Maximum Sum (二维)
108 - Maximum Sum
Quite similar to 10074 and 10667.
[edit]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.
[edit]Explanation
- First, calculate the vertical prefix sum (cumulative sum) for all columns (an O(n2) algorithm).
- Second, assume that the maximum sub-array will be between row a and row b, inclusive. There are only O(n2) 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(n3).
/*************************************************
* 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++ *
**************************************************/
#include<iostream>
#include<list>
#include <algorithm>
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);
/*
** input Matrix
**/
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;
/*
**sumMatrix:
** sumMatrix[i][j] gives the sum of column j till the ith row.
*/
int sumMatrix[10][10] = {{0}};
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; j<n; j++) {
/*r_ij matrix.
* k th element in the matrix is the sum of all elements in
* column k from row i to j in inputMatrix.
*/
int r_ij[10] = {0};
cout<<"R"<<i<<","<<j<<": ";
for(int k=0; k<m; k++) {
r_ij[k] = sumMatrix[j+1][k] - sumMatrix[i][k];
cout<<r_ij[k]<<" ";
}
cout<<endl;
int maxStartForRij, maxTailForRij, maxSumRij;
maxSumRij = maximumSumSubArray(r_ij, m, &maxStartForRij, &maxTailForRij);
//update with tne new maxSum
if(maxSumRij>maxSum) {
maxSum = maxSumRij;
maxRowStart = i;
maxRowTail = j;
maxColStart = maxStartForRij;
maxColTail = maxTailForRij;
}
}
}
cout<<"\n\nAnswer:\n"<<"maxSum:"<<maxSum<<endl;
printMatrix(*inputMatrix,maxRowStart,maxRowTail,maxColStart,maxColTail);
getNextInputSet();
}
}
/**
* This solves the sub problem of maximumSumSubArray.
* i.e. returns the sub array with the maximum sum.
*
* refer: Kadane's algorithm.
*/
int maximumSumSubArray(int array[], int size, int* maxStartRetVal, int* maxTailRetVal) {
int currentStart = 0, maxStart = 0, maxTail =0, currentSum = array[0], maxSum = array[0] ;
for(int currentTail = 1;currentTail < size; currentTail++) {
if(currentSum > 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;
}
/*
* Utility functions.
*/
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<m;i++) {
cout<<"Enter row " << i <<endl;
for(int j=0; j<n; j++)
cin>>inputMatrix[i][j];
}
}
}
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;
}
}
[edit]Notes
An O(n4) algorithm can get an AC on UVa Online Judge. But anything above that won't.
[edit]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]." ]
[edit]Output
-1 18589 15