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Commitffe2aee

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‎README.md

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|338|[Counting Bits](https://leetcode.com/problems/counting-bits/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_338.java)| O(nlogn)|O(h)| Medium|
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|337|[House Robber III](https://leetcode.com/problems/house-robber-iii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_337.java)| O(n)|O(n)| Medium | DP
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|336|[Palindrome Pairs](https://leetcode.com/problems/palindrome-pairs/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_336.java)| O(n^2)|O(n)| Hard|
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|335|[Self Crossing](https://leetcode.com/problems/self-crossing/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_335.java)| O(n)|O(1) | Hard| Math
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|334|[Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_334.java)| O(n^2)|O(1)| Medium|
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|333|[Largest BST Subtree](https://leetcode.com/problems/largest-bst-subtree/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_333.java)| O(n)|O(n) | Medium| Tree
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|332|[Reconstruct Itinerary](https://leetcode.com/problems/reconstruct-itinerary/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_332.java)| O(n)|O(n) | Medium| Graph, DFS
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packagecom.fishercoder.solutions;
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/**
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* 335. Self Crossing
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*
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* You are given an array x of n positive numbers.
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* You start at point (0,0) and moves x[0] metres to the north,
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* then x[1] metres to the west,
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* x[2] metres to the south,
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* x[3] metres to the east and so on.
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* In other words, after each move your direction changes counter-clockwise.
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Write a one-pass algorithm with O(1) extra space to determine, if your path crosses itself, or not.
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Example 1:
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Given x =
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[2, 1, 1, 2]
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,
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?????
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? ?
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???????>
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?
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Return true (self crossing)
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Example 2:
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Given x =
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[1, 2, 3, 4]
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,
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????????
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? ?
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?
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?
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?????????????>
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Return false (not self crossing)
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Example 3:
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Given x =
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[1, 1, 1, 1]
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,
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?????
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? ?
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?????>
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Return true (self crossing)
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*/
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publicclass_335 {
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/**reference: https://discuss.leetcode.com/topic/38014/java-oms-with-explanation/2*/
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publicbooleanisSelfCrossing(int[]x) {
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intl =x.length;
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if (l <=3)returnfalse;
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for (inti =3;i <l;i++) {
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if (x[i] >=x[i -2] &&x[i -1] <=x[i -3]) {
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returntrue;//Fourth line crosses first line and onward
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}
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if (i >=4) {
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if (x[i -1] ==x[i -3] &&x[i] +x[i -4] >=x[i -2]) {
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returntrue;// Fifth line meets first line and onward
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}
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}
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if (i >=5) {
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if (x[i -2] -x[i -4] >=0 &&x[i] >=x[i -2] -x[i -4] &&x[i -1] >=x[i -3] -x[i -5] &&x[i -1] <=x[i -3]) {
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returntrue;// Sixth line crosses first line and onward
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}
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}
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}
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returnfalse;
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}
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}

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