Amazon Interview Question for Software Engineers


Country: United States
Interview Type: In-Person




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

This is a 'merge set' question. Given a graph, figure out which nodes belong to the same connected component and put them into a set.
Since the input comes in as an edge set, UNION FIND will be a good way to solve this.

Initially every node sources to itself. As we read the statement X = Y, we point the source of Y to the source of X so that they join the same set. After all connected components are sorted out. We check the unequal statements X != Y. If any of the X, Y pairs do share the same source, then X != Y contradicts with the equal statements.

public class CheckStatements {
    public boolean validStatements(char[][] equal, char[][] unequal) {
        int[] sets = new int[26];
        for(int i = 0; i < 26; i++) {
            sets[i] = i;
        }

        for(char[] pair: equal) {
            mergeSets(sets, pair[0] - 'A', pair[1] - 'A');
        }
        for(int i = 0; i < 26; i++) {
            findSrc(sets, i);
        }

        for(char[] pair: unequal) {
            if(sets[pair[0] - 'A'] == sets[pair[1] - 'A']) return false;
        }
        return true;
    }

    private int findSrc(int[] sets, int i) {
        int src = i;
        while(src != sets[src]) {
            src = sets[src];
        }
        int tmp;
        while (i != sets[i]) {
            tmp = sets[i];
            sets[i] = src;
            i = tmp;
        }
        return src;
    }

    private void mergeSets(int[] sets, int a, int b) {
        int srcA = findSrc(sets, a);
        int srcB = findSrc(sets, b);
        sets[srcB] = srcA;
    }
}

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- aonecoding4 December 25, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public bool CheckStatements(char[,] seta, char[,] setb){

            List<HashSet<char>> setGroups = new List<HashSet<char>>();

            int len_seta_row = seta.GetLength(0);

            Dictionary<char, char> map = new Dictionary<char, char>();

            // Group the first set into the groups.
            for(int i = 0; i < len_seta_row; i++){
                
                // if both the char's are new, then add both to dictionary.
                // if A = B add map.Add('A','A'); map.Add('B', 'A');
                if(!map.ContainsKey(seta[i, 0]) && !map.ContainsKey(seta[i,1])){
                    map.Add(seta[i,0], seta[i,0]);
                    map.Add(seta[i,1], seta[i,0]);
                }else if (map.ContainsKey(seta[i,0]) && !map.ContainsKey(seta[i,1]))
                {
                    //if  B = C, then map.Add('C', 'A');
                    var key = seta[i,1];
                    var value = map[seta[i,0]];
                    map.Add(key,value);
                }else{ //if D = C, then map.Add('D', 'A');
                    var key = seta[i,0];
                    var value = map[seta[i,1]];
                    map.Add(key,value);
                }
                
            }

            //Check for the setb if the pairing is correct.

            int len2_setb_row = setb.GetLength(0);

            for(int i = 0; i < len2_setb_row; i++){

                if(map[setb[i,0]] == map[setb[i,1]]){
                    return false;
                }
            }

            return true;
        }

- ssimhadri December 29, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class ValidateEquations {

private static int[] parent = new int[27];

private static int[] rank = new int[27];
static {
for (int i = 0; i < 26; i++)
parent[i] = i;
}

public static int getParent(int x) {
if (parent[x] != x)
parent[x] = getParent(parent[x]);

return parent[x];
}

public static void union(int x, int y) {
int rootx = parent[x];
int rooty = parent[y];
if (rootx != rooty) {
if (rank[rootx] > rank[rooty]) {
parent[rooty] = rootx;
} else if (rank[rooty] >= rank[rootx]) {
parent[rootx] = rooty;
}
} else {
parent[rooty] = rootx;
rank[rooty] += 1;
}
}

public static boolean validStatements(char[][] equal, char[][] unequal) {

for (char[] pair : equal) {
union(pair[0] - 'A' + 1, pair[1] - 'A' + 1);
// System.out.println((pair[0] - 'A') + " , " + (pair[1] - 'A'));
}

for (char[] opp : unequal) {
if (getParent(opp[0] - 'A' + 1) == getParent(opp[1] - 'A' + 1))
return false;
}

return true;
}

public static void main(String[] args) {
char[][] equal = { { 'A', 'B' }, { 'B', 'D' }, { 'C', 'D' }, { 'F', 'G' }, { 'E', 'H' }, { 'H', 'C' } };
char[][] unequal = { { 'A', 'N' } };

boolean res = validStatements(equal, unequal);
System.out.println(res);
}

- Anonymous January 01, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static boolean validStatements(String[][] equal, String[][] unequal) {
        HashMap<String,Integer> hm = new HashMap<String,Integer>();
        for(String[] pair: equal) {
            hm.put(pair[0]+pair[1],1);
            hm.put(pair[1]+pair[0],1);
        }
        for(String[] pair: unequal) {
            if (hm.containsKey(pair[1]+pair[0]) || hm.containsKey(pair[0]+pair[1])) {
               return false;
            }
        }
        return true;
    }

- John Doe January 05, 2019 | Flag Reply


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