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make-a-square-with-the-same-color | Beats 100% | Python | Easy to Understand | Complex to Write 🌚🕺🦹♂️ | beats-100-python-easy-to-understand-comp-22pv | IntuitionFOR THE LACKWITS!!!
There are only 9 cells. Square is made up of 4 cells. There are only 4 squares. So, why not hard code it?!!ApproachComplexity
Time | aman-sagar | NORMAL | 2025-01-27T17:29:34.781555+00:00 | 2025-01-27T17:29:34.781555+00:00 | 40 | false | # Intuition
**FOR THE LACKWITS!!!**
There are only 9 cells. Square is made up of 4 cells. There are only 4 squares. So, why not hard code it?!!
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O(1)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complex... | 1 | 0 | ['Array', 'Matrix', 'Enumeration', 'Python3'] | 1 |
make-a-square-with-the-same-color | 0ms 100% beat in java very easiest code 😊. | 0ms-100-beat-in-java-very-easiest-code-b-delo | Code | Galani_jenis | NORMAL | 2025-01-08T11:43:54.375573+00:00 | 2025-01-08T11:43:54.375573+00:00 | 114 | false | 
# Code
```java []
class Solution {
public boolean canMakeSquare(char[][] grid) {
byte countB = 0, countW = 0;
for (int i = 0; i <= 1;... | 1 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Easy Solution in C | easy-solution-in-c-by-sathurnithy-kbjo | Code | Sathurnithy | NORMAL | 2025-01-05T13:55:01.964327+00:00 | 2025-01-05T13:55:01.964327+00:00 | 21 | false |
# Code
```c []
bool canMakeSquare(char** grid, int gridSize, int* gridColSize) {
for(int i=0; i<2; i++){
for(int j=0; j<2; j++){
int gridSumWhite = (grid[i][j] == 'W' ? 1:0) + (grid[i][j+1] == 'W' ? 1:0) +
(grid[i+1][j] == 'W' ? 1:0) + (grid[i+1][j+1] == 'W' ? 1:0);
... | 1 | 0 | ['C'] | 0 |
make-a-square-with-the-same-color | C | c-by-pavithrav25-m8yn | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | pavithrav25 | NORMAL | 2025-01-02T16:15:35.614695+00:00 | 2025-01-02T16:15:35.614695+00:00 | 29 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['C'] | 0 |
make-a-square-with-the-same-color | Simple Kotlin solution | simple-kotlin-solution-by-tamhuynhit-twwd | Intuition\nCheck each 2x2 square, if it has 3 \'B\' or 3 \'W\' then it\'s valid and return True immediately.\n\nThere is special cases where there is an already | tamhuynhit | NORMAL | 2024-08-24T06:35:19.096910+00:00 | 2024-08-24T06:35:19.096931+00:00 | 4 | false | # Intuition\nCheck each 2x2 square, if it has 3 \'B\' or 3 \'W\' then it\'s valid and return True immediately.\n\nThere is special cases where there is an already full \'B\'/\'W\' 2x2 square in the grid, so changing any other cell around it still make it valid.\n\n# Approach\nCheck every 2x2 square\nUse a variable to c... | 1 | 0 | ['Array', 'Kotlin'] | 0 |
make-a-square-with-the-same-color | Check color count in all 2x2 grids | check-color-count-in-all-2x2-grids-by-an-qm3m | Intuition\nWe can check how many possible 2x2 grids can be made of 3x3 grid and then check the the B & W count in each block.\n\n# Approach\n4 such 2x2 grids ca | ankit88 | NORMAL | 2024-06-22T04:56:21.284021+00:00 | 2024-06-22T04:56:21.284051+00:00 | 21 | false | # Intuition\nWe can check how many possible 2x2 grids can be made of 3x3 grid and then check the the B & W count in each block.\n\n# Approach\n4 such 2x2 grids can be made. Left-Top, Right-Top, Left-Bottom, Right-Bottom.\n* We iterate over 3x3 grid and just check in each grid how many black boxes are there.(We just nee... | 1 | 0 | ['Java'] | 1 |
make-a-square-with-the-same-color | Fashionable Approach | fashionable-approach-by-baghyawati-senu | Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n# Code\n\nclass Solution:\n def canMakeSquare(self, g: List[List[str]]) -> bool:\n | Baghyawati | NORMAL | 2024-05-08T11:42:38.219046+00:00 | 2024-05-08T11:42:38.219105+00:00 | 52 | false | # Complexity\n- Time complexity: $$O(1)$$\n\n- Space complexity: $$O(1)$$\n\n# Code\n```\nclass Solution:\n def canMakeSquare(self, g: List[List[str]]) -> bool:\n return any((c:=Counter(g[i][j:j+2]+g[i+1][j:j+2]))[\'W\']<2 or c[\'B\']<2 for i in range(2) for j in range(2))\n``` | 1 | 0 | ['Array', 'Python3'] | 1 |
make-a-square-with-the-same-color | BEST C++ SOLUTION ✅✅ | best-c-solution-by-rishu_raj5938-unv7 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Rishu_Raj5938 | NORMAL | 2024-05-03T04:22:00.891277+00:00 | 2024-05-03T04:22:00.891308+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Simple and intuitive thought. Time Complexity O(N) and Space Complexity O(1) | simple-and-intuitive-thought-time-comple-zuwo | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ankurjeesingh310 | NORMAL | 2024-04-30T19:57:18.692083+00:00 | 2024-04-30T19:57:18.692114+00:00 | 266 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Check Four Squares - Fastest way ( 39 ms ) | check-four-squares-fastest-way-39-ms-by-e61h7 | javascript\nvar canMakeSquare = function(grid) {\n if (checkSquare(0, 0) >= 3) return true\n if (checkSquare(1, 0) >= 3) return true\n if (checkSquare( | zemamba | NORMAL | 2024-04-29T14:46:53.140990+00:00 | 2024-04-29T14:46:53.141015+00:00 | 62 | false | ```javascript\nvar canMakeSquare = function(grid) {\n if (checkSquare(0, 0) >= 3) return true\n if (checkSquare(1, 0) >= 3) return true\n if (checkSquare(0, 1) >= 3) return true\n if (checkSquare(1, 1) >= 3) return true\n return false\n\n function checkSquare(a, b) {\n black = 0, white = 0\n ... | 1 | 0 | ['JavaScript'] | 0 |
make-a-square-with-the-same-color | Beat 100% easy condition based solution | beat-100-easy-condition-based-solution-b-qyrs | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Rsharma_09 | NORMAL | 2024-04-28T14:20:05.906343+00:00 | 2024-04-28T14:20:05.906430+00:00 | 161 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O... | 1 | 0 | ['Matrix', 'C++'] | 0 |
make-a-square-with-the-same-color | Simple Python solution | simple-python-solution-by-antarab-o57q | \nclass Solution:\n def canMakeSquare(self, grid: List[List[str]]) -> bool:\n import numpy as np\n matrix = np.array(grid)\n for i in ra | antarab | NORMAL | 2024-04-27T21:12:56.930633+00:00 | 2024-04-27T21:12:56.930677+00:00 | 370 | false | ```\nclass Solution:\n def canMakeSquare(self, grid: List[List[str]]) -> bool:\n import numpy as np\n matrix = np.array(grid)\n for i in range(0, len(matrix[0])):\n for j in range(0, len(matrix)):\n if i<2 and j<2:\n res=matrix[i:i+2,j:j+2]\n ... | 1 | 0 | ['Matrix', 'Python3'] | 1 |
make-a-square-with-the-same-color | [Python] check 4 squares | python-check-4-squares-by-pbelskiy-nrqn | \nclass Solution:\n def canMakeSquare(self, grid: List[List[str]]) -> bool:\n \n def check(y, x):\n b = w = 0\n \n | pbelskiy | NORMAL | 2024-04-27T20:13:21.100639+00:00 | 2024-04-27T20:13:21.100670+00:00 | 18 | false | ```\nclass Solution:\n def canMakeSquare(self, grid: List[List[str]]) -> bool:\n \n def check(y, x):\n b = w = 0\n \n for dy, dx in ((x, y), (x + 1, y), (x, y + 1), (x + 1, y +1)):\n if grid[dy][dx] == \'B\':\n b += 1\n e... | 1 | 0 | ['Python'] | 0 |
make-a-square-with-the-same-color | Python3 || 1 line Solution | python3-1-line-solution-by-gbaian10-qwdc | \n# Code\npython\nclass Solution:\n def canMakeSquare(self, grid: List[List[str]]) -> bool:\n return any([grid[i][j], grid[i+1][j], grid[i][j+1], grid | gbaian10 | NORMAL | 2024-04-27T18:41:08.525520+00:00 | 2024-04-27T18:45:49.361275+00:00 | 156 | false | \n# Code\n```python\nclass Solution:\n def canMakeSquare(self, grid: List[List[str]]) -> bool:\n return any([grid[i][j], grid[i+1][j], grid[i][j+1], grid[i+1][j+1]].count("B") != 2 for i in range(2) for j in range(2))\n \n``` | 1 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | |Brute Force || General Solution || C++ | brute-force-general-solution-c-by-sujalg-85ya | \n\n# Code\n\n\nclass Solution {\npublic:\n bool hasValidSquare(vector<vector<char>>& grid) {\n for (int i = 0; i < 2; ++i) {\n for (int j | sujalgupta09 | NORMAL | 2024-04-27T18:39:15.216490+00:00 | 2024-04-27T18:39:15.216518+00:00 | 112 | false | \n\n# Code\n```\n\nclass Solution {\npublic:\n bool hasValidSquare(vector<vector<char>>& grid) {\n for (int i = 0; i < 2; ++i) {\n for (int j = 0; j < 2; ++j) {\n if (grid[i][j] == grid[i+1][j] && grid[i][j] == grid[i][j+1] && grid[i][j] == grid[i+1][j+1]) {\n retu... | 1 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Easy Java Solution || 0ms exec time | easy-java-solution-0ms-exec-time-by-kari-uujx | \n\n# Complexity\n- Time complexity:O(1)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e.g. O(n) \n\n | karishmaagrawal | NORMAL | 2024-04-27T17:33:34.790142+00:00 | 2024-04-27T17:33:34.790160+00:00 | 97 | false | \n\n# Complexity\n- Time complexity:O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public boolean canMakeSquare(char[][] grid) {\n for(int row=0;row<2;row++){\n for(int... | 1 | 0 | ['Java'] | 2 |
make-a-square-with-the-same-color | C 0ms | c-0ms-by-fredo30400-dkm7 | Code\n\nbool canMakeSquare(char** grid, int gridSize, int* gridColSize) {\n for(int i=0;i<2;i++){\n for(int j=0;j<2;j++){\n int cnt = 0;\n | fredo30400 | NORMAL | 2024-04-27T16:59:25.039297+00:00 | 2024-04-27T16:59:25.039316+00:00 | 50 | false | # Code\n```\nbool canMakeSquare(char** grid, int gridSize, int* gridColSize) {\n for(int i=0;i<2;i++){\n for(int j=0;j<2;j++){\n int cnt = 0;\n if (grid[j][i]==\'W\'){cnt++;}\n if (grid[j+1][i]==\'W\'){cnt++;}\n if (grid[j+1][i+1]==\'W\'){cnt++;}\n if (gr... | 1 | 0 | ['C'] | 1 |
make-a-square-with-the-same-color | 📌📌✅ Beats 100.00% of users with Java💥💥 | beats-10000-of-users-with-java-by-abinay-sizs | Brute Force Approach\n\n\n\n# Code\n\nclass Solution {\n public boolean canMakeSquare(char[][] grid) {\n int w=0,i,j,r=grid.length,c=grid[0].length;\n | Abinayaprakash | NORMAL | 2024-04-27T16:20:52.494387+00:00 | 2024-04-27T16:20:52.494416+00:00 | 19 | false | # Brute Force Approach\n\n\n\n# Code\n```\nclass Solution {\n public boolean canMakeSquare(char[][] grid) {\n int w=0,i,j,r=grid.length,c=grid[0].length;\n int b=0;\n for(i=0;i<=1;i... | 1 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | [C++] Simple Check on 4 Sub-Squares, 6ms, 19.4MB | c-simple-check-on-4-sub-squares-6ms-194m-xfrv | This problem is rather easy to pin down to a few base cases - basically we want that either the first, second, third or fourth possible inner square have an amo | Ajna2 | NORMAL | 2024-04-27T16:19:16.306590+00:00 | 2024-04-28T19:24:55.137373+00:00 | 28 | false | This problem is rather easy to pin down to a few base cases - basically we want that either the first, second, third or fourth possible inner square have an amount of a specific `!= 2`, since that is the only case in which we cannot achieve a `4` changing AT MOST one other element.\n\nWe can check that rather brute-for... | 1 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | [C++][Time & Space = O(1)][Detailed Array Directional Solution] | ctime-space-o1detailed-array-directional-btwd | Intuition\n\n- Think about currentCell as one corner of square (2 X 2)\n- Consider all possible 2 X 2 square can be made from that current cell\n- Now apply log | karnalrohit | NORMAL | 2024-04-27T16:10:37.986855+00:00 | 2024-04-27T16:10:37.986875+00:00 | 55 | false | # Intuition\n\n- Think about currentCell as one corner of square (2 X 2)\n- Consider all possible 2 X 2 square can be made from that current cell\n- Now apply logic on all possible square that any square have all value as same or not.\n\n# Approach\n\n- Take currentCell as one cell of square\n- Now from oneCell you nee... | 1 | 1 | ['C++'] | 0 |
make-a-square-with-the-same-color | ✅Beats 100.00% 🔥 || Java || C++ || C# || Easy Explanation || Beginner Friendly🔥🔥🔥 | beats-10000-java-c-c-easy-explanation-be-pgsn | \n# Intuition\nThe problem aims to determine whether it is possible to form a square using the given grid of characters. A square can be formed if there exists | Sayan98 | NORMAL | 2024-04-27T16:07:29.375701+00:00 | 2024-04-27T16:14:48.534470+00:00 | 76 | false | \n# Intuition\nThe problem aims to determine whether it is possible to form a square using the given grid of characters. A square can be formed if there exists a 2x2 subgrid in the given grid where all the ... | 1 | 0 | ['Math', 'C++', 'Java', 'C#'] | 0 |
make-a-square-with-the-same-color | Java Clean Solution | java-clean-solution-by-shree_govind_jee-f49k | Complexity\n- Time complexity:O(2*2)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e.g. O(n) \n\n# Co | Shree_Govind_Jee | NORMAL | 2024-04-27T16:07:05.668913+00:00 | 2024-04-27T16:07:05.668948+00:00 | 250 | false | # Complexity\n- Time complexity:$$O(2*2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public boolean canMakeSquare(char[][] grid) {\n int black=0, white=0;\n for(int i=0... | 1 | 0 | ['Array', 'Math', 'Matrix', 'Java'] | 0 |
make-a-square-with-the-same-color | BEAT 100% with JAVA | beat-100-with-java-by-hoang_soict-lcm1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | hoang_soict | NORMAL | 2024-04-27T16:01:44.574632+00:00 | 2024-04-27T16:01:44.574661+00:00 | 186 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Simple C++ code, no hardcoding. | simple-c-code-no-hardcoding-by-coolpeng0-dzgg | IntuitionI saw that as long all of the rows or all of the columns have alternating cells, then you cannot make a 2x2 matching area. No matter whether they start | coolpeng0 | NORMAL | 2025-04-08T00:45:43.942141+00:00 | 2025-04-08T00:45:43.942141+00:00 | 4 | false | # Intuition
I saw that as long all of the rows or all of the columns have alternating cells, then you cannot make a 2x2 matching area. No matter whether they start with white or black, there are 2 white cells and 2 black cells in each 2x2 area when the cells are alternating in each line.
I'm pretty sure this solution ... | 0 | 0 | ['C++'] | 1 |
make-a-square-with-the-same-color | Simple Swift Solution | simple-swift-solution-by-felisviridis-2h2k | Code | Felisviridis | NORMAL | 2025-04-03T08:01:50.051312+00:00 | 2025-04-03T08:01:50.051312+00:00 | 1 | false | 
# Code
```swift []
class Solution {
func canMakeSquare(_ grid: [[Character]]) -> Bool {
for a in 0...1 {
for b in 0...1 {
var fill =... | 0 | 0 | ['Array', 'Swift', 'Matrix'] | 0 |
make-a-square-with-the-same-color | CPP beats 100% | cpp-beats-100-by-hareeshghk-2iuf | IntuitionWe just need to check 4 squares. Luckily start position points for squares are same as direction we need to go from starting point. So used same direti | hareeshghk | NORMAL | 2025-03-29T19:29:16.477711+00:00 | 2025-03-29T19:29:16.477711+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We just need to check 4 squares. Luckily start position points for squares are same as direction we need to go from starting point. So used same diretionb 2D array for both.
# Approach
<!-- Describe your approach to solving the problem. -->... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Runtime 100% beats, memory 100% beats solution in Kotlin | runtime-100-beats-memory-100-beats-solut-zc6u | IntuitionI found four possible square and solved the problemComplexity
Time complexity:
O(1)
Space complexity:
O(1)
Code | ahmadali_ok | NORMAL | 2025-03-27T06:50:37.569965+00:00 | 2025-03-27T06:50:37.569965+00:00 | 1 | false | # Intuition
I found four possible square and solved the problem
# Complexity
- Time complexity:
O(1)
- Space complexity:
O(1)
# Code
```kotlin []
class Solution {
fun canMakeSquare(grid: Array<CharArray>): Boolean {
var bHelper = 0
if (grid[0][0] == 'B') bHelper++
if (grid[0][1] == 'B') bHelper++
... | 0 | 0 | ['Kotlin'] | 0 |
make-a-square-with-the-same-color | Java; math approach; 0ms beats 100%; 41.4 MB beats 97% | java-math-approach-0ms-beats-100-414-mb-6y8x9 | IntuitionThe first thing I thought was that I must have only 1 element of a color on a 2x2 "subsquare" (it implies we will have 3 elements of the other).
If the | osmarcf | NORMAL | 2025-03-19T23:45:54.537274+00:00 | 2025-03-19T23:45:54.537274+00:00 | 4 | false | # Intuition
The first thing I thought was that I must have only 1 element of a color on a 2x2 "subsquare" (it implies we will have 3 elements of the other).
If there's only 1, then I can already return "true". If not, I have to check the other subsquares. For this specific problem, we have only 4 subsquares.
Basically,... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Easy to understand solution in Java. Beats 100 % | easy-to-understand-solution-in-java-beat-lglk | Complexity
Time complexity:
O(1)
Space complexity:
O(1)
Code | Khamdam | NORMAL | 2025-03-16T06:18:31.115225+00:00 | 2025-03-16T06:18:31.115225+00:00 | 6 | false | # Complexity
- Time complexity:
O(1)
- Space complexity:
O(1)
# Code
```java []
class Solution {
public boolean canMakeSquare(char[][] grid) {
return isValid(grid, 0, 0) || isValid(grid, 0, 1) ||
isValid(grid, 1, 0) || isValid(grid, 1, 1);
}
private boolean isValid(char[][] grid, i... | 0 | 0 | ['Array', 'Matrix', 'Java'] | 0 |
make-a-square-with-the-same-color | Simple Counting | simple-counting-by-beken-waf6 | IntuitionSimple CountingApproachSimple CountingComplexity
Time complexity:O(1)
Space complexity:O(1)
Code | beken | NORMAL | 2025-02-22T15:36:04.112398+00:00 | 2025-02-22T15:36:04.112398+00:00 | 7 | false | # Intuition
Simple Counting
# Approach
Simple Counting
# Complexity
- Time complexity:
O(1)
- Space complexity:
O(1)
# Code
```java []
class Solution {
public boolean canMakeSquare(char[][] grid) {
for(int y = 0; y < grid.length-1; y++) {
for(int x = 0; x < grid[0].length-1; x++) {
... | 0 | 0 | ['Array', 'Matrix', 'Java'] | 0 |
make-a-square-with-the-same-color | Brute Force Solution, T = O(1) and S = O(1) | brute-force-solution-t-o1-and-s-o1-by-pu-iej8 | IntuitionIf the no. of 'B' or 'W' is greater than 2 in any of the square then return true.ApproachIn all the squares we compare if the no. of B or W is greater | PuruMohite | NORMAL | 2025-02-20T16:38:08.154172+00:00 | 2025-02-20T16:38:08.154172+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
If the no. of 'B' or 'W' is greater than 2 in any of the square then return true.
# Approach
<!-- Describe your approach to solving the problem. -->
In all the squares we compare if the no. of B or W is greater than 2. If for a square it is... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Simple Java solution -> 0 ms Beats 100.00% | simple-java-solution-0-ms-beats-10000-by-jdkf | IntuitionApproachComplexity
Time complexity:
O(1)
Space complexity:
O(1)
Code | DevelopersUsername | NORMAL | 2025-02-10T08:10:18.195107+00:00 | 2025-02-10T08:10:18.195107+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(1)
- Space complexity:
O(1)
# Code
```java []
class Solution {
public boolean canMakeSquare(char[][] grid) {
int center = ... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Java solution | java-solution-by-java_developer-6u7p | null | Java_Developer | NORMAL | 2025-02-02T21:36:40.274630+00:00 | 2025-02-02T21:36:40.274630+00:00 | 10 | false |
```java []
class Solution {
public boolean canMakeSquare(char[][] grid) {
for (int r = 0; r < 2; r++) {
for (int c = 0; c < 2; c++) {
int whiteTiles = 0;
whiteTiles += grid[r][c] == 'W' ? 1 : 0;
whiteTiles += grid[r][c + 1] == 'W' ? 1 : 0;
... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Java&JS&TS Solution (JW) | javajsts-solution-jw-by-specter01wj-awhw | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | specter01wj | NORMAL | 2025-01-28T07:19:58.492614+00:00 | 2025-01-28T07:19:58.492614+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Array', 'Matrix', 'Enumeration', 'Java', 'TypeScript', 'JavaScript'] | 0 |
make-a-square-with-the-same-color | 100% Beat || Rust | 100-beat-rust-by-nocabris-3t1k | Intuitionsplit the 3x3 matrix in 4 2x2 matrices and just count if we got more than 3 'W' or less than 2 'W'ApproachComplexityCode | nocabris | NORMAL | 2025-01-23T13:35:34.380731+00:00 | 2025-01-23T13:35:34.380731+00:00 | 5 | false | # Intuition
split the 3x3 matrix in 4 2x2 matrices and just count if we got more than 3 'W' or less than 2 'W'
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
# Code
```rust []
impl Solution {
pub fn can_make_square(grid: Vec<Vec<char>>) -> bool {
for i in 0..4 {
... | 0 | 0 | ['Rust'] | 0 |
make-a-square-with-the-same-color | Solution in C | solution-in-c-by-praveenkesava-rnos | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | praveenkesava | NORMAL | 2025-01-21T09:53:06.216265+00:00 | 2025-01-21T09:53:06.216265+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C'] | 0 |
make-a-square-with-the-same-color | Typescript ✅ 100% | typescript-100-by-kay-79-0fan | Code | Kay-79 | NORMAL | 2025-01-18T17:41:57.456816+00:00 | 2025-01-18T17:41:57.456816+00:00 | 3 | false | # Code
```typescript []
function canMakeSquare(grid: string[][]): boolean {
for (let i = 0; i < 2; i++) {
for (let j = 0; j < 2; j++) {
let countW = 0;
let countB = 0;
grid[i][j] === "B" ? countB++ : countW++;
grid[i][j + 1] === "W" ? countW++ : countB++;
... | 0 | 0 | ['TypeScript'] | 0 |
make-a-square-with-the-same-color | scala oneliner | scala-oneliner-by-vititov-w75a | null | vititov | NORMAL | 2025-01-18T16:08:50.227687+00:00 | 2025-01-18T16:08:50.227687+00:00 | 1 | false | ```scala []
object Solution {
def canMakeSquare(grid: Array[Array[Char]]): Boolean =
( for{i <- grid.indices.init; j <- grid.head.indices.init}
yield { Iterator(grid(i)(j), grid(i)(j+1), grid(i+1)(j), grid(i+1)(j+1))
.map(_.compare('B').sign).sum}
).filterNot(_ == 2).nonEmpty
}
``` | 0 | 0 | ['Scala'] | 0 |
make-a-square-with-the-same-color | CPP | cpp-by-algoace_84-dr89 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Chandan_84 | NORMAL | 2025-01-15T17:32:16.457393+00:00 | 2025-01-15T17:32:16.457393+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | make-a-square-with-the-same-color in c | make-a-square-with-the-same-color-in-c-b-r6wk | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Udaya_V | NORMAL | 2025-01-15T09:56:48.907320+00:00 | 2025-01-15T09:56:48.907320+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C'] | 0 |
make-a-square-with-the-same-color | Make-a-square-with-the-same-color (In Java) | make-a-square-with-the-same-color-in-jav-83g4 | Code | nishanthinimani525 | NORMAL | 2025-01-12T16:18:55.355338+00:00 | 2025-01-12T16:18:55.355338+00:00 | 7 | false | # Code
```java []
class Solution {
public boolean canMakeSquare(char[][] grid) {
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
int w = 0, b=0;
if(grid[i][j] == 'B'){
b++;
}
else{
w++;
... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Make a square with the Same Color - python3 solution | make-a-square-with-the-same-color-python-25a8 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | ChaithraDee | NORMAL | 2025-01-12T15:35:56.338476+00:00 | 2025-01-12T15:35:56.338476+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(1)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 0 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | Check Balanced Square probability in Grid Matrix | check-balanced-square-probability-in-gri-hpkw | IntuitionApproachI divided the 3x3 grid into four 2x2 subgrids and checked if any of them already contain all cells of the same color or can be made uniform by | traezeeofor | NORMAL | 2025-01-12T09:31:59.098524+00:00 | 2025-01-12T09:31:59.098524+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
I divided the 3x3 grid into four 2x2 subgrids and checked if any of them already contain all cells of the same color or can be made uniform by changing at most one cell. I checked for conditions where three cells are the same co... | 0 | 0 | ['TypeScript'] | 0 |
make-a-square-with-the-same-color | Can make the grid larger size(not only 3*3) | can-make-the-grid-larger-sizenot-only-33-54p1 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | linda2024 | NORMAL | 2025-01-10T23:03:44.766708+00:00 | 2025-01-10T23:03:44.766708+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C#'] | 0 |
make-a-square-with-the-same-color | C++ | Easy to understand | c-easy-to-understand-by-aman786-lbee | Complexity
Time complexity: O(1)
Space complexity: O(1)
Code | Aman786 | NORMAL | 2025-01-10T18:36:49.500037+00:00 | 2025-01-10T18:36:49.500037+00:00 | 1 | false |
# Complexity
- Time complexity: $$O(1)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
bool canMakeSquare(vector<vector<char>>& grid) {
int n = grid.size(), m=grid[0].size... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Make a square with same color Solution in C | make-a-square-with-same-color-solution-i-f2dk | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(1)
Code | Darsan20 | NORMAL | 2025-01-07T17:15:35.541004+00:00 | 2025-01-07T17:15:35.541004+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(n)
- Space complexity:
O(1)
# Code
```c []
bool canMakeSquare(char** grid, int gridSize, int* gridColSize) {
*gridColSize = 3;
... | 0 | 0 | ['C'] | 0 |
make-a-square-with-the-same-color | 3127. Make a Square with the Same Color | 3127-make-a-square-with-the-same-color-b-c31q | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-07T07:07:45.405019+00:00 | 2025-01-07T07:07:45.405019+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | Easy Solution in Java | easy-solution-in-java-by-sathurnithy-zavo | Code | Sathurnithy | NORMAL | 2025-01-05T14:00:55.939227+00:00 | 2025-01-05T14:00:55.939227+00:00 | 7 | false |
# Code
```java []
class Solution {
public boolean canMakeSquare(char[][] grid) {
for(int i=0; i<2; i++){
for(int j=0; j<2; j++){
int gridSumWhite = (grid[i][j] == 'W' ? 1:0) + (grid[i][j+1] == 'W' ? 1:0) +
(grid[i+1][j] == 'W' ? 1:0) + (grid[i+1][... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | super easy solution / python / beats 100% ദ്ദി(˵ •̀ ᴗ - ˵ ) ✧ | super-easy-solution-python-beats-100-ddi-951s | Code | Orin_M | NORMAL | 2025-01-05T10:41:47.022188+00:00 | 2025-01-05T10:41:47.022188+00:00 | 2 | false | 
# Code
```python []
class Solution(object):
def canMakeSquare(self, grid):
"""
:type grid: List[List[str]]
:rtype: bool
"""
for r in range(2):
for c ... | 0 | 0 | ['Python'] | 0 |
make-a-square-with-the-same-color | Easy Java Solution 100% beating in time-complexity 🧑💻 | easy-java-solution-100-beating-in-time-c-r0ro | Intuition
Identify Squares: The core idea is to efficiently check for existing squares or potential squares within the grid.
Color Dominance: Focus on the major | yuvsingh7650 | NORMAL | 2025-01-04T12:57:10.363061+00:00 | 2025-01-04T12:57:10.363061+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
- **Identify Squares**: The core idea is to efficiently check for existing squares or potential squares within the grid.
- **Color Dominance**: Focus on the majority color within each 2x2 sub-grid. If a sub-grid has three or four of the sam... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | <<easy c++ solution -beats 100%>> | easy-c-solution-beats-100-by-shaivya2918-n9nm | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | shaivya291872 | NORMAL | 2024-12-24T10:19:28.699009+00:00 | 2024-12-24T10:19:28.699009+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Easy solution || java || 100% beat | easy-solution-java-100-beat-by-subhash_k-k63f | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Subhash_kumar143 | NORMAL | 2024-12-21T08:28:53.713891+00:00 | 2024-12-21T08:28:53.713891+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Java | Easy Solution | O(1) | 0ms | Beats 100% | java-easy-solution-o1-0ms-beats-100-by-a-uh5t | ApproachDivide the 3x3 grid into 4 possible 2x2 sub-grids:
Top-left: (0,0), (0,1), (1,0), (1,1)
Top-right: (0,1), (0,2), (1,1), (1,2)
Bottom-left: (1,0), (1,1), | arti8190 | NORMAL | 2024-12-15T10:18:45.806207+00:00 | 2024-12-15T10:18:45.806207+00:00 | 8 | false | \n# Approach\n\n**Divide the 3x3 grid into 4 possible 2x2 sub-grids:**\n\n* Top-left: (0,0), (0,1), (1,0), (1,1)\n* Top-right: (0,1), (0,2), (1,1), (1,2)\n* Bottom-left: (1,0), (1,1), (2,0), (2,1)\n* Bottom-right: (1,1), (1,2), (2,1), (2,2)\n\n**Check if one of these 2x2 grids can be converted to a single-color grid:**... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Easiest Solution Beats 100% | easiest-solution-beats-100-by-mayankbish-16br | null | MayankBisht8630 | NORMAL | 2024-12-11T09:55:27.047137+00:00 | 2024-12-11T09:55:27.047137+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Easy & Simple Solution to Check If Same Color 2x2 Square Is Possible | easy-simple-solution-to-check-if-same-co-455h | null | hassan21kh1996 | NORMAL | 2024-12-10T17:17:22.257336+00:00 | 2024-12-10T17:17:22.257336+00:00 | 3 | false | # Intuition\nWe need to check all 2x2 subgrids within the 3x3 grid to see if it\'s possible to make a uniform 2x2 square (all \'B\' or all \'W\') by changing at most one cell. If a subgrid contains 3 or more cells of the same color, we can change the remaining cell to make it uniform.\n\n# Approach\n1. **Iterate throug... | 0 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | Beats 100% - Super Simple Solution | beats-100-super-simple-solution-by-brend-gbof | null | brendanc490 | NORMAL | 2024-12-10T02:11:47.192402+00:00 | 2024-12-10T02:11:47.192402+00:00 | 4 | false | # Intuition\nIn order to make a 2X2 square of all the same color, we need at least 3 of the tiles to be the same color in the 2x2 area. Here are the cases:\n\nO O\nO X\n\nO X\nO O\n\nO O\nX O\n\nX O\nO O\n\nYou might notice a pattern here, which is that they all form an L like shape. Therefore, we just need to look for... | 0 | 0 | ['Python'] | 0 |
make-a-square-with-the-same-color | Understandable Python for 0ms (beside 17.41Mb) | understandable-python-for-0ms-beside-174-2m7s | Intuition\n Describe your first thoughts on how to solve this problem. \n\nYou only have to check the 4 2x2 square and count the number of color\n\n# Approach\n | Clums | NORMAL | 2024-12-05T12:38:42.263908+00:00 | 2024-12-05T12:38:42.263953+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nYou only have to check the 4 2x2 square and count the number of color\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nTo simplify it you see that it\'s about any color and it needs at most 1 move (cause with 2 ... | 0 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | Solves for any size grid | solves-for-any-size-grid-by-rezaman-h0vd | Intuition\nFind any 2x2 squares that are majority the same color.\n\n# Approach\n- Split the grid into multiple 2x2 grids.\n- Supports arbitrarily sized square | rezaman | NORMAL | 2024-12-03T19:21:18.787178+00:00 | 2024-12-03T19:23:44.833182+00:00 | 0 | false | # Intuition\nFind any 2x2 squares that are majority the same color.\n\n# Approach\n- Split the grid into multiple 2x2 grids.\n- Supports arbitrarily sized square grid\n- Check each individual grid for majority same color\n\n\n# Complexity\n- Time complexity: N-squared\n\n- Space complexity: $$2n$$\n\n# Code\n```ruby []... | 0 | 0 | ['Ruby'] | 0 |
make-a-square-with-the-same-color | Iterate & Split | iterate-split-by-rezaman-492b | Intuition\nFind any 2x2 squares that are majority the same color.\n\n# Approach\n- Split the grid into multiple 2x2 grids. \n- Check each individual grid for ma | rezaman | NORMAL | 2024-12-03T19:16:18.374650+00:00 | 2024-12-03T19:16:18.374683+00:00 | 0 | false | # Intuition\nFind any 2x2 squares that are majority the same color.\n\n# Approach\n- Split the grid into multiple 2x2 grids. \n- Check each individual grid for majority same color\n\n# Complexity\n- Time complexity: N-squared\n\n- Space complexity: $$2n$$\n\n# Code\n```ruby []\n# @param {Character[][]} grid\n# @return ... | 0 | 0 | ['Ruby'] | 0 |
make-a-square-with-the-same-color | Submission beat 100% of other submissions' runtime. | submission-beat-100-of-other-submissions-xpui | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Vishva-mitra | NORMAL | 2024-11-26T10:41:31.344952+00:00 | 2024-11-26T10:41:31.344987+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | [Java] ✅ 0MS ✅ 100% ✅ FASTEST ✅ BEST ✅ CLEAN CODE | java-0ms-100-fastest-best-clean-code-by-o3fq5 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. As you have 9 cells (W and B), you can make a special submatrix (2x2) | StefanelStan | NORMAL | 2024-11-15T05:00:26.299422+00:00 | 2024-11-15T05:00:26.299452+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. As you have 9 cells (W and B), you can make a special submatrix (2x2) only if that matrix does not have 2 W or 2B. \n2. Any other combination (0-4, 1-3, 3-1, 4-0) will make a special submatrix.\n\n# Complexity\n- Time com... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Build-up pairs of pairs of sum of consecutive Bs and are any not 2? | build-up-pairs-of-pairs-of-sum-of-consec-5ice | Count up consecutive pairs of Bs\n2. Transpose so that ...\n3. Sum up consecutive counts of Bs and ...\n4. Is there any sum that isnt 2 (which would be a 2x2 su | czrpb | NORMAL | 2024-11-15T00:24:21.250773+00:00 | 2024-11-15T00:24:21.250807+00:00 | 0 | false | 1. Count up consecutive pairs of Bs\n2. Transpose so that ...\n3. Sum up consecutive counts of Bs and ...\n4. Is there any sum that isnt 2 (which would be a 2x2 sub-grid of half Bs, half Ws)\n\n# Code\n```racket []\n(define/contract (can-make-square grid (n 2) (m 2))\n (-> (listof (listof char?)) boolean?)\n\n (defin... | 0 | 0 | ['Racket'] | 0 |
make-a-square-with-the-same-color | 3127. Make a Square with the Same Color | 3127-make-a-square-with-the-same-color-b-1767 | Complexity\nTime: O(N^2)\nSpace: O(1)\n\n# Code\npython []\nclass Solution(object):\n def canMakeSquare(self, grid):\n """\n :type grid: List[L | pasha_iden | NORMAL | 2024-11-12T00:46:30.757208+00:00 | 2024-11-12T00:46:30.757242+00:00 | 3 | false | # Complexity\nTime: $$O(N^2)$$\nSpace: $$O(1)$$\n\n# Code\n```python []\nclass Solution(object):\n def canMakeSquare(self, grid):\n """\n :type grid: List[List[str]]\n :rtype: bool\n """\n q=1\n for x in range(len(grid)-1):\n for y in range(len(grid[x])-1):\n ... | 0 | 0 | ['Python'] | 0 |
make-a-square-with-the-same-color | Does a 2x2 view not have 2 Bs? | does-a-2x2-view-not-have-2-bs-by-czrpb-ddih | Create a 2x2 lens\n2. Whip it thru the matrix; lets call these views\n3. Exit with #t if a view is found to not have 2 Bs\n\n# Code\nracket []\n(define/contract | czrpb | NORMAL | 2024-11-11T23:00:49.301848+00:00 | 2024-11-11T23:00:49.301903+00:00 | 2 | false | 1. Create a 2x2 lens\n2. Whip it thru the matrix; lets call these *view*s\n3. Exit with `#t` if a *view* is found to not have 2 `B`s\n\n# Code\n```racket []\n(define/contract (can-make-square grid (n 2) (m 2))\n (-> (listof (listof char?)) boolean?)\n\n (define is-B? (curry char=? #\\B))\n\n (define [build-lens (n n... | 0 | 0 | ['Racket'] | 0 |
make-a-square-with-the-same-color | Easy C++ Code (Beats 100%) | easy-c-code-beats-100-by-ai1a_2310812-nwv0 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ai1a_2310812 | NORMAL | 2024-11-10T09:03:11.843378+00:00 | 2024-11-10T09:03:11.843405+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Simple solution using python | simple-solution-using-python-by-priya_re-8ydu | Time complexity of O(1):\nThere is a simple trick, if the 2x2 matrix contains equal number of "B" and "W" then we can\'t change the matrix as perfect square.\n\ | Priya_Reka_S | NORMAL | 2024-11-06T12:56:44.830133+00:00 | 2024-11-06T12:56:44.830166+00:00 | 4 | false | Time complexity of O(1):\nThere is a simple trick, if the 2x2 matrix contains equal number of "B" and "W" then we can\'t change the matrix as perfect square.\n\n# Code\n```python3 []\nclass Solution:\n def canMakeSquare(self, grid: List[List[str]]) -> bool:\n l1=[grid[0][1],grid[0][0],grid[1][0],grid[1][1]]\n... | 0 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | Simple PHP solution | simple-php-solution-by-file2k-ym9e | Intuition\nCreate 2x2 squares array from 3x3 square. Loop thru 2x2 squares and check weather it\'s possible for any of them to have all 4 cells with the same co | file2k | NORMAL | 2024-11-04T20:47:49.111193+00:00 | 2024-11-04T20:47:49.111220+00:00 | 3 | false | # Intuition\nCreate 2x2 squares array from 3x3 square. Loop thru 2x2 squares and check weather it\'s possible for any of them to have all 4 cells with the same color. \n\n# Approach\nOuter loops with iterators "i" and "j" are used for navigating 2x2 squares. Inner loops with iterators "k" and "l" are used for navigatin... | 0 | 0 | ['PHP'] | 0 |
make-a-square-with-the-same-color | Java 0ms beats 100%, check all 2x2 subgrids and early return upon success | java-0ms-beats-100-check-all-2x2-subgrid-0sl3 | Intuition\n Describe your first thoughts on how to solve this problem. \nIf a 2x2 subgrid has all one color, or 3 of one color, then we can have a 2x2 of 1 colo | texastim | NORMAL | 2024-11-04T06:34:03.871096+00:00 | 2024-11-04T06:34:03.871158+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf a 2x2 subgrid has all one color, or 3 of one color, then we can have a 2x2 of 1 color with at most one change. If 2x2 subgrid has 2 of each color, we can\'t.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSyste... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | Beat 100% with Cpp | beat-100-with-cpp-by-carfel14-09l1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | carfel14 | NORMAL | 2024-11-03T03:42:27.874779+00:00 | 2024-11-03T03:42:27.874825+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | POO Solution in Python | poo-solution-in-python-by-astros-5x57 | Intuition\nThe solution is overcomplicated, but it is elegant :-P\n\n# Code\npython3 []\nclass Colors:\n def __init__(self, white: int, black: int):\n | Astros | NORMAL | 2024-10-30T13:15:08.628909+00:00 | 2024-10-30T13:15:08.628935+00:00 | 2 | false | # Intuition\nThe solution is overcomplicated, but it is elegant :-P\n\n# Code\n```python3 []\nclass Colors:\n def __init__(self, white: int, black: int):\n self.white = white\n self.black = black\n \n def is_possible(self) -> bool:\n return self.white in [3, 4] or self.black in [3, 4]\n\nc... | 0 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | Easy | easy-by-shanku1999-q3ip | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | shanku1999 | NORMAL | 2024-10-23T09:46:53.971648+00:00 | 2024-10-23T09:46:53.971681+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
make-a-square-with-the-same-color | a new one | a-new-one-by-ianard-z332 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | IanArd | NORMAL | 2024-10-21T19:15:33.480506+00:00 | 2024-10-21T19:15:33.480537+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Rust'] | 0 |
make-a-square-with-the-same-color | Customizable Solution | C++ | customizable-solution-c-by-joao_ferrao-f29a | Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n# Code\nC++ []\nclass Solution {\npublic:\n bool canMakeSquare(vector<vector<char>>& grid) | Joao_Ferrao | NORMAL | 2024-10-16T15:07:51.447908+00:00 | 2024-10-16T15:07:51.447935+00:00 | 0 | false | # Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n# Code\n```C++ []\nclass Solution {\npublic:\n bool canMakeSquare(vector<vector<char>>& grid) {\n const int squareLength = 2;\n for(int i = 0; i < squareLength; ++i){\n for(int j = 0; j < squareLength; ++j){\n ... | 0 | 0 | ['C++'] | 0 |
make-a-square-with-the-same-color | Brute Force Solution || JAVA | brute-force-solution-java-by-ashwinmurug-gufv | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ashwinmurugan46 | NORMAL | 2024-10-16T04:53:28.418856+00:00 | 2024-10-16T04:53:28.418901+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
make-a-square-with-the-same-color | C# simple solution | c-simple-solution-by-nzspider-ftiw | Code\ncsharp []\npublic class Solution {\n public bool CanMakeSquare(char[][] grid) {\n int row=3,col=3,r=0,c=0,b=0,w=0;\n var pos = new List<( | nzspider | NORMAL | 2024-10-13T22:35:32.261739+00:00 | 2024-10-13T22:35:32.261838+00:00 | 7 | false | # Code\n```csharp []\npublic class Solution {\n public bool CanMakeSquare(char[][] grid) {\n int row=3,col=3,r=0,c=0,b=0,w=0;\n var pos = new List<(int,int)>{(0,0),(0,1),(1,0),(1,1)};\n \n foreach(var (x, y) in pos){\n for(r=y;r<2+y;r++){\n for(c=x;c<2+x;c++){\n ... | 0 | 0 | ['C#'] | 0 |
make-a-square-with-the-same-color | Python3 O(N ^ 4) Solution with counting (97.76% Runtime) | python3-on-4-solution-with-counting-9776-ogqy | Intuition\n Describe your first thoughts on how to solve this problem. \n\n\n\n# Approach\n Describe your approach to solving the problem. \n- Iteratively check | missingdlls | NORMAL | 2024-10-10T11:51:09.848105+00:00 | 2024-10-10T11:51:09.848128+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Iteratively check 4 adjacent cells and see if i... | 0 | 0 | ['Array', 'Matrix', 'Counting', 'Python', 'Python3'] | 0 |
binary-tree-preorder-traversal | Accepted iterative solution in Java using stack. | accepted-iterative-solution-in-java-usin-9nab | Note that in this solution only right children are stored to stack.\n\n public List preorderTraversal(TreeNode node) {\n\t\tList list = new LinkedList();\n\t | pavel-shlyk | NORMAL | 2014-12-29T15:03:10+00:00 | 2018-10-21T02:03:29.238021+00:00 | 77,976 | false | Note that in this solution only right children are stored to stack.\n\n public List<Integer> preorderTraversal(TreeNode node) {\n\t\tList<Integer> list = new LinkedList<Integer>();\n\t\tStack<TreeNode> rights = new Stack<TreeNode>();\n\t\twhile(node != null) {\n\t\t\tlist.add(node.val);\n\t\t\tif (node.right != null... | 347 | 7 | [] | 47 |
binary-tree-preorder-traversal | Very simple iterative Python solution | very-simple-iterative-python-solution-by-rhhu | Classical usage of stack's LIFO feature, very easy to grasp:\n\n \n def preorderTraversal(self, root):\n ret = []\n stack = [root]\n | clue | NORMAL | 2015-01-27T04:17:27+00:00 | 2018-10-15T14:36:23.147933+00:00 | 36,928 | false | Classical usage of stack's LIFO feature, very easy to grasp:\n\n \n def preorderTraversal(self, root):\n ret = []\n stack = [root]\n while stack:\n node = stack.pop()\n if node:\n ret.append(node.val)\n stack.append(node.right)\n ... | 283 | 3 | ['Iterator', 'Python'] | 33 |
binary-tree-preorder-traversal | C++ Iterative, Recursive and Morris Traversal | c-iterative-recursive-and-morris-travers-5lbv | There are three solutions to this problem.\n\n 1. Iterative solution using stack --- O(n) time and O(n) space;\n 2. Recursive solution --- O(n) time and O(n) sp | jianchao-li | NORMAL | 2015-05-21T16:31:34+00:00 | 2018-10-02T16:31:25.908094+00:00 | 27,836 | false | There are three solutions to this problem.\n\n 1. Iterative solution using stack --- `O(n)` time and `O(n)` space;\n 2. Recursive solution --- `O(n)` time and `O(n)` space (function call stack);\n 3. Morris traversal --- `O(n)` time and `O(1)` space.\n\n**Iterative solution using stack**\n\n```cpp\nclass Solution {\npu... | 229 | 2 | ['Stack', 'Recursion', 'Binary Tree', 'Iterator', 'C++'] | 18 |
binary-tree-preorder-traversal | Solution | solution-by-deleted_user-v7rj | C++ []\nclass Solution {\npublic:\n vector<int> preorderTraversal(TreeNode* root) {\n vector<int> preorder;\n stack<TreeNode*> stack;\n | deleted_user | NORMAL | 2023-01-09T16:00:31.516361+00:00 | 2023-03-07T10:18:41.372462+00:00 | 15,432 | false | ```C++ []\nclass Solution {\npublic:\n vector<int> preorderTraversal(TreeNode* root) {\n vector<int> preorder;\n stack<TreeNode*> stack;\n if (root == NULL)\n return preorder;\n stack.push(root);\n while(!stack.empty()) {\n TreeNode* curr = stack.top();\n ... | 221 | 0 | ['C++', 'Java', 'Python3'] | 3 |
binary-tree-preorder-traversal | 3 Different Solutions | 3-different-solutions-by-fabrizio3-1xgm | Recursive method with List as returning value:\n\n \tpublic List preorderTraversal(TreeNode root) {\n \t\tList pre = new LinkedList();\n \t\tif(root==n | fabrizio3 | NORMAL | 2015-04-22T07:50:47+00:00 | 2018-10-19T02:07:21.808809+00:00 | 43,107 | false | Recursive method with List as returning value:\n\n \tpublic List<Integer> preorderTraversal(TreeNode root) {\n \t\tList<Integer> pre = new LinkedList<Integer>();\n \t\tif(root==null) return pre;\n \t\tpre.add(root.val);\n \t\tpre.addAll(preorderTraversal(root.left));\n \t\tpre.addAll(preorderTraversal... | 220 | 4 | ['Recursion', 'Iterator', 'Java'] | 23 |
binary-tree-preorder-traversal | Python solutions (recursively and iteratively). | python-solutions-recursively-and-iterati-8emk | \n # recursively\n def preorderTraversal1(self, root):\n res = []\n self.dfs(root, res)\n return res\n \n def dfs(self, | oldcodingfarmer | NORMAL | 2015-08-13T15:14:56+00:00 | 2018-09-24T23:32:41.072858+00:00 | 22,638 | false | \n # recursively\n def preorderTraversal1(self, root):\n res = []\n self.dfs(root, res)\n return res\n \n def dfs(self, root, res):\n if root:\n res.append(root.val)\n self.dfs(root.left, res)\n self.dfs(root.right, res)\n \n # i... | 170 | 1 | ['Recursion', 'Python'] | 14 |
binary-tree-preorder-traversal | 【Video】Recursive Solution and Stack Solution | video-recursive-solution-and-stack-solut-pygs | Intuition\nWe should know how to implement inorder, preorder and postorder.\n\n# Solution Video\n\nhttps://youtu.be/9dTgOlzLDa0\n\n\u25A0 Timeline\n0:09 Explain | niits | NORMAL | 2024-12-07T03:51:56.880619+00:00 | 2024-12-07T03:51:56.880643+00:00 | 8,862 | false | # Intuition\nWe should know how to implement inorder, preorder and postorder.\n\n# Solution Video\n\nhttps://youtu.be/9dTgOlzLDa0\n\n\u25A0 Timeline\n`0:09` Explain Recursive Solution\n`1:02` Coding for Recursive Solution\n`2:08` Time Complexity and Space Complexity for Recursive Solution\n`2:27` Explain Stack Solution... | 98 | 1 | ['Stack', 'Tree', 'Depth-First Search', 'Binary Tree', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
binary-tree-preorder-traversal | ✅[C++] EASY|| Beats100% || 3 Approach With explaination✅ | c-easy-beats100-3-approach-with-explaina-fxde | \uD83C\uDFA5\uD83D\uDD25 Exciting News! Join my Coding Journey! Subscribe Now! \uD83D\uDD25\uD83C\uDFA5\n\n\uD83D\uDD17 Link in the leetcode profile \n\nNew cod | vishnoi29 | NORMAL | 2023-01-09T00:29:48.816065+00:00 | 2023-06-14T02:11:40.159956+00:00 | 12,531 | false | \uD83C\uDFA5\uD83D\uDD25 Exciting News! Join my Coding Journey! Subscribe Now! \uD83D\uDD25\uD83C\uDFA5\n\n\uD83D\uDD17 Link in the leetcode profile \n\nNew coding channel alert! \uD83D\uDE80\uD83D\uDCBB Subscribe to unlock amazing coding content and tutorials. Help me reach 1K subs to start posting more videos! Join n... | 97 | 2 | ['Stack', 'Recursion', 'C++'] | 4 |
binary-tree-preorder-traversal | Accepted code. Explaination with Algo. | accepted-code-explaination-with-algo-by-i69xy | Create an empty stack, Push root node to the stack.\n 2. Do following while stack is not empty.\n\n 2.1. pop an item from the stack and print it.\n \n 2.2. push | deepalaxmi | NORMAL | 2014-08-23T21:24:07+00:00 | 2018-10-09T18:57:57.389248+00:00 | 15,584 | false | 1. Create an empty stack, Push root node to the stack.\n 2. Do following while stack is not empty.\n\n 2.1. pop an item from the stack and print it.\n \n 2.2. push the right child of popped item to stack.\n\n 2.3. push the left child of popped item to stack.\n\n \n\n\n> class Solution {\n> public:\n> v... | 78 | 0 | [] | 10 |
binary-tree-preorder-traversal | 4 solutions in c++ | 4-solutions-in-c-by-gogo93-3vsc | \n // recursive, but it's trivial...\n vector preorderTraversal(TreeNode root) {\n vector v;\n preTraversal(root, v);\n return v;\n | gogo93 | NORMAL | 2015-08-20T07:25:45+00:00 | 2018-08-30T09:40:50.439971+00:00 | 9,756 | false | \n // recursive, but it's trivial...\n vector<int> preorderTraversal(TreeNode* root) {\n vector<int> v;\n preTraversal(root, v);\n return v;\n }\n void preTraversal(TreeNode* root, vector<int>& v){\n if(!root) return;\n v.push_back(root->val);\n preTraversal(root->l... | 60 | 0 | ['C++'] | 3 |
binary-tree-preorder-traversal | Preorder Traversal Java solution both iteration and recursion | preorder-traversal-java-solution-both-it-0aqe | // recursive\n public class Solution {\n public List<Integer> preorderTraversal(TreeNode root) {\n List<Integer> result = new ArrayList<Int | king_yl | NORMAL | 2015-10-20T02:13:17+00:00 | 2015-10-20T02:13:17+00:00 | 10,588 | false | // recursive\n public class Solution {\n public List<Integer> preorderTraversal(TreeNode root) {\n List<Integer> result = new ArrayList<Integer>();\n if (root != null){\n result.add(root.val);\n result.addAll(preorderTraversal(root.left));\n ... | 60 | 0 | ['Java'] | 3 |
binary-tree-preorder-traversal | Python 3 || 1-10 lines, 3 versions w/example || T/S: 98% / 99% | python-3-1-10-lines-3-versions-wexample-uai3r | \nThe iterative version, with example:\n\nclass Solution:\n def preorderTraversal(self, root: TreeNode) -> list[int]:\n\n | Spaulding_ | NORMAL | 2023-01-09T01:19:38.642836+00:00 | 2024-05-22T21:09:30.701958+00:00 | 7,861 | false | \nThe iterative version, with example:\n```\nclass Solution:\n def preorderTraversal(self, root: TreeNode) -> list[int]:\n\n # Ex: root = [1, 2,None, 3,4]\n if not root: return [] # __1\n stack, ans = [root], [] # /\n ... | 58 | 1 | ['Python', 'Python3'] | 2 |
binary-tree-preorder-traversal | Easy C++ solution using Stack | easy-c-solution-using-stack-by-yulingtia-mo3d | class Solution {\n public:\n vector<int> preorderTraversal(TreeNode *root) {\n if (root==NULL) {\n return vector<int>();\n }\n | yulingtianxia | NORMAL | 2014-12-06T07:34:56+00:00 | 2018-10-17T15:12:18.679707+00:00 | 12,694 | false | class Solution {\n public:\n vector<int> preorderTraversal(TreeNode *root) {\n if (root==NULL) {\n return vector<int>();\n }\n vector<int> result;\n stack<TreeNode *> treeStack;\n treeStack.push(root);\n while (!treeStack.empty()) {\n TreeNode *t... | 54 | 0 | [] | 8 |
binary-tree-preorder-traversal | Java Solution with Explanation | java-solution-with-explanation-by-sartha-qa38 | \n\n# Approach and Explanation ( Recursive )\n Describe your approach to solving the problem. \n1. A preorder traversal is a traversal order in which the root n | Sarthak_Singh_ | NORMAL | 2023-01-09T04:16:13.742817+00:00 | 2023-01-09T04:44:42.444406+00:00 | 4,566 | false | \n\n# Approach and Explanation ( Recursive )\n<!-- Describe your approach to solving the problem. -->\n1. A preorder traversal is a traversal order in which the root node is visited first, followed by the left subtree, and then the right subtree. `The traversal order for a node is: root, left, right.`\n\n2. The code fi... | 53 | 1 | ['Java'] | 2 |
binary-tree-preorder-traversal | 132ms in JavaScript | 132ms-in-javascript-by-linfongi-ktad | var preorderTraversal = function(root) {\n if (!root) return [];\n var result = [];\n var stack = [root];\n \n while(stack.length) {\n | linfongi | NORMAL | 2015-05-21T01:03:21+00:00 | 2015-05-21T01:03:21+00:00 | 4,975 | false | var preorderTraversal = function(root) {\n if (!root) return [];\n var result = [];\n var stack = [root];\n \n while(stack.length) {\n var node = stack.pop();\n result.push(node.val);\n if (node.right) stack.push(node.right);\n if (node.left) stack.push(node.left... | 48 | 1 | ['JavaScript'] | 4 |
binary-tree-preorder-traversal | 3 Iterative Solutions: Stack And Morris Traversal (Complexity Explained) | 3-iterative-solutions-stack-and-morris-t-ugod | 1 Stack-based:\nStack-based: \njava\npublic class Solution {\n public List<Integer> preorderTraversal(TreeNode root) {\n List<Integer> res = new Array | vegito2002 | NORMAL | 2017-08-13T21:41:34.002000+00:00 | 2018-10-12T03:20:40.021825+00:00 | 3,635 | false | ### 1 Stack-based:\nStack-based: \n```java\npublic class Solution {\n public List<Integer> preorderTraversal(TreeNode root) {\n List<Integer> res = new ArrayList<>();\n if (root == null)\n return res;\n Stack<TreeNode> s = new Stack<>();\n s.push(root);\n while (!s.isEmp... | 41 | 1 | [] | 6 |
binary-tree-preorder-traversal | Java recursive and iterative solutions. | java-recursive-and-iterative-solutions-b-qunu | \n // recursively\n public List<Integer> preorderTraversal1(TreeNode root) {\n List<Integer> ret = new ArrayList<>();\n dfs(root, ret);\ | oldcodingfarmer | NORMAL | 2016-05-05T15:16:15+00:00 | 2016-05-05T15:16:15+00:00 | 4,020 | false | \n // recursively\n public List<Integer> preorderTraversal1(TreeNode root) {\n List<Integer> ret = new ArrayList<>();\n dfs(root, ret);\n return ret;\n }\n \n private void dfs(TreeNode root, List<Integer> ret) {\n if (root != null) {\n ret.add(root.val);\n ... | 33 | 0 | ['Recursion', 'Iterator', 'Java'] | 6 |
binary-tree-preorder-traversal | Preorder\u3001inorder\u3001postorder iterative solution by c++ | preorderu3001inorderu3001postorder-itera-spe4 | preorder:\n\n vector preorderTraversal(TreeNode root) {\n \tvector res;\n \tstd::stack temp;\n \twhile (root || !temp.empty()) {\n \t\twhile (roo | hustlx1 | NORMAL | 2016-05-03T13:22:17+00:00 | 2018-09-17T12:33:35.324240+00:00 | 4,216 | false | preorder:\n\n vector<int> preorderTraversal(TreeNode* root) {\n \tvector<int> res;\n \tstd::stack<TreeNode*> temp;\n \twhile (root || !temp.empty()) {\n \t\twhile (root) {\n \t\t\ttemp.push(root);\n \t\t\tres.push_back(root->val);\n \t\t\troot = root->left;\n \t\t}\n \t\troot = temp.top();... | 31 | 0 | [] | 2 |
binary-tree-preorder-traversal | C++ Solution || 100% faster || Iterative Solution | c-solution-100-faster-iterative-solution-2rmz | \nclass Solution {\npublic:\n vector<int> preorderTraversal(TreeNode* root) {\n vector<int> preorder;\n stack<TreeNode*> stack;\n if (ro | AlankarSaxena2712 | NORMAL | 2021-07-13T11:12:58.592023+00:00 | 2021-07-13T11:13:29.709375+00:00 | 3,445 | false | ```\nclass Solution {\npublic:\n vector<int> preorderTraversal(TreeNode* root) {\n vector<int> preorder;\n stack<TreeNode*> stack;\n if (root == NULL)\n return preorder;\n stack.push(root);\n while(!stack.empty()) {\n TreeNode* curr = stack.top();\n ... | 26 | 0 | ['Stack', 'C', 'Iterator', 'C++'] | 5 |
binary-tree-preorder-traversal | 3 Simple Python solutions | 3-simple-python-solutions-by-shraddhapp-ggnc | Solution 1: Preorder traversal with stack\n\n\nclass Solution:\n def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:\n ans = []\n | shraddhapp | NORMAL | 2021-08-14T17:29:22.400501+00:00 | 2021-08-14T17:30:04.534860+00:00 | 3,067 | false | ##### Solution 1: Preorder traversal with stack\n\n```\nclass Solution:\n def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:\n ans = []\n stack = [root]\n \n while stack:\n temp = stack.pop()\n \n if temp:\n ans.append(temp.... | 25 | 1 | ['Stack', 'Recursion', 'Python', 'Python3'] | 2 |
binary-tree-preorder-traversal | Python recursive and iterative solutions | python-recursive-and-iterative-solutions-z34v | Please see and vote for my solutions for these similar problems.\n94. Binary Tree Inorder Traversal\n144. Binary Tree Preorder Traversal\n145. Binary Tree Posto | otoc | NORMAL | 2019-07-11T05:19:34.651622+00:00 | 2019-07-11T05:52:34.499313+00:00 | 3,457 | false | Please see and vote for my solutions for these similar problems.\n[94. Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/discuss/332283/Python-recursive-and-iterative-solutions)\n[144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/d... | 25 | 0 | [] | 3 |
binary-tree-preorder-traversal | ✅ 🔥 0 ms Runtime Beats 100% User🔥|| Code Idea ✅ || Algorithm & Solving Step ✅ || | 0-ms-runtime-beats-100-user-code-idea-al-5stm | \n\n\u2705 IF YOU LIKE THIS SOLUTION, PLEASE UPVOTE AT THE END \u2705 :\n# I prefer Iterative Approach:\n\n### Intuition:\nPreorder traversal of a binary tree | Letssoumen | NORMAL | 2024-12-04T00:23:01.820310+00:00 | 2024-12-04T00:23:01.820350+00:00 | 4,096 | false | \n\n\u2705 IF YOU LIKE THIS SOLUTION, PLEASE UPVOTE AT THE END \u2705 :\n# I prefer Iterative Approach:\n\n### **Intuition:**\nPreorder traversal of a binary tree visits ... | 23 | 1 | ['Stack', 'Tree', 'C++', 'Java', 'Python3'] | 0 |
binary-tree-preorder-traversal | C++ recursive simple solution || 0 ms, 100% faster | c-recursive-simple-solution-0-ms-100-fas-bht7 | C++ :\n\n\nvoid preorderTraversalHelper(TreeNode* root, vector<int> &res) {\n\tif(root)\n\t{\n\t\tres.push_back(root -> val); \n\n\t\tif(root -> left) | TovAm | NORMAL | 2021-10-07T10:54:16.705950+00:00 | 2021-10-07T10:54:16.705984+00:00 | 1,997 | false | **C++ :**\n\n```\nvoid preorderTraversalHelper(TreeNode* root, vector<int> &res) {\n\tif(root)\n\t{\n\t\tres.push_back(root -> val); \n\n\t\tif(root -> left)\n\t\t\tpreorderTraversalHelper(root -> left, res);\n\n\t\tif(root -> right)\n\t\t\tpreorderTraversalHelper(root -> right, res);\n\t}\n\n\treturn;\n}\n\n... | 23 | 1 | ['Recursion', 'C', 'C++'] | 3 |
binary-tree-preorder-traversal | Easy C++ Solution with Explaination | easy-c-solution-with-explaination-by-lon-5jvg | AN UPVOTE WOULD BE HIGHLY APPRECIATED !!!\n\nOverload the comments section with doubts and praises if you have.!!!\n\nTime Complexity: O(n)\nSpace Complexity: O | Longhorn_65 | NORMAL | 2021-05-29T12:48:49.170617+00:00 | 2021-05-29T12:49:46.326568+00:00 | 1,155 | false | **AN UPVOTE WOULD BE HIGHLY APPRECIATED !!!**\n\n*Overload the comments section with doubts and praises if you have.!!!*\n\nTime Complexity: O(n)\nSpace Complexity: O(n)\n\n```\nclass Solution {\npublic:\n// Declairing global variable *ans* so that we don\'t declare it again and again and affect the SC by some factors\... | 21 | 1 | ['C', 'C++'] | 2 |
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