How do I go through all the diagonals in the matrix? There were similar questions, like this one, but they just calculate the sum. I need to perform certain operations going through all the diagonals, not just summing. To be clear, I need to traverse through it in way like in the picture:
I've came up with this solution, but it is awful:
for(int j = 0; j < m; j++) {
for(int i = 0; i < n && i + j < m; i++) {
cout << matrix[i][i + j] << " ";
}
cout << endl;
}
for(int i = 1; i < n; i++) {
for(int j = 0; j < m && i + j < n; j++) {
cout << matrix[j + i][j] << " ";
}
cout << endl;
}
For the matrix n*m I first go through every diagonal right from the main one, and then left from the main one, but this solution seems ugly to me.
You can use a simpler iteration at the cost of a modulo operation:
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cout << a[(i+j)%n][j] << " ";
}
cout << endl;
}
You need a first loop to find the starting point of each diagonal, and a second loop to follow the diagonal.
Here is a pseudo-code, assuming the point (0, 0) is the lower one.
(i_start, j_start) = (1, 0)
While (1)
If (i_start, j_start) out of bound: break
(i, j) = (i_start, j_start)
While (1)
If (i, j) out of bound: break
Write matrix[i][j]
i--, j++
End while
If (i_start not max) i_start++
Else j_start++
End while
Let's say we have given an array of digits, A, and a positive number, B. The problem is to generate all the possible B-digit numbers combined of A's elements.
For example, if A = [0,1,2,3] and B = 2, then the output must be,
[10,11,12,13,20,21,22,23,30,31,32,33]
Generate all possible combinations of 2 digit numbers by multiplying and adding the elements of the array in nested for loops.
Check if the generated numbers are greater than 10 to be a valid two digit number.
`
#include<iostream>
#include<cmath>
int main () {
int A[4] = {0,1,2,3}; int B = 2; int k;
for(size_t i = 0; i < sizeof(A)/sizeof(A[0]); i++)
{
for(size_t j = 0; j < sizeof(A)/sizeof(A[0]); j++)
{
k = (A[i] * pow(10, B-1) + j);
if(k / 10 > 0)
std::cout << k << '\n';
}
}
}
We are given a maze in which we need to visit as many rooms as possible. The specialty of the maze is that once you enter any room it will only lead you to rooms with a higher tag in the direction you move . B and C decide to move in opposite directions trying their luck to maximize the number of rooms they search .(They can start with any room , need not be the same)
We need to find out the maximum number of rooms that can be searched.
1. Access to any room with a higher tag is allowed, not just adjacent rooms or the next room with a higher tag.
2. Tags are unique.
So given the input:
12 11 10 1 2 3 4 13 6 7 8 5 9
the answer is 12: (1,2,3,4,6,7,8,9) for B and (5,10,11,12) for C.
I thought of solving this using longest increasing sub sequence first from right and then from left.And the count of unique elements in above two sub sequence would be the answer.
But my logic seems to fail,how can this be done?
My program below computes the maximum number of rooms searched. This has time complexity of O(n^3). I modified the DP algorithm for computing the longest increasing sequence available online to solve OP's problem. This also addresses OP's concerns on arrays like {1,4,6,2,5}. I rightly get the max value as 5 for the previous example. So, I used the idea from #BeyelerStudios that we need to compute the longest increasing subsequence from both left to right and from right to left. But, there is a caveat. If we compute the Left to right max sequence, the sequence from right to left should be on the remaining elements. Example:
For the array {1, 4, 6, 2, 5}. If the forward rooms selected are {1, 4, 5 }, then the reverse longest increasing sequence should be computed on the left out elements {6, 2}.
Below is my program:
#include <iostream>
using namespace std;
// compute the max increasing sequence from right to left.
int r2lRooms (int arr[], int n)
{
int dp[n];
int i =0, j = 0;
int max = 0;
for ( i = 0; i < n; i++ ) {
dp[i] = 1;
}
for (i = n-2; i >= 0; i--) {
for ( j = n-1; j > i; j-- ) {
if ( arr[i] > arr[j] && dp[i] < dp[j] + 1) {
dp[i] = dp[j] + 1;
}
}
}
for ( i = 0; i < n; i++ ) {
if ( max < dp[i] ) {
max = dp[i];
}
}
return max;
}
// compute max rooms.
int maxRooms( int arr[], int n )
{
int dp[n], revArray[n];
int i =0, j = 0, k = 0;
int currentMax = 0;
int forwardMax = 0, reverseMax = 0;
for ( i = 0; i < n; i++ ) {
dp[i] = 1;
}
// First case is that except for first elem, all others are in revArray
for (i=1; i < n; i++, k++) {
revArray[k] = arr[i];
}
reverseMax = r2lRooms (revArray, k);
forwardMax = 1;
if (currentMax < (forwardMax + reverseMax)) {
currentMax = forwardMax + reverseMax;
}
cout << "forwardmax revmax and currentmax are: " << forwardMax << " " << reverseMax << " " << currentMax << endl;
cout << endl;
for ( i = 1; i < n; i++ ) {
k = 0;
forwardMax = 1;
reverseMax = 0;
cout << "Forward elems for arr[" << i << "]=" << arr[i] << endl;
for ( j = 0; j < i; j++ ) {
if ( arr[i] > arr[j] && dp[i] < dp[j] + 1) {
dp[i] = dp[j] + 1;
forwardMax = dp[i];
cout << arr[j] << " ";
}
else {
// element was not in DP calculation, so put in revArray.
revArray[k] = arr[j];
k++;
}
}
// copy the remaining elements in revArray.
for ( j = i+1; j < n; j++ ) {
revArray[k] = arr[j];
k++;
}
cout << endl;
reverseMax = r2lRooms (revArray, k);
if (currentMax < (forwardMax + reverseMax)) {
currentMax = forwardMax + reverseMax;
}
cout << "forwardmax revmax and currentmax are: " << forwardMax << " " << reverseMax << " " << currentMax << endl;
cout << endl;
}
cout << " Max rooms searched " << currentMax << endl;
return currentMax;
}
int main (void) {
int arr[] = {12, 11, 10, 1, 2, 3, 4, 13, 6, 7, 8, 5, 9 };
int size = sizeof(arr) / sizeof(int);
cout << maxRooms (arr, size);
}
I think the trick is at the intersection, where B and C might share a value or there's options to go around that (say the sequence is 12 11 10 1 2 3 4 <3 5> 13 6 7 8 9 The extra numbers here adds 1 to the solution, but doesn't change the result for either longest increasing sub-sequences.
So the only problem is the one room in the middle, since on both side the values chosen diverge.
What I would do is this: do the longest subsequence in one direction, figure out a solution (any solution), take out the numbers in the solution and do the longest subsequence in the other direction. This way if there's a way around the crossing room in the middle the second pass will prefer it, unless that's the chosen number is really needed. To check for that do the same thing, but build the first subsequence in the opposite direction and the second one (after removing the solution) in the direction chosen initially.
Complexity remains O(N) but with a slightly higher constant factor.
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.