Leetcode Self Crossing problem solution

YASH PAL, 31 July 202421 January 2026

In this Leetcode Self Crossing problem solution, You are given an array of integers distance.

You start at point (0,0) on an X-Y plane and you move distance[0] meters to the north, then distance[1] meters to the west, distance[2] meters to the south, distance[3] meters to the east, and so on. In other words, after each move, your direction changes counter-clockwise. Return true if your path crosses itself, and false if it does not.

Leetcode Self Crossing problem solution

Leetcode Self Crossing problem solution in Python.

class Solution:
    def isSelfCrossing(self, x):
        """
        :type x: List[int]
        :rtype: bool
        """
        if len(x) < 4:
            return False
        x.append(0)
        x.append(0)
        for i in range(2,len(x)-2):
            if x[i] <= x[i-2]:
                tmp = x[i-2] - x[i-4]
                if tmp > 0:
                    if x[i]>=tmp:
                        if x[i+1] != 0 and x[i+1] >= x[i-1]-x[i-3]:
                            return True
                    else:
                        if x[i+1] >= x[i-1]:
                            return True
                else:
                    if x[i+1] >= x[i-1]:
                        return True
        return False

Self Crossing problem solution in Java.

public boolean isSelfCrossingExpand(int[] x, int start){
        for (int i=start;i x[i-2]){
                continue;
            } else {
                int bar = (i>=4) ? x[i-4] : 0;
                if (x[i] <= x[i-2] && x[i] >= (x[i-2] - bar)){
                    if ( i+1 == x.length){
                        return false;
                    }
                    if (x[i+1] >= (x[i-1] - x[i-3])){
                        return true;
                    }
                    return isSelfCrossingShrink(x,i+2);
                } else {
                    return isSelfCrossingShrink(x,i+1);
                }
            }        
        }
        return false;
    }

    public boolean isSelfCrossingShrink(int[] x, int start){
        for (int i=start; i= x[i-2])
                return true;
        }
        return false;
    }
    
    public boolean isSelfCrossing(int[] x) {
        if (x.length<=3) return false;

        if (x[2] <= x[0]) return isSelfCrossingShrink(x,3);        
        return isSelfCrossingExpand(x,3);
        
    }
}

Problem solution in C++.

class Solution {
public:
    bool isSelfCrossing(vector<int>& x) {
        
        if(x.size() < 4)
            return false;
        
        for(int i = 3; i < x.size(); i++) {
            if(x[i] >= x[i-2] and x[i-1] <= x[i-3])
                return true;
            if(i >= 4) {
                if(x[i-1] == x[i-3] and x[i] >= x[i-2]-x[i-4])
                    return true;
            }
            if(i >= 5) {
                if(x[i-2]-x[i-4] >= 0 and x[i] >= x[i-2]-x[i-4] and
                   x[i-1] >= x[i-3]-x[i-5] and x[i-1] <= x[i-3]) 
                    return true;
            }
        }
        return false;
    }
};

Problem solution in C.

bool isSelfCrossing(int* x, int xSize) {

int i;
for (i = 2; i < xSize && x[i] > x[i-2]; i++);
 
if ((++i < xSize) && ((i > 4 && x[i-1]+x[i-5]>=x[i-3]) || (i==4 && x[i-1]==x[i-3])) && x[i]+x[i-4] >= x[i-2]) return true;

for (; i < xSize && x[i] < x[i-2]; i++);

return (i < xSize);
}

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