grid;
for (int i = 0; i < n; ++i) {
vector<int>tmp;
for (int j = 0; j < m; ++j) {
int x;
cin>>x;
tmp.push_back(x);
}
grid.push_back(tmp);
}
cout<<endl;
cout<< solve(grid, n, m) <<endl;
return 0;
}
This is a follow-up to my previous question. I still find it very interesting problem and as there is one algorithm which deserves more attention I'm posting it here.
From Wikipedia: For the case that each xi is positive and bounded by the same constant, Pisinger found a linear time algorithm.
There is a different paper which seems to describe the same algorithm but it is a bit difficult to read for me so please - does anyone know how to translate the pseudo-code from page 4 (balsub) into working implementation?
Here are couple of pointers I collected so far:
http://www.diku.dk/~pisinger/95-6.ps (the paper)
https://stackoverflow.com/a/9952759/1037407
http://www.diku.dk/hjemmesider/ansatte/pisinger/codes.html
PS: I don't really insist on precisely this algorithm so if you know of any other similarly performant algorithm please feel free to suggest it bellow.
Edit
This is a Python version of the code posted bellow by oldboy:
class view(object):
def __init__(self, sequence, start):
self.sequence, self.start = sequence, start
def __getitem__(self, index):
return self.sequence[index + self.start]
def __setitem__(self, index, value):
self.sequence[index + self.start] = value
def balsub(w, c):
'''A balanced algorithm for Subset-sum problem by David Pisinger
w = weights, c = capacity of the knapsack'''
n = len(w)
assert n > 0
sum_w = 0
r = 0
for wj in w:
assert wj > 0
sum_w += wj
assert wj <= c
r = max(r, wj)
assert sum_w > c
b = 0
w_bar = 0
while w_bar + w[b] <= c:
w_bar += w[b]
b += 1
s = [[0] * 2 * r for i in range(n - b + 1)]
s_b_1 = view(s[0], r - 1)
for mu in range(-r + 1, 1):
s_b_1[mu] = -1
for mu in range(1, r + 1):
s_b_1[mu] = 0
s_b_1[w_bar - c] = b
for t in range(b, n):
s_t_1 = view(s[t - b], r - 1)
s_t = view(s[t - b + 1], r - 1)
for mu in range(-r + 1, r + 1):
s_t[mu] = s_t_1[mu]
for mu in range(-r + 1, 1):
mu_prime = mu + w[t]
s_t[mu_prime] = max(s_t[mu_prime], s_t_1[mu])
for mu in range(w[t], 0, -1):
for j in range(s_t[mu] - 1, s_t_1[mu] - 1, -1):
mu_prime = mu - w[j]
s_t[mu_prime] = max(s_t[mu_prime], j)
solved = False
z = 0
s_n_1 = view(s[n - b], r - 1)
while z >= -r + 1:
if s_n_1[z] >= 0:
solved = True
break
z -= 1
if solved:
print c + z
print n
x = [False] * n
for j in range(0, b):
x[j] = True
for t in range(n - 1, b - 1, -1):
s_t = view(s[t - b + 1], r - 1)
s_t_1 = view(s[t - b], r - 1)
while True:
j = s_t[z]
assert j >= 0
z_unprime = z + w[j]
if z_unprime > r or j >= s_t[z_unprime]:
break
z = z_unprime
x[j] = False
z_unprime = z - w[t]
if z_unprime >= -r + 1 and s_t_1[z_unprime] >= s_t[z]:
z = z_unprime
x[t] = True
for j in range(n):
print x[j], w[j]
// Input:
// c (capacity of the knapsack)
// n (number of items)
// w_1 (weight of item 1)
// ...
// w_n (weight of item n)
//
// Output:
// z (optimal solution)
// n
// x_1 (indicator for item 1)
// ...
// x_n (indicator for item n)
#include <algorithm>
#include <cassert>
#include <iostream>
#include <vector>
using namespace std;
int main() {
int c = 0;
cin >> c;
int n = 0;
cin >> n;
assert(n > 0);
vector<int> w(n);
int sum_w = 0;
int r = 0;
for (int j = 0; j < n; ++j) {
cin >> w[j];
assert(w[j] > 0);
sum_w += w[j];
assert(w[j] <= c);
r = max(r, w[j]);
}
assert(sum_w > c);
int b;
int w_bar = 0;
for (b = 0; w_bar + w[b] <= c; ++b) {
w_bar += w[b];
}
vector<vector<int> > s(n - b + 1, vector<int>(2 * r));
vector<int>::iterator s_b_1 = s[0].begin() + (r - 1);
for (int mu = -r + 1; mu <= 0; ++mu) {
s_b_1[mu] = -1;
}
for (int mu = 1; mu <= r; ++mu) {
s_b_1[mu] = 0;
}
s_b_1[w_bar - c] = b;
for (int t = b; t < n; ++t) {
vector<int>::const_iterator s_t_1 = s[t - b].begin() + (r - 1);
vector<int>::iterator s_t = s[t - b + 1].begin() + (r - 1);
for (int mu = -r + 1; mu <= r; ++mu) {
s_t[mu] = s_t_1[mu];
}
for (int mu = -r + 1; mu <= 0; ++mu) {
int mu_prime = mu + w[t];
s_t[mu_prime] = max(s_t[mu_prime], s_t_1[mu]);
}
for (int mu = w[t]; mu >= 1; --mu) {
for (int j = s_t[mu] - 1; j >= s_t_1[mu]; --j) {
int mu_prime = mu - w[j];
s_t[mu_prime] = max(s_t[mu_prime], j);
}
}
}
bool solved = false;
int z;
vector<int>::const_iterator s_n_1 = s[n - b].begin() + (r - 1);
for (z = 0; z >= -r + 1; --z) {
if (s_n_1[z] >= 0) {
solved = true;
break;
}
}
if (solved) {
cout << c + z << '\n' << n << '\n';
vector<bool> x(n, false);
for (int j = 0; j < b; ++j) x[j] = true;
for (int t = n - 1; t >= b; --t) {
vector<int>::const_iterator s_t = s[t - b + 1].begin() + (r - 1);
vector<int>::const_iterator s_t_1 = s[t - b].begin() + (r - 1);
while (true) {
int j = s_t[z];
assert(j >= 0);
int z_unprime = z + w[j];
if (z_unprime > r || j >= s_t[z_unprime]) break;
z = z_unprime;
x[j] = false;
}
int z_unprime = z - w[t];
if (z_unprime >= -r + 1 && s_t_1[z_unprime] >= s_t[z]) {
z = z_unprime;
x[t] = true;
}
}
for (int j = 0; j < n; ++j) {
cout << x[j] << '\n';
}
}
}
great code man, but it sometimes crashed in this codeblock
for (mu = w[t]; mu >= 1; --mu)
{
for (int j = s_t[mu] - 1; j >= s_t_1[mu]; --j)
{
if (j >= w.size())
{ // !!! PROBLEM !!!
}
int mu_prime = mu - w[j];
s_t[mu_prime] = max(s_t[mu_prime], j);
}
}
...
Input: A 2-dimensional array NxN - Matrix - with positive and negative elements.Output: A submatrix of any size such that its summation is the maximum among all possible submatrices.
Requirement: Algorithm complexity to be of O(N^3)
History: With the help of the Algorithmist, Larry and a modification of Kadane's Algorithm, i managed to solve the problem partly which is determining the summation only - below in Java.
Thanks to Ernesto who managed to solve the rest of the problem which is determining the boundaries of the matrix i.e. top-left, bottom-right corners - below in Ruby.
Here's an explanation to go with the posted code. There are two key tricks to make this work efficiently: (I) Kadane's algorithm and (II) using prefix sums. You also need to (III) apply the tricks to the matrix.
Part I: Kadane's algorithm
Kadane's algorithm is a way to find a contiguous subsequence with maximum sum. Let's start with a brute force approach for finding the max contiguous subsequence and then consider optimizing it to get Kadane's algorithm.
Suppose you have the sequence:
-1, 2, 3, -2
For the brute force approach, walk along the sequence generating all possible subsequences as shown below. Considering all possibilities, we can start, extend, or end a list with each step.
At index 0, we consider appending the -1
-1, 2, 3, -2
^
Possible subsequences:
-1 [sum -1]
At index 1, we consider appending the 2
-1, 2, 3, -2
^
Possible subsequences:
-1 (end) [sum -1]
-1, 2 [sum 1]
2 [sum 2]
At index 2, we consider appending the 3
-1, 2, 3, -2
^
Possible subsequences:
-1, (end) [sum -1]
-1, 2 (end) [sum -1]
2 (end) [sum 2]
-1, 2, 3 [sum 4]
2, 3 [sum 5]
3 [sum 3]
At index 3, we consider appending the -2
-1, 2, 3, -2
^
Possible subsequences:
-1, (end) [sum -1]
-1, 2 (end) [sum 1]
2 (end) [sum 2]
-1, 2 3 (end) [sum 4]
2, 3 (end) [sum 5]
3, (end) [sum 3]
-1, 2, 3, -2 [sum 2]
2, 3, -2 [sum 3]
3, -2 [sum 1]
-2 [sum -2]
For this brute force approach, we finally pick the list with the best sum, (2, 3), and that's the answer. However, to make this efficient, consider that you really don't need to keep every one of the lists. Out of the lists that have not ended, you only need to keep the best one, the others cannot do any better. Out of the lists that have ended, you only might need to keep the best one, and only if it's better than ones that have not ended.
So, you can keep track of what you need with just a position array and a sum array. The position array is defined like this: position[r] = s keeps track of the list which ends at r and starts at s. And, sum[r] gives a sum for the subsequence ending at index r. This is optimized approach is Kadane's algorithm.
Running through the example again keeping track of our progress this way:
At index 0, we consider appending the -1
-1, 2, 3, -2
^
We start a new subsequence for the first element.
position[0] = 0
sum[0] = -1
At index 1, we consider appending the 2
-1, 2, 3, -2
^
We choose to start a new subsequence because that gives a higher sum than extending.
position[0] = 0 sum[0] = -1
position[1] = 1 sum[1] = 2
At index 2, we consider appending the 3
-1, 2, 3, -2
^
We choose to extend a subsequence because that gives a higher sum than starting a new one.
position[0] = 0 sum[0] = -1
position[1] = 1 sum[1] = 2
position[2] = 1 sum[2] = 5
Again, we choose to extend because that gives a higher sum that starting a new one.
-1, 2, 3, -2
^
position[0] = 0 sum[0] = -1
position[1] = 1 sum[1] = 2
position[2] = 1 sum[2] = 5
positions[3] = 3 sum[3] = 3
Again, the best sum is 5 and the list is from index 1 to index 2, which is (2, 3).
Part II: Prefix sums
We want to have a way to compute the sum along a row, for any start point to any endpoint. I want to compute that sum in O(1) time rather than just adding, which takes O(m) time where m is the number of elements in the sum. With some precomputing, this can be achieved. Here's how. Suppose you have a matrix:
a d g
b e h
c f i
You can precompute this matrix:
a d g
a+b d+e g+h
a+b+c d+e+f g+h+i
Once that is done you can get the sum running along any column from any start to endpoint in the column just by subtracting two values.
Part III: Bringing tricks together to find the max submatrix
Assume that you know the top and bottom row of the max submatrix. You could do this:
Ignore rows above your top row and ignore rows below your bottom
row.
With what matrix remains, consider the using sum of each column to
form a sequence (sort of like a row that represents multiple rows).
(You can compute any element of this sequence rapidly with the prefix
sums approach.)
Use Kadane's approach to figure out best subsequence in this
sequence. The indexes you get will tell you the left and right
positions of the best submatrix.
Now, what about actually figuring out the top and bottom row? Just try all possibilities. Try putting the top anywhere you can and putting the bottom anywhere you can, and run the Kadane-base procedure described previously for every possibility. When you find a max, you keep track of the top and bottom position.
Finding the row and column takes O(M^2) where M is the number of rows. Finding the column takes O(N) time where N is the number of columns. So total time is O(M^2 * N). And, if M=N, the time required is O(N^3).
About recovering the actual submatrix, and not just the maximum sum, here's what I got. Sorry I do not have time to translate my code to your java version, so I'm posting my Ruby code with some comments in the key parts
def max_contiguous_submatrix_n3(m)
rows = m.count
cols = rows ? m.first.count : 0
vps = Array.new(rows)
for i in 0..rows
vps[i] = Array.new(cols, 0)
end
for j in 0...cols
vps[0][j] = m[0][j]
for i in 1...rows
vps[i][j] = vps[i-1][j] + m[i][j]
end
end
max = [m[0][0],0,0,0,0] # this is the result, stores [max,top,left,bottom,right]
# these arrays are used over Kadane
sum = Array.new(cols) # obvious sum array used in Kadane
pos = Array.new(cols) # keeps track of the beginning position for the max subseq ending in j
for i in 0...rows
for k in i...rows
# Kadane over all columns with the i..k rows
sum.fill(0) # clean both the sum and pos arrays for the upcoming Kadane
pos.fill(0)
local_max = 0 # we keep track of the position of the max value over each Kadane's execution
# notice that we do not keep track of the max value, but only its position
sum[0] = vps[k][0] - (i==0 ? 0 : vps[i-1][0])
for j in 1...cols
value = vps[k][j] - (i==0 ? 0 : vps[i-1][j])
if sum[j-1] > 0
sum[j] = sum[j-1] + value
pos[j] = pos[j-1]
else
sum[j] = value
pos[j] = j
end
if sum[j] > sum[local_max]
local_max = j
end
end
# Kadane ends here
# Here's the key thing
# If the max value obtained over the past Kadane's execution is larger than
# the current maximum, then update the max array with sum and bounds
if sum[local_max] > max[0]
# sum[local_max] is the new max value
# the corresponding submatrix goes from rows i..k.
# and from columns pos[local_max]..local_max
# the array below contains [max_sum,top,left,bottom,right]
max = [sum[local_max], i, pos[local_max], k, local_max]
end
end
end
return max # return the array with [max_sum,top,left,bottom,right]
end
Some notes for clarification:
I use an array to store all the values pertaining to the result for convenience. You can just use five standalone variables: max, top, left, bottom, right. It's just easier to assign in one line to the array and then the subroutine returns the array with all the needed information.
If you copy and paste this code in a text-highlight-enabled editor with Ruby support you'll obviously understand it better. Hope this helps!
There are already plenty of answers, but here is another Java implementation I wrote. It compares 3 solutions:
Naïve (brute force) - O(n^6) time
The obvious DP solution - O(n^4) time and O(n^3) space
The more clever DP solution based on Kadane's algorithm - O(n^3) time and O(n^2) space
There are sample runs for n = 10 thru n = 70 in increments of 10 with a nice output comparing run time and space requirements.
Code:
public class MaxSubarray2D {
static int LENGTH;
final static int MAX_VAL = 10;
public static void main(String[] args) {
for (int i = 10; i <= 70; i += 10) {
LENGTH = i;
int[][] a = new int[LENGTH][LENGTH];
for (int row = 0; row < LENGTH; row++) {
for (int col = 0; col < LENGTH; col++) {
a[row][col] = (int) (Math.random() * (MAX_VAL + 1));
if (Math.random() > 0.5D) {
a[row][col] = -a[row][col];
}
//System.out.printf("%4d", a[row][col]);
}
//System.out.println();
}
System.out.println("N = " + LENGTH);
System.out.println("-------");
long start, end;
start = System.currentTimeMillis();
naiveSolution(a);
end = System.currentTimeMillis();
System.out.println(" run time: " + (end - start) + " ms no auxiliary space requirements");
start = System.currentTimeMillis();
dynamicProgammingSolution(a);
end = System.currentTimeMillis();
System.out.println(" run time: " + (end - start) + " ms requires auxiliary space for "
+ ((int) Math.pow(LENGTH, 4)) + " integers");
start = System.currentTimeMillis();
kadane2D(a);
end = System.currentTimeMillis();
System.out.println(" run time: " + (end - start) + " ms requires auxiliary space for " +
+ ((int) Math.pow(LENGTH, 2)) + " integers");
System.out.println();
System.out.println();
}
}
// O(N^2) !!!
public static void kadane2D(int[][] a) {
int[][] s = new int[LENGTH + 1][LENGTH]; // [ending row][sum from row zero to ending row] (rows 1-indexed!)
for (int r = 0; r < LENGTH + 1; r++) {
for (int c = 0; c < LENGTH; c++) {
s[r][c] = 0;
}
}
for (int r = 1; r < LENGTH + 1; r++) {
for (int c = 0; c < LENGTH; c++) {
s[r][c] = s[r - 1][c] + a[r - 1][c];
}
}
int maxSum = Integer.MIN_VALUE;
int maxRowStart = -1;
int maxColStart = -1;
int maxRowEnd = -1;
int maxColEnd = -1;
for (int r1 = 1; r1 < LENGTH + 1; r1++) { // rows 1-indexed!
for (int r2 = r1; r2 < LENGTH + 1; r2++) { // rows 1-indexed!
int[] s1 = new int[LENGTH];
for (int c = 0; c < LENGTH; c++) {
s1[c] = s[r2][c] - s[r1 - 1][c];
}
int max = 0;
int c1 = 0;
for (int c = 0; c < LENGTH; c++) {
max = s1[c] + max;
if (max <= 0) {
max = 0;
c1 = c + 1;
}
if (max > maxSum) {
maxSum = max;
maxRowStart = r1 - 1;
maxColStart = c1;
maxRowEnd = r2 - 1;
maxColEnd = c;
}
}
}
}
System.out.print("KADANE SOLUTION | Max sum: " + maxSum);
System.out.print(" Start: (" + maxRowStart + ", " + maxColStart +
") End: (" + maxRowEnd + ", " + maxColEnd + ")");
}
// O(N^4) !!!
public static void dynamicProgammingSolution(int[][] a) {
int[][][][] dynTable = new int[LENGTH][LENGTH][LENGTH + 1][LENGTH + 1]; // [row][col][height][width]
int maxSum = Integer.MIN_VALUE;
int maxRowStart = -1;
int maxColStart = -1;
int maxRowEnd = -1;
int maxColEnd = -1;
for (int r = 0; r < LENGTH; r++) {
for (int c = 0; c < LENGTH; c++) {
for (int h = 0; h < LENGTH + 1; h++) {
for (int w = 0; w < LENGTH + 1; w++) {
dynTable[r][c][h][w] = 0;
}
}
}
}
for (int r = 0; r < LENGTH; r++) {
for (int c = 0; c < LENGTH; c++) {
for (int h = 1; h <= LENGTH - r; h++) {
int rowTotal = 0;
for (int w = 1; w <= LENGTH - c; w++) {
rowTotal += a[r + h - 1][c + w - 1];
dynTable[r][c][h][w] = rowTotal + dynTable[r][c][h - 1][w];
}
}
}
}
for (int r = 0; r < LENGTH; r++) {
for (int c = 0; c < LENGTH; c++) {
for (int h = 0; h < LENGTH + 1; h++) {
for (int w = 0; w < LENGTH + 1; w++) {
if (dynTable[r][c][h][w] > maxSum) {
maxSum = dynTable[r][c][h][w];
maxRowStart = r;
maxColStart = c;
maxRowEnd = r + h - 1;
maxColEnd = c + w - 1;
}
}
}
}
}
System.out.print(" DP SOLUTION | Max sum: " + maxSum);
System.out.print(" Start: (" + maxRowStart + ", " + maxColStart +
") End: (" + maxRowEnd + ", " + maxColEnd + ")");
}
// O(N^6) !!!
public static void naiveSolution(int[][] a) {
int maxSum = Integer.MIN_VALUE;
int maxRowStart = -1;
int maxColStart = -1;
int maxRowEnd = -1;
int maxColEnd = -1;
for (int rowStart = 0; rowStart < LENGTH; rowStart++) {
for (int colStart = 0; colStart < LENGTH; colStart++) {
for (int rowEnd = 0; rowEnd < LENGTH; rowEnd++) {
for (int colEnd = 0; colEnd < LENGTH; colEnd++) {
int sum = 0;
for (int row = rowStart; row <= rowEnd; row++) {
for (int col = colStart; col <= colEnd; col++) {
sum += a[row][col];
}
}
if (sum > maxSum) {
maxSum = sum;
maxRowStart = rowStart;
maxColStart = colStart;
maxRowEnd = rowEnd;
maxColEnd = colEnd;
}
}
}
}
}
System.out.print(" NAIVE SOLUTION | Max sum: " + maxSum);
System.out.print(" Start: (" + maxRowStart + ", " + maxColStart +
") End: (" + maxRowEnd + ", " + maxColEnd + ")");
}
}
Here is a Java version of Ernesto implementation with some modifications:
public int[][] findMaximumSubMatrix(int[][] matrix){
int dim = matrix.length;
//computing the vertical prefix sum for columns
int[][] ps = new int[dim][dim];
for (int i = 0; i < dim; i++) {
for (int j = 0; j < dim; j++) {
if (j == 0) {
ps[j][i] = matrix[j][i];
} else {
ps[j][i] = matrix[j][i] + ps[j - 1][i];
}
}
}
int maxSum = matrix[0][0];
int top = 0, left = 0, bottom = 0, right = 0;
//Auxiliary variables
int[] sum = new int[dim];
int[] pos = new int[dim];
int localMax;
for (int i = 0; i < dim; i++) {
for (int k = i; k < dim; k++) {
// Kadane over all columns with the i..k rows
reset(sum);
reset(pos);
localMax = 0;
//we keep track of the position of the max value over each Kadane's execution
// notice that we do not keep track of the max value, but only its position
sum[0] = ps[k][0] - (i==0 ? 0 : ps[i-1][0]);
for (int j = 1; j < dim; j++) {
if (sum[j-1] > 0){
sum[j] = sum[j-1] + ps[k][j] - (i==0 ? 0 : ps[i-1][j]);
pos[j] = pos[j-1];
}else{
sum[j] = ps[k][j] - (i==0 ? 0 : ps[i-1][j]);
pos[j] = j;
}
if (sum[j] > sum[localMax]){
localMax = j;
}
}//Kadane ends here
if (sum[localMax] > maxSum){
/* sum[localMax] is the new max value
the corresponding submatrix goes from rows i..k.
and from columns pos[localMax]..localMax
*/
maxSum = sum[localMax];
top = i;
left = pos[localMax];
bottom = k;
right = localMax;
}
}
}
System.out.println("Max SubMatrix determinant = " + maxSum);
//composing the required matrix
int[][] output = new int[bottom - top + 1][right - left + 1];
for(int i = top, k = 0; i <= bottom; i++, k++){
for(int j = left, l = 0; j <= right ; j++, l++){
output[k][l] = matrix[i][j];
}
}
return output;
}
private void reset(int[] a) {
for (int index = 0; index < a.length; index++) {
a[index] = 0;
}
}
With the help of the Algorithmist and Larry and a modification of Kadane's Algorithm, here is my solution:
int dim = matrix.length;
//computing the vertical prefix sum for columns
int[][] ps = new int[dim][dim];
for (int i = 0; i < dim; i++) {
for (int j = 0; j < dim; j++) {
if (j == 0) {
ps[j][i] = matrix[j][i];
} else {
ps[j][i] = matrix[j][i] + ps[j - 1][i];
}
}
}
int maxSoFar = 0;
int min , subMatrix;
//iterate over the possible combinations applying Kadane's Alg.
for (int i = 0; i < dim; i++) {
for (int j = i; j < dim; j++) {
min = 0;
subMatrix = 0;
for (int k = 0; k < dim; k++) {
if (i == 0) {
subMatrix += ps[j][k];
} else {
subMatrix += ps[j][k] - ps[i - 1 ][k];
}
if(subMatrix < min){
min = subMatrix;
}
if((subMatrix - min) > maxSoFar){
maxSoFar = subMatrix - min;
}
}
}
}
The only thing left is to determine the submatrix elements, i.e: the top left and the bottom right corner of the submatrix. Anyone suggestion?
this is my implementation of 2D Kadane algorithm. I think it is more clear. The concept is based on just kadane algorithm. The first and second loop of the main part (that is in the bottom of the code) is to pick every combination of the rows and 3rd loop is to use 1D kadane algorithm by every following column sum (that can be computed in const time because of preprocessing of matrix by subtracting values from two picked (from combintation) rows). Here is the code:
int [][] m = {
{1,-5,-5},
{1,3,-5},
{1,3,-5}
};
int N = m.length;
// summing columns to be able to count sum between two rows in some column in const time
for (int i=0; i<N; ++i)
m[0][i] = m[0][i];
for (int j=1; j<N; ++j)
for (int i=0; i<N; ++i)
m[j][i] = m[j][i] + m[j-1][i];
int total_max = 0, sum;
for (int i=0; i<N; ++i) {
for (int k=i; k<N; ++k) { //for each combination of rows
sum = 0;
for (int j=0; j<N; j++) { //kadane algorithm for every column
sum += i==0 ? m[k][j] : m[k][j] - m[i-1][j]; //for first upper row is exception
total_max = Math.max(sum, total_max);
}
}
}
System.out.println(total_max);
I am going to post an answer here and can add actual c++ code if it is requested because I had recently worked through this. Some rumors of a divide and conqueror that can solve this in O(N^2) are out there but I haven't seen any code to support this. In my experience the following is what I have found.
O(i^3j^3) -- naive brute force method
o(i^2j^2) -- dynamic programming with memoization
O(i^2j) -- using max contiguous sub sequence for an array
if ( i == j )
O(n^6) -- naive
O(n^4) -- dynamic programming
O(n^3) -- max contiguous sub sequence
Have a look at JAMA package; I believe it will make your life easier.
Here is the C# solution. Ref: http://www.algorithmist.com/index.php/UVa_108
public static MaxSumMatrix FindMaxSumSubmatrix(int[,] inMtrx)
{
MaxSumMatrix maxSumMtrx = new MaxSumMatrix();
// Step 1. Create SumMatrix - do the cumulative columnar summation
// S[i,j] = S[i-1,j]+ inMtrx[i-1,j];
int m = inMtrx.GetUpperBound(0) + 2;
int n = inMtrx.GetUpperBound(1)+1;
int[,] sumMatrix = new int[m, n];
for (int i = 1; i < m; i++)
{
for (int j = 0; j < n; j++)
{
sumMatrix[i, j] = sumMatrix[i - 1, j] + inMtrx[i - 1, j];
}
}
PrintMatrix(sumMatrix);
// Step 2. Create rowSpans starting each rowIdx. For these row spans, create a 1-D array r_ij
for (int x = 0; x < n; x++)
{
for (int y = x; y < n; y++)
{
int[] r_ij = new int[n];
for (int k = 0; k < n; k++)
{
r_ij[k] = sumMatrix[y + 1,k] - sumMatrix[x, k];
}
// Step 3. Find MaxSubarray of this r_ij. If the sum is greater than the last recorded sum =>
// capture Sum, colStartIdx, ColEndIdx.
// capture current x as rowTopIdx, y as rowBottomIdx.
MaxSum currMaxSum = KadanesAlgo.FindMaxSumSubarray(r_ij);
if (currMaxSum.maxSum > maxSumMtrx.sum)
{
maxSumMtrx.sum = currMaxSum.maxSum;
maxSumMtrx.colStart = currMaxSum.maxStartIdx;
maxSumMtrx.colEnd = currMaxSum.maxEndIdx;
maxSumMtrx.rowStart = x;
maxSumMtrx.rowEnd = y;
}
}
}
return maxSumMtrx;
}
public static void PrintMatrix(int[,] matrix)
{
int endRow = matrix.GetUpperBound(0);
int endCol = matrix.GetUpperBound(1);
PrintMatrix(matrix, 0, endRow, 0, endCol);
}
public static void PrintMatrix(int[,] matrix, int startRow, int endRow, int startCol, int endCol)
{
StringBuilder sb = new StringBuilder();
for (int i = startRow; i <= endRow; i++)
{
sb.Append(Environment.NewLine);
for (int j = startCol; j <= endCol; j++)
{
sb.Append(string.Format("{0} ", matrix[i,j]));
}
}
Console.WriteLine(sb.ToString());
}
// Given an NxN matrix of positive and negative integers, write code to find the sub-matrix with the largest possible sum
public static MaxSum FindMaxSumSubarray(int[] inArr)
{
int currMax = 0;
int currStartIndex = 0;
// initialize maxSum to -infinity, maxStart and maxEnd idx to 0.
MaxSum mx = new MaxSum(int.MinValue, 0, 0);
// travers through the array
for (int currEndIndex = 0; currEndIndex < inArr.Length; currEndIndex++)
{
// add element value to the current max.
currMax += inArr[currEndIndex];
// if current max is more that the last maxSum calculated, set the maxSum and its idx
if (currMax > mx.maxSum)
{
mx.maxSum = currMax;
mx.maxStartIdx = currStartIndex;
mx.maxEndIdx = currEndIndex;
}
if (currMax < 0) // if currMax is -ve, change it back to 0
{
currMax = 0;
currStartIndex = currEndIndex + 1;
}
}
return mx;
}
struct MaxSum
{
public int maxSum;
public int maxStartIdx;
public int maxEndIdx;
public MaxSum(int mxSum, int mxStart, int mxEnd)
{
this.maxSum = mxSum;
this.maxStartIdx = mxStart;
this.maxEndIdx = mxEnd;
}
}
class MaxSumMatrix
{
public int sum = int.MinValue;
public int rowStart = -1;
public int rowEnd = -1;
public int colStart = -1;
public int colEnd = -1;
}
Here is my solution. It's O(n^3) in time and O(n^2) space.
https://gist.github.com/toliuweijing/6097144
// 0th O(n) on all candidate bottoms #B.
// 1th O(n) on candidate tops #T.
// 2th O(n) on finding the maximum #left/#right match.
int maxRect(vector<vector<int> >& mat) {
int n = mat.size();
vector<vector<int> >& colSum = mat;
for (int i = 1 ; i < n ; ++i)
for (int j = 0 ; j < n ; ++j)
colSum[i][j] += colSum[i-1][j];
int optrect = 0;
for (int b = 0 ; b < n ; ++b) {
for (int t = 0 ; t <= b ; ++t) {
int minLeft = 0;
int rowSum[n];
for (int i = 0 ; i < n ; ++i) {
int col = t == 0 ? colSum[b][i] : colSum[b][i] - colSum[t-1][i];
rowSum[i] = i == 0? col : col + rowSum[i-1];
optrect = max(optrect, rowSum[i] - minLeft);
minLeft = min(minLeft, rowSum[i]);
}
}
}
return optrect;
}
I would just parse the NxN array removing the -ves whatever remains is the highest sum of a sub matrix.
The question doesn't say you have to leave the original matrix intact or that the order matters.