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488. Zuma Game Free
Sign up488. Zuma Game Online
In the sequel to the original ball-blasting classic, shoot colored balls from your stone frog idol's mouth to make sets of three, but don't let the balls reach the golden skull!
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488. Zuma Game Download
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| packagecom.fishercoder.solutions; |
| /** |
| * 488. Zuma Game |
| * |
| * Think about Zuma Game. You have a row of balls on the table, colored red(R), yellow(Y), blue(B), green(G), and white(W). |
| * You also have several balls in your hand. |
| Each time, you may choose a ball in your hand, and insert it into the row (including the leftmost place and rightmost place). |
| Then, if there is a group of 3 or more balls in the same color touching, remove these balls. Keep doing this until no more balls can be removed. |
| Find the minimal balls you have to insert to remove all the balls on the table. If you cannot remove all the balls, output -1. |
| Examples: |
| Input: 'WRRBBW', 'RB' |
| Output: -1 |
| Explanation: WRRBBW -> WRR[R]BBW -> WBBW -> WBB[B]W -> WW |
| Input: 'WWRRBBWW', 'WRBRW' |
| Output: 2 |
| Explanation: WWRRBBWW -> WWRR[R]BBWW -> WWBBWW -> WWBB[B]WW -> WWWW -> empty |
| Input:'G', 'GGGGG' |
| Output: 2 |
| Explanation: G -> G[G] -> GG[G] -> empty |
| Input: 'RBYYBBRRB', 'YRBGB' |
| Output: 3 |
| Explanation: RBYYBBRRB -> RBYY[Y]BBRRB -> RBBBRRB -> RRRB -> B -> B[B] -> BB[B] -> empty |
| Note: |
| You may assume that the initial row of balls on the table won’t have any 3 or more consecutive balls with the same color. |
| The number of balls on the table won't exceed 20, and the string represents these balls is called 'board' in the input. |
| The number of balls in your hand won't exceed 5, and the string represents these balls is called 'hand' in the input. |
| Both input strings will be non-empty and only contain characters 'R','Y','B','G','W'. |
| */ |
| publicclass_488 { |
| publicstaticclassSolution1 { |
| /** |
| * credit: https://discuss.leetcode.com/topic/79820/short-java-solution-beats-98 |
| * Two layer of recursion, pretty cool! |
| */ |
| int maxcount =6; // the max balls you need will not exceed 6 since 'The number of balls in your hand won't exceed 5' |
| publicintfindMinStep(Stringboard, Stringhand) { |
| int[] handCount =newint[26]; |
| for (int i =0; i < hand.length(); ++i) { |
| ++handCount[hand.charAt(i) -'A']; |
| } |
| int result = dfs(board +'#', handCount); // append a '#' to avoid special process while jboard.length, make the code shorter. |
| return result maxcount ?-1: result; |
| } |
| privateintdfs(Strings, int[] handCount) { |
| s = removeConsecutive(s); |
| if (s.equals('#')) { |
| return0; |
| } |
| int result = maxcount; |
| int need =0; |
| for (int i =0, j =0; j < s.length(); ++j) { |
| if (s.charAt(j) s.charAt(i)) { |
| continue; |
| } |
| need =3- (j - i); //balls need to remove current consecutive balls. |
| if (handCount[s.charAt(i) -'A'] >= need) { |
| handCount[s.charAt(i) -'A'] -= need; |
| result =Math.min(result, need + dfs(s.substring(0, i) + s.substring(j), handCount)); |
| handCount[s.charAt(i) -'A'] += need; |
| } |
| i = j; |
| } |
| return result; |
| } |
| //remove consecutive balls longer than 3 |
| privateStringremoveConsecutive(Stringboard) { |
| for (int i =0, j =0; j < board.length(); ++j) { |
| if (board.charAt(j) board.charAt(i)) { |
| continue; |
| } |
| if (j - i >=3) { |
| return removeConsecutive(board.substring(0, i) + board.substring(j)); |
| } else { |
| i = j; |
| } |
| } |
| return board; |
| } |
| } |
| } |
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