UVa 108 - Maximum Sum

Summary
Given an n by n two-dimensional array arr ($$1 <= n <= 100$$) of arbitrary integers, find the maximum sub-array. Maximum sub-array is defined to be the sub-array whose sum of integer elements are the maximum possible.

Explanation

 * First, calculate the vertical prefix sum (cumulative sum) for all columns (an $$O(n^2)$$ algorithm).
 * Second, assume that the maximum sub-array will be between row a and row b, inclusive. There are only $$O(n^2)$$ a, b pairs such that $$a < b$$. Try each of them.
 * Since we already have the vertical prefix sum for all columns, the sum of elements in $$arr[a..b][c]$$ for column c can be computed in $$O(1)$$ time. This allows us to imagine each column sum as if it is a single element of a one-dimensional array across all columns (one dimensional array with one row and n columns).
 * There's an $$O(n)$$ algorithm to compute the maximum sub-array for a one-dimensional array, known as Kadane's Algorithm.
 * Applying the Kadane's algorithm inside each a and b combination gives the total complexity of $$O(n^3)$$.

/*************************************************
 * Amazon Interview Question:                    *
 * Problem: Maximum Sum Sub Matrix               *
 * Visit: http://programming-puzzles.blogspot.in/ *
 * https://gist.github.com/ramprasadgopal/5364342 *
 * Author : Ramprasad Gopalakrishnan             *
 * Language : C++                                *


 * 1) include
 * 2) include
 * 3) include

using namespace std;

void getNextInputSet; void printMatrix(int* matrix, int m1, int m2, int n1, int n2); int maximumSumSubArray(int array[], int size, int* maxStartRetVal, int* maxTailRetVal);

/* int inputMatrix[10][10] = { {2,-1,2,-1,4,-5},                     {2,8,2,-1,4,-5},                      {2,-1,2,-1,4,-5},                      {2,-1,2,-1,4,-5},                      {2,-1,2,-1,4,-5},                      {-2,-1,-2,-1,4,-5}                     }; int m = 6, n = 6;
 * input Matrix

/* int sumMatrix[10][10] = ;
 * sumMatrix:
 * sumMatrix[i][j] gives the sum of column j till the ith row.

int main { while(n > 0 && m >0) {

for(int i = 1; i<=m;i++) { for(int j=0; j<n; j++) { sumMatrix[i][j] = sumMatrix[i-1][j] + inputMatrix[i-1][j]; }        }

cout<<"SumMatrix:\n"; printMatrix(*sumMatrix,0,m,0,n-1);

int maxSum = sumMatrix[0][0], maxRowStart = 0, maxRowTail = 0, maxColStart =0, maxColTail = 0; cout<<"\n\nRij Matrix:\n"; for(int i=0; i<n; i++) { for(int j=i; jmaxSum) { maxSum = maxSumRij; maxRowStart = i;                    maxRowTail = j;                     maxColStart = maxStartForRij; maxColTail = maxTailForRij; }            }         }

cout<<"\n\nAnswer:\n"<<"maxSum:"< 0) { currentSum += array[currentTail]; } else { currentStart = currentTail; currentSum = array[currentTail]; }       if(currentSum > maxSum) { maxSum = currentSum; maxTail = currentTail; maxStart = currentStart; }   }    *maxStartRetVal = maxStart; *maxTailRetVal = maxTail; return maxSum; }
 * This solves the sub problem of maximumSumSubArray.
 * i.e. returns the sub array with the maximum sum.
 * refer: Kadane's algorithm.
 * refer: Kadane's algorithm.

/* void getNextInputSet { //get next set of inputs cout<<"\n\nEnter M N (0 0 to quit):"; cin>>m>>n; if(m>0 && n>0) { for(int i = 0; i>inputMatrix[i][j]; }   } }
 * Utility functions.

void printMatrix(int* matrix, int m1, int m2, int n1, int n2) { for(int i = 0; i<=m;i++) { for(int j=0; j<n; j++) { cout<<sumMatrix[i][j]<<" "; }        cout<<endl; } }

Input
4 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15 -16 5 1 -61 5126 612 6 41 6 7 2 -7 1 73 -62 678 1 7 -616136 61 -83 724 -151 6247 872 2517 8135 4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8  0 -2

[ Be careful! This doesn't seem a valid sample. Problem statement says "The numbers in the array will be in the range [-127, 127]."]

Output
-1 18589 15

[This algo will return 0 for the first example (a nil matrix) instead of -1]

Solution

 * C++: http://acm-solution.blogspot.co.uk/2012/05/acm-uva-108-maximum-sum.html