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;
}
}
												

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