LeetCode 723: Candy Crush Solution
Master LeetCode problem 723 (Candy Crush), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
723. Candy Crush
Problem Explanation
Explanation:
To solve this problem, we can use a brute-force approach where we iterate through the board to find and remove any contiguous groups of 3 or more same candies. We can use a flag to mark cells that need to be removed, then perform gravity to move candies down and fill in empty spaces.
- Iterate through the board and mark cells with the same candies in vertical or horizontal directions.
- If any cell is marked, set it to 0.
- Perform gravity by moving candies down and fill in empty spaces.
- Repeat the process until no more candies can be crushed.
Time Complexity:
The time complexity of this approach is O((R * C)^2), where R is the number of rows and C is the number of columns in the board.
Space Complexity:
The space complexity is O(1) as we are modifying the input board in place without using any additional data structures.
:
Solution Code
class Solution {
public int[][] candyCrush(int[][] board) {
int R = board.length, C = board[0].length;
boolean found = true;
while (found) {
found = false;
for (int r = 0; r < R; r++) {
for (int c = 0; c < C - 2; c++) {
int val = Math.abs(board[r][c]);
if (val != 0 && val == Math.abs(board[r][c + 1]) && val == Math.abs(board[r][c + 2])) {
board[r][c] = board[r][c + 1] = board[r][c + 2] = -val;
found = true;
}
}
}
for (int r = 0; r < R - 2; r++) {
for (int c = 0; c < C; c++) {
int val = Math.abs(board[r][c]);
if (val != 0 && val == Math.abs(board[r + 1][c]) && val == Math.abs(board[r + 2][c])) {
board[r][c] = board[r + 1][c] = board[r + 2][c] = -val;
found = true;
}
}
}
if (found) {
for (int c = 0; c < C; c++) {
int idx = R - 1;
for (int r = R - 1; r >= 0; r--) {
if (board[r][c] > 0) {
board[idx--][c] = board[r][c];
}
}
while (idx >= 0) {
board[idx--][c] = 0;
}
}
}
}
return board;
}
}Try It Yourself
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