## Amazon student Interview Question for Students

Country: Germany
Interview Type: Phone Interview

Comment hidden because of low score. Click to expand.
1
of 1 vote

``````private int getPath(int start, int end, boolean[] vis, int[] pathLen)
{
if(start == end) return 0;
if(pathLen[start] != -1) return pathLen[start];

int len = m_Matrix.length + 1;

vis[start] = true;
for(int i = 0; i < m_Matrix.length; ++i){
if(vis[m_Matrix[start][i]]) continue;
len = Math.min(len, 1 + getPath(i, end, vis, pathLen));
}
vis[start] = false;

return pathLen[start] = len;
}
public int shortestPath(int start, int end)
{
if(start == end) return 0;
if(m_Matrix[start][end]) return 1;

boolean[] vis = new boolean[m_Matrix.length];
Arrays.fill(vis, false);
int pathLen = new int[m_Matrix.length];
Arrays.fill(pathLen, -1);

int len = getPath(start, end, vis);
return len > m_Matrix.length ? -1 : len;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

shouldn't we get a pair of ints for start, and another pair for end?

Comment hidden because of low score. Click to expand.
0
of 0 vote

by adding int[] parent, you can run BFS from the start. Set all nodes' parent to -infinity, and start to 0. Whenever you visit a node because it is adjacent to another already visited node and its parent is not set, set the parent to the current node.

At the end, by finding end's parent, end's parent's parent's , .... you have the path backward. Reverse it.

Name:

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