Amazon interview question:
Given a 2-dimensional array with arbitrary sizes and contains random positive values, you are required to move from the first element [0][0] to the last element [n][n] using the path which will yield the maximum sum of all the elements traversed. You can only move right and down; NOT left and up.
With brute force,this question can be solved by our thought but not computer,because time complexity is exponential.Actually it's a typical DP question,and we should try our best to keep track of something useful to save CPU while we are running in the matrix.I mean for each step we take,we should make sure that it's the optimized choice,which can be used to make choices later.So what does it mean by "later"?This is the point of every DP problem.Most of the time,when we figure out this core problem,we are just near the final solution.Check the code below.
1 /*******************************************
2 Author:Zhou You
3 Time:2014.09.07
4 Feature:finding the optimized path in an matrix
5 *******************************************/
6 #include <iostream>
7 #include <cstdio>
8 #include <algorithm>
9
10 using namespace std;
11
12 void BuildMatrix(int *** pmaze,unsigned row_num,unsigned column_num)
13 {
14 *pmaze = new int*[row_num];
15 for(unsigned i=0;i<row_num;++i){
16 (*pmaze)[i] = new int[column_num];
17 }
18 }
19
20 void ReleaseMatrix(int ***pmaze,unsigned row_num)
21 {
22 if(!pmaze) return;
23
24 for(unsigned i=0;i<row_num;++i){
25 delete [](*pmaze)[i];
26 }
27
28 delete [](*pmaze);
29 }
30
31 void CoreSolve(int ***ppDistanceMatrix,unsigned matrix_size)
32 {
33 for(int i=0;i<matrix_size;++i){
34 for(int j=i;j<matrix_size;++j){
35 if(i-1>=0&&j-1>=0){
36 (*ppDistanceMatrix)[i][j] += max((*ppDistanceMatrix)[i-1][j],(*ppDistanceMatrix)[i][j-1]);
37 }else if(i-1>=0){
38 (*ppDistanceMatrix)[i][j] += (*ppDistanceMatrix)[i-1][j];
39 }else if(j-1>=0){
40 (*ppDistanceMatrix)[i][j] += (*ppDistanceMatrix)[i][j-1];
41 }
42 }
43
44 for(int k=i+1;k<matrix_size;++k){
45 if(k-1>=0&&i-1>=0){
46 (*ppDistanceMatrix)[k][i] += max((*ppDistanceMatrix)[k-1][i],(*ppDistanceMatrix)[k][i-1]);
47 }else if(k-1>=0){
48 (*ppDistanceMatrix)[k][i] += (*ppDistanceMatrix)[k-1][i];
49 }else if(i-1>=0){
50 (*ppDistanceMatrix)[k][i] += (*ppDistanceMatrix)[k][i-1];
51 }
52 }
53 }
54 }
55
56 void Solve()
57 {
58 unsigned matrix_size = 0;
59 int **ppmatrix = NULL;
60 cin>>matrix_size;
61 BuildMatrix(&ppmatrix,matrix_size,matrix_size);
62 for(unsigned i=0;i<matrix_size;++i){
63 for(unsigned j=0;j<matrix_size;++j){
64 cin>>ppmatrix[i][j];
65 }
66 }
67
68 int **ppDistanceMatrix = NULL;
69 BuildMatrix(&ppDistanceMatrix,matrix_size,matrix_size);
70 for(unsigned i=0;i<matrix_size;++i){
71 for(unsigned j=0;j<matrix_size;++j){
72 ppDistanceMatrix[i][j]=ppmatrix[i][j];
73 }
74 }
75
76 CoreSolve(&ppDistanceMatrix,matrix_size);
77 cout<<ppDistanceMatrix[matrix_size-1][matrix_size-1];
78
79 ReleaseMatrix(&ppmatrix,matrix_size);
80 ReleaseMatrix(&ppDistanceMatrix,matrix_size);
81 }
82
83 int main()
84 {
85 freopen("data.in","r",stdin);
86 freopen("data.out","w",stdout);
87
88 unsigned case_num = 0;
89 cin>>case_num;
90
91 for(unsigned i=1;i<=case_num;++i){
92 cout<<"Case #"<<i<<": ";
93 Solve();
94 cout<<endl;
95 }
96
97 return 0;
98 }
Cases in data.in file
3
3
1 2 8
7 20 8
5 3 8
3
1 2 8
7 0 8
5 3 8
2
1 2
3 4
output in data.out file
Case #1: 44
Case #2: 27
Case #3: 8
Pls let me know if you find any mistakes above.Thx.