1. Description
You are given an array x of n positive numbers. You start at point (0,0) and moves x[0] metres to the north, then x[1] metres to the west, x[2] metres to the south, x[3] metres to the east and so on. In other words, after each move your direction changes counter-clockwise.
Write a one-pass algorithm with O(1) extra space to determine, if your path crosses itself, or not.
Example 1:
Given x =
[2, 1, 1, 2]
,
┌───┐
│ │
└───┼──>
│
Return true (self crossing)
Example 2:
Given x =
[1, 2, 3, 4]
,
┌──────┐
│ │
│
│
└────────────>
Return false (not self crossing)
Example 3:
Given x =
[1, 1, 1, 1]
,
┌───┐
│ │
└───┼>
Return true (self crossing)
2. Answer
public class Solution { public boolean isSelfCrossing(int[] x) { // Check for initial four values manually. if (x.length < 4) { for (int el : x) { if (el == 0) return true; } return false; } for (int i = 3; i < x.length; i++) { int cur = x[i]; if (cur == 0) return true; // At any point of time, i-1 has to be less than i-3 in order to // intersect. Draw few figures to realize this. if (x[i - 1] <= x[i - 3]) { // Basic case. Straight forward intersection. // ___ // |___|__ // | // if (cur >= x[i - 2]) { return true; } // Special case. if (i >= 5) { // if i-2 edge is less than i-4 th edge then it cannot // intersect no matter what if i < i-2 th edge. // ____ // | _ | // |__| | // | if (x[i - 2] < x[i - 4]) continue; // the intersecting case. // ____ // ___| | // | | | // | | | // |_______| // if ((x[i] + x[i - 4] >= x[i - 2]) && (x[i - 1] + x[i - 5] >= x[i - 3])) return true; } } // equals case // ___ // | | // |___| // if (i >= 4) if (x[i - 1] == x[i - 3] && cur + x[i - 4] == x[i - 2]) return true; } return false; } }
