The trick here is to name the state correctly, then let the implementation follow. For 1072. Flip Columns For Maximum Number of Equal Rows, the solution in this repository is mainly a straightforward implementation solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: straightforward implementation.
The notes already sitting in the source point us in the right direction:
- TLE
Guide
When?
This is the kind of solution you want when the problem has structure hiding inside a messy-looking input. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are all0, all1, allSame, maxEqualRowsAfterFlips, sum.
Guide
Why?
The solution works because it narrows the problem until every update has a clear reason to exist.
- A map keeps the lookup side cheap; the code pays a little memory to avoid repeated searching.
- A set is doing the membership or uniqueness work, which keeps the main loop readable.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Read the setup variables first.
- Follow the main loop or recursive helper next.
- Watch where invalid states get skipped.
- Check which value survives to the return statement.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//TLE
02
03class Solution {
04public:
05 bool all0(vector<int>& arr){
06 return accumulate(arr.begin(), arr.end(), 0) == 0;
07 };
08
09 bool all1(vector<int>& arr){
10 return accumulate(arr.begin(), arr.end(), 1, multiplies<int>()) == 1;
11 };
12
13 bool allSame(vector<int>& arr1, vector<int>& arr2){
14 bool flag = true;
15 for(int i = 0; i < arr1.size(); i++){
16 if(arr1[i] != arr2[i]){
17 flag = false;
18 break;
19 }
20 }
21 return flag;
22 };
23
24 int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
25 int m = matrix.size(), n = matrix[0].size();
26 vector<int> sum(n), reverted(n), ones(n, 1);
27 int ans = 0;
28
29 for(int i = 0; i < m; i++){
30 set<vector<int>> group;
31 int row_friend = 0;
32 for(int j = 0; j < m; j++){
33 if(j == i) continue;
34 //sum two vector element-wise
35 transform(matrix[i].begin(), matrix[i].end(),
36 matrix[j].begin(), sum.begin(), plus<int>());
37 //if their sum is all 0 or all 1,
38 //row_friend++
39 if(allSame(matrix[i], matrix[j]) || all1(sum)){
40 // if(all1(sum)){
41 //ensure the 0th element is 0 and then record it into the set
42 reverted = matrix[i];
43 if(matrix[i][0] != 0){
44 transform(ones.begin(), ones.end(), matrix[i].begin(), reverted.begin(), minus<int>());
45 }
46 if(group.find(reverted) == group.end()){
47 row_friend += 2;
48 }else{
49 row_friend++;
50 }
51 group.insert(reverted);
52 }
53 }
54 ans = max(ans, row_friend);
55 }
56
57 if(ans == 0){
58 for(int i = 0; i < m; i++){
59 if(all0(matrix[i]) || all1(matrix[i])){
60 // if(all1(matrix[i])){
61 ans++;
62 }
63 }
64 }
65
66 //we can always do custom operations for a specific row
67 if(ans == 0) ans = 1;
68
69 return ans;
70 }
71};
72
73//https://leetcode.com/problems/flip-columns-for-maximum-number-of-equal-rows/discuss/303897/Java-easy-solution-%2B-explanation
74//Runtime: 1620 ms, faster than 5.34% of C++ online submissions for Flip Columns For Maximum Number of Equal Rows.
75//Memory Usage: 26.7 MB, less than 100.00% of C++ online submissions for Flip Columns For Maximum Number of Equal Rows.
76class Solution {
77public:
78 bool all1(vector<int>& arr){
79 return accumulate(arr.begin(), arr.end(), 1, multiplies<int>()) == 1;
80 };
81
82 bool allSame(vector<int>& a, vector<int>& b){
83 bool flag = true;
84 for(int i = 0; i < a.size(); i++){
85 if(a[i] != b[i]){
86 flag = false;
87 break;
88 }
89 }
90 return flag;
91 };
92
93 int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
94 int m = matrix.size(), n = matrix[0].size();
95 int ans = 0;
96 vector<int> sum(n);
97 vector<int> reversed(n);
98
99 for(int i = 0; i < m; i++){
100 int row_friend = 0;
101 for(int j = 0; j < m; j++){
102 // transform(matrix[i].begin(), matrix[i].end(), matrix[j].begin(), sum.begin(), plus<int>());
103 for(int k = 0; k < n; k++){
104 reversed[k] = 1 - matrix[j][k];
105 }
106 // if(allSame(matrix[i], matrix[j]) || all1(sum)){
107 // if(allSame(matrix[i], matrix[j]) || allSame(matrix[i], reversed)){
108 if(matrix[i] == matrix[j] || matrix[i] == reversed){
109 row_friend++;
110 }
111 }
112 //row_friend's minimum is 1
113 ans = max(ans, row_friend);
114 }
115
116 return ans;
117 }
118};
119
120//speed the solution above with a set
121//Runtime: 632 ms, faster than 9.71% of C++ online submissions for Flip Columns For Maximum Number of Equal Rows.
122//Memory Usage: 27.4 MB, less than 100.00% of C++ online submissions for Flip Columns For Maximum Number of Equal Rows.
123class Solution {
124public:
125 int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
126 int m = matrix.size(), n = matrix[0].size();
127 int ans = 0;
128 vector<int> sum(n);
129 vector<int> reversed(n);
130 set<int> grouped;
131
132 for(int i = 0; i < m; i++){
133 int row_friend = 0;
134 //skip rows that can be grouped to earlier rows
135 if(grouped.find(i) != grouped.end()) continue;
136 for(int j = 0; j < m; j++){
137 for(int k = 0; k < n; k++){
138 reversed[k] = 1 - matrix[j][k];
139 }
140 if(matrix[i] == matrix[j] || matrix[i] == reversed){
141 row_friend++;
142 grouped.insert(i);
143 grouped.insert(j);
144 }
145 }
146 //row_friend's minimum is 1
147 ans = max(ans, row_friend);
148 }
149
150 return ans;
151 }
152};
153
154//use a map
155//https://leetcode.com/problems/flip-columns-for-maximum-number-of-equal-rows/discuss/303752/Python-1-Line
156//Runtime: 212 ms, faster than 50.49% of C++ online submissions for Flip Columns For Maximum Number of Equal Rows.
157//Memory Usage: 44.7 MB, less than 100.00% of C++ online submissions for Flip Columns For Maximum Number of Equal Rows.
158class Solution {
159public:
160 int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
161 int m = matrix.size(), n = matrix[0].size();
162 map<vector<int>, int> counter;
163 vector<int> ones(n, 1), reversed(n);
164
165 for(int i = 0; i < matrix.size(); i++){
166 transform(ones.begin(), ones.end(), matrix[i].begin(), reversed.begin(), minus<int>());
167 counter[matrix[i]]++;
168 counter[reversed]++;
169 }
170
171 auto it = max_element(counter.begin(), counter.end(),
172 [](const pair<vector<int>, int> & p1, const pair<vector<int>, int> & p2){
173 return p1.second < p2.second;
174 });
175
176 return it->second;
177 }
178};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.