I am currently stuck with the classic longest increasing subsequence problem, but there is a slight twist to it. Instead of just finding the longest increasing subsequence, I need to find the largest sum of all increasing subsequences that are of length k.
I have the following pseudo code implemented:
input = [4,13,5,14] k = 2
n = size of input
opt = array of size n which stores the highest increasing subsequence sum up to this index
counts = array of size n which stores the amount of values in the subsequence up to this index
highestSum = -1
FOR i in range(0, n)
high = new data object(value = 0, sum = 0, count = 0)
FOR j in range(i-1, 0, -1)
IF high.sum < opt[j] AND opt[j] < opt[i] AND counts[j] < k
high.value = input[j]
high.sum = opt[j]
high.count = counts[j]
opt[i] = high.sum + input[i]
counts[i] = high.count + 1
IF counts[i] == k
highestSum = higher value between (highestSum, opt[i])
return highestSum
This dynamic programming approach works in most cases, but for the list I outlined above it does not return the optimal subsequence sum. The optimal subsequence sum with length 2 should be 27 (13-14), but 18 is returned (4-14). This is due to the opt and counts array looking like this:
k = 2
input: 0 4 13 5 14
opt: 0 4 17 9 18
counts: 0 1 2 2 2
Due to 13 already having a subsequence of 4-13, and thus its count value (2) is no longer less than k, 14 is unable to accept 13 as a correct subsequence due to its count value.
Are there any suggestions as to what I can change?
You'll need k+1 sorted data structures, one for each possible length of subsequence currently found.
Each structure contains, by the last entry in an optimal subsequence, the current sum. That is, we only care about a subsequence that can lead to the best possible solution. (Technical note. Of those that can lead to the best solution, pick the one whose positions are lexicographically first.) Which will be sorted by increasing last entry, and decreasing sum.
In pseudocode it works like this.
initialize optimal[0..k]
optimal[0][min(sequence) - 1] = 0 # empty set.
for entry in sequence:
for i in k..1:
entry_prev = biggest < entry in optimal[i-1]
if entry_prev is not None:
this_sum = optimal[i-1][entry_prev] + entry
entry_smaller = biggest <= entry in optimal[i-1]
if entry_smaller is None or optimal[i][entry_smaller] < this_sum:
delete (e, v) from optimal[i] where entry <= e and v <= this_sum
insert (entry, this_sum) into optimal[i]
return optimal[k][largest entry in optimal[k]]
But you need this kind of 2-d structure to keep track of what might happen from here.
The total memory needed is O(k n) and running time will be O(k n log(n)).
It is possible to also reconstruct the optimal subsequence, but that requires a more complex data structure.
Here is a working solution in C++ that runs in O(logn * n * k) time with O(n*k) space. I think you can not make it faster but let me know if you find a faster solution. This is a modification of the solution for from https://stackoverflow.com/questions/16402854/number-of-increasing-subsequences-of-length-k. The key difference here is that we keep track of the maximum sum for each subsequences of different legths instead of accumulating the number of subsequences and we are iterating from the back of the array (since for increasing subsequences that have length larger than k the best k-length subarray will be at the end).
An other trick is that we use the array sums to map index + length combinations to maximum sums.
maxSumIncreasingKLenSeqDP function is the simple dynamic programming solution with O(n * n * k) time complexity.
#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <limits.h>
using namespace std;
#include <random>
int maxSumIncreasingKLenSeq(int arr[], size_t n, int k){
// inverse compression: assign N-1, N-2, ... , 1 to smallest, ..., largest
size_t N = 1;
size_t compArr[n];
{
for(size_t i = 0; i<n; ++i)
compArr[i] = arr[i];
// descending order
sort(compArr, compArr + n, greater<int>());
unordered_map<int, size_t> compMap;
for(int val : compArr){
if(compMap.find(val) == compMap.end()){
compMap[val] = N;
++N;
}
}
for(size_t i = 0; i<n; ++i)
compArr[i] = compMap[arr[i]];
}
int sums[n * (k - 1) + n]; // key is combined from index and length by n * (length - 1) + index
for(size_t i = 0; i < n * (k - 1) + n; ++i)
sums[i] = -1;
for(size_t i = 0; i < n; ++i)
sums[i] = arr[i]; // i, 1
int BIT[N];
for(size_t len = 2; len <= k; ++len){
for(size_t i = 0; i<N; ++i)
BIT[i] = INT_MIN;
for(size_t i = 0; i < len - 1; ++i)
sums[n * (len - 1) + i] = INT_MIN;
for(int i = n - len; i >= 0; --i){
int val = sums[n * (len - 2) + i + 1]; // i + 1, len - 1
int idx = compArr[i + 1];
while(idx <= N){
BIT[idx] = max(val, BIT[idx]);
idx += (idx & (-idx));
}
// it does this:
//BIT[compArr[i + 1]] = sums[n * (len - 2) + i + 1];
idx = compArr[i] - 1;
int maxSum = INT_MIN;
while(idx > 0){
maxSum = max(BIT[idx], maxSum);
idx -= (idx & (-idx));
}
sums[n * (len - 1) + i] = maxSum;
// it does this:
//for(int j = 0; j < compArr[i]; ++j)
// sums[n * (len - 1) + i] = max(sums[n * (len - 1) + i], BIT[j]);
if(sums[n * (len - 1) + i] > INT_MIN)
sums[n * (len - 1) + i] += arr[i];
}
}
int maxSum = INT_MIN;
for(int i = n - k; i >= 0; --i)
maxSum = max(maxSum, sums[n * (k - 1) + i]); // i, k
return maxSum;
}
int maxSumIncreasingKLenSeqDP(int arr[], int n, int k){
int sums[n * (k - 1) + n]; // key is combined from index and length by n * (length - 1) + index
for(size_t i = 0; i < n; ++i)
sums[i] = arr[i]; // i, 1
for(int i = 2; i <= k; ++i)
sums[n * (i - 1) + n - 1] = INT_MIN; // n - 1, i
// moving backward since for increasing subsequences it will be the last k items
for(int i = n - 2; i >= 0; --i){
for(size_t len = 2; len <= k; ++len){
int idx = n * (len - 1) + i; // i, length
sums[idx] = INT_MIN;
for(int j = n - 1; j > i; --j){
if(arr[i] < arr[j])
sums[idx] = max(sums[idx], sums[n * (len - 2) + j]); // j, length - 1
}
if(sums[idx] > INT_MIN)
sums[idx] += arr[i];
}
}
int maxSum = INT_MIN;
for(int i = n - k; i >= 0; --i)
maxSum = max(maxSum, sums[n * (k - 1) + i]); // i, k
return maxSum;
}
int main(){
std::random_device dev;
std::mt19937 rng(dev());
std::uniform_int_distribution<std::mt19937::result_type> dist(1,10);
for(int len = 3; len < 10; ++len){
for(int i = 0; i < 10000; ++i){
int arr[100];
for(int n = 0; n < 100; ++n)
arr[n] = dist(rng);
int res = maxSumIncreasingKLenSeqDP(arr, 100, len);
int fastRes = maxSumIncreasingKLenSeq(arr, 100, len);
if(res != fastRes)
cout << "failed" << endl;
else
cout << "passed" << endl;
}
}
return 0;
}
I am trying to implement Knuth's optimal binary search tree which can run in O(n^2) time. I have the code which is running in O(n^3).
float P[N + 1] = {0, .13, .12, .15, .05, .12, .10, .08, .09, .03, .13};
float sum[N + 1] = {0, .13, .25, .40, .45, .57, .67, .75, .84, .87, 1.00};
float M[N + 1][N + 1];
int root[N + 1][N + 1];
int s, i, j;
float temp;
for (s = 0; s <= N; s++){
for (i = 0; i <= N; i++){
M[s][i] = 0;
root[s][i] = 0;
}
}
for (s = 2; s <= N; s++){
for (i = 1; i <= N - s + 1; i++){
M[s][i] = N;
for (j = i; j <= i + s - 1; j++){
temp = M[j - i][i] + M[i + s - j - 1][j + 1]+ sum[i + s - 1] - sum[i - 1] - P[j];
if (M[s][i] > temp){
M[s][i] = temp;
root[s][i] = j;
}
}
}
}
M is the array of costs. P is the probability of each node. I get some ideas from: Dynamic Programming: Why Knuth's improvement to Optimal Binary Search Tree O(n^2)?. In my case, I tried to modify the third loop from for (j = i; j <= i + s - 1; j++) to for (j = root[s+1][i]; j <= root[s][i-1]; j++). But it doesn't work. Could someone give me some clue on this problem?
You're supposed to be computing the costs of the optimal subtrees in non-decreasing order by size, so when you're filling in M[s][i] --- the minimum cost of a subtree of size s whose leftmost key has index i --- you haven't filled in M[s+1][i] or root[s+1][i] yet.
Cheers,
Travis
PS "j <= root[s][i-1]" isn't quite right either.
I need help with a problem problem c,Although I have a solution But i want to ask If we want to minimize the sweets then
after assigning these ans += b[i] * M sweets we can now say that this is minimum number .
But after that we are assigning again here again ans += g[i] - b[N]
Here b[N] is the max number in group of boys
Please help!
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int N, M;
long long g[maxn], b[maxn];
int main() {
scanf("%d%d", &N, &M);
for(int i = 1; i <= N; i ++)
scanf("%lld", &b[i]);
for(int i = 1; i <= M; i ++) {
scanf("%lld", &g[i]);
}
sort(b + 1, b + 1 + N);
sort(g + 1, g + 1 + M);
if(b[N] > g[1]) printf("-1\n");
else {
long long ans = 0;
for(int i = N; i >= 1; i --)
ans += b[i] * M;
for(int i = 2; i <= M; i ++)
ans += g[i] - b[N];
if(g[1] != b[N]) ans += g[1] - b[N - 1];
printf("%lld\n", ans);
}
return 0;
}
First we sort the girls and boys by min given/max get sweets.
There are 3 steps then:
1 - Each boys give their minimum sweets to each girls.
for(int i = N; i >= 1; i --)
ans += b[i] * M;
boys' conditions are fulfilled.
2 - Each girls except the first one get their maximum sweets from the last boy. (Who has the largest minimum, because this way we can minimize the sweets given in this step.) But he had already given his minimum to all of the girls in the previous step, so that's why he gives only the remaining g[i]-b[N] sweets here.
Notice, he cannot add the max to the first girl too, because he needs to add his minimum to somebody.
for(int i = 2; i <= M; i ++)
ans += g[i] - b[N];
girls' conditions (except the first one) are fulfilled.
3 - somebody needs to add the maximum sweets to the first girl. The optimum solution is if the boy with the 2nd largest minimum gives them to her. But it's only needed if the first girl's maximum is greater than the lasts' boy's minimum, otherwise her condition was already fulfilled.
if(g[1] != b[N])
ans += g[1] - b[N - 1];
firs't girl's condition is fulfilled.
Maybe it's more understandable in this format:
for(int i = 1; i < N; i ++)
ans += b[i] * M;
for(int i = 2; i <= M; i ++)
ans += g[i];
ans += b[N];
if(g[1] != b[N])
ans += g[1] - b[N - 1];
... however, there are a few corner-cases which are not handled in this algorithm: What is there is only 1 boy or/and 1 girl? (But that's only a few extra if/else branches.)
We are given a sequence of n positive integers, which I will denote as a0, a1, …, an-1. We are also given an integer k, and our task is to:
find a subsequence of length exactly k (denoted as b0, b1, …, bk-1), such that abs(b1 - b0) + abs(b2 - b1) + … + abs(bk-1 - bk-2) is maximal; and
output the sum (no need to output the entire subsequence).
I have been trying to solve this using a dynamic programming approach but all my efforts have been futile.
EDIT: k <= n. The elements in the sequence b must appear in the same order as they appear in a (otherwise, this would be solved by simply finding max, min, ... or min, max, ...).
Example input:
n = 10
k = 3
1 9 2 3 6 1 3 2 1 3
Output:
16 (the subsequence is 1 9 1, and abs(9 - 1) + abs(1 - 9) = 8 + 8 = 16)
Any help / hints would be greatly appreciated.
I managed to solve this problem. Here's the full code:
#include <stdio.h>
#include <stdlib.h>
int abs(int a)
{
return (a < 0) ? -a : a;
}
int solve(int *numbers, int N, int K)
{
int **storage = malloc(sizeof(int *) * N);
int i, j, k;
int result = 0;
for (i = 0; i < N; ++i)
*(storage + i) = calloc(K, sizeof(int));
// storage[i][j] keeps the optimal result where j + 1 elements are taken (K = j + 1) with numbers[i] appearing as the last element.
for (i = 1; i < N; ++i) {
for (j = 1; j < K; ++j) {
for (k = j - 1; k < i; ++k) {
if (storage[i][j] < storage[k][j - 1] + abs(numbers[k] - numbers[i]))
storage[i][j] = storage[k][j - 1] + abs(numbers[k] - numbers[i]);
if (j == K - 1 && result < storage[i][j])
result = storage[i][j];
}
}
}
for (i = 0; i < N; ++i)
free(*(storage + i));
free(storage);
return result;
}
int main()
{
int N, K;
scanf("%d %d", &N, &K);
int *numbers = malloc(sizeof(int) * N);
int i;
for (i = 0; i < N; ++i)
scanf("%d", numbers + i);
printf("%d\n", solve(numbers, N, K));
return 0;
}
The idea is simple (thanks to a friend of mine for hinting me at it). As mentioned in the comment, storage[i][j] keeps the optimal result where j + 1 elements are taken (K = j + 1) with numbers[i] appearing as the last element. Then, we simply try out each element appearing as the last one, taking each possible number of 1, 2, ..., K elements out of all. This solution works in O(k * n^2).
I first tried a 0-1 Knapsack approach where I kept the last element I had taken in each [i][j] index. This solution did not give a correct result in just a single test case, but it worked in O(k * n). I think I can see where it would yield a suboptimal solution, but if anyone is interested, I can post that code, too (it is rather messy though).
The code posted here passed on all test cases (if you can detect some possible errors, feel free to state them).
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.