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README.md

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|652|[Find Duplicate Subtrees](https://leetcode.com/problems/find-duplicate-subtrees/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_652.java) | O(n) |O(n) | Medium | Tree
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|651|[4 Keys Keyboard](https://leetcode.com/problems/4-keys-keyboard/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_651.java) | O(n^2) |O(n) | Medium | DP
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|650|[2 Keys Keyboard](https://leetcode.com/problems/2-keys-keyboard/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_650.java) | O(n^2) |O(n) | Medium | DP
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|649|[Dota2 Senate](https://leetcode.com/problems/dota2-senate/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_649.java) | O(n) |O(n) | Medium | Greedy
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|648|[Replace Words](https://leetcode.com/problems/replace-words/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_648.java) | O(n) |O(n) | Medium | Trie
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|647|[Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_647.java) | O(n^2) |O(1) | Medium | DP
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|646|[Maximum Length of Pair Chain](https://leetcode.com/problems/maximum-length-of-pair-chain/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_646.java) | O(nlogn) |O(1) | Medium | DP
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package com.fishercoder.solutions;
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import java.util.LinkedList;
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import java.util.Queue;
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/**
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* 649. Dota2 Senate
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*
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* In the world of Dota2, there are two parties: the Radiant and the Dire.
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The Dota2 senate consists of senators coming from two parties.
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Now the senate wants to make a decision about a change in the Dota2 game.
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The voting for this change is a round-based procedure.
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In each round, each senator can exercise one of the two rights:
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1. Ban one senator's right:
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A senator can make another senator lose all his rights in this and all the following rounds.
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2. Announce the victory:
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If this senator found the senators who still have rights to vote are all from the same party,
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he can announce the victory and make the decision about the change in the game.
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Given a string representing each senator's party belonging.
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The character 'R' and 'D' represent the Radiant party and the Dire party respectively.
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Then if there are n senators, the size of the given string will be n.
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The round-based procedure starts from the first senator to the last senator in the given order.
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This procedure will last until the end of voting.
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All the senators who have lost their rights will be skipped during the procedure.
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Suppose every senator is smart enough and will play the best strategy for his own party,
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you need to predict which party will finally announce the victory and make the change in the Dota2 game.
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The output should be Radiant or Dire.
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Example 1:
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Input: "RD"
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Output: "Radiant"
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Explanation: The first senator comes from Radiant and he can just ban the next senator's right in the round 1.
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And the second senator can't exercise any rights any more since his right has been banned.
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And in the round 2, the first senator can just announce the victory since he is the only guy in the senate who can vote.
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Example 2:
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Input: "RDD"
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Output: "Dire"
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Explanation:
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The first senator comes from Radiant and he can just ban the next senator's right in the round 1.
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And the second senator can't exercise any rights anymore since his right has been banned.
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And the third senator comes from Dire and he can ban the first senator's right in the round 1.
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And in the round 2, the third senator can just announce the victory since he is the only guy in the senate who can vote.
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Note:
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The length of the given string will in the range [1, 10,000].
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*/
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public class _649 {
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public String predictPartyVictory(String senate) {
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Queue<Integer> radiantQ = new LinkedList<>();
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Queue<Integer> direQ = new LinkedList<>();
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int len = senate.length();
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for (int i = 0; i < len; i++) {
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if (senate.charAt(i) == 'R') radiantQ.offer(i);
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else direQ.offer(i);
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}
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while (!radiantQ.isEmpty() && !direQ.isEmpty()) {
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int radiantIndex = radiantQ.poll();
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int direIndex = direQ.poll();
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if (radiantIndex < direIndex) {
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/**Radiant will ban Dire in this case, so we'll add radiant index back to the queue plus n*/
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radiantQ.offer(radiantIndex + len);
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} else {
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direQ.offer(direIndex + len);
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}
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}
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return radiantQ.isEmpty() ? "Dire" : "Radiant";
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}
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}
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package com.fishercoder;
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import com.fishercoder.solutions._649;
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import org.junit.BeforeClass;
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import org.junit.Test;
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import static org.junit.Assert.assertEquals;
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/**
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* Created by fishercoder on 5/8/17.
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*/
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public class _649Test {
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private static _649 test;
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@BeforeClass
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public static void setup(){
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test = new _649();
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}
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@Test
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public void test1(){
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assertEquals("Dire", test.predictPartyVictory("RDDDR"));
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}
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@Test
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public void test2(){
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assertEquals("Radiant", test.predictPartyVictory("RD"));
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}
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@Test
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public void test3(){
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assertEquals("Dire", test.predictPartyVictory("RDD"));
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}
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@Test
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public void test4(){
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assertEquals("Radiant", test.predictPartyVictory("RDDR"));
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}
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@Test
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public void test5(){
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assertEquals("Dire", test.predictPartyVictory("RDDRD"));
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}
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@Test
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public void test6(){
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assertEquals("Dire", test.predictPartyVictory("RDDDDDRR"));
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}
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@Test
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public void test7(){
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assertEquals("Dire", test.predictPartyVictory("RDDDDRR"));
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}
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}

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