## Amazon Interview Question for SDE-2s

Country: United States
Interview Type: In-Person

Comment hidden because of low score. Click to expand.
0
of 2 vote

``````#include<iostream>
#include<queue>

using namespace std;

class node{
public:
int data;
node *left;
node *right;
node(int d);
bool isLeaf();
};

node::node(int d){
data = d;
left = NULL;
right = NULL;
}

bool node::isLeaf(){
return ((this->right == NULL) && (this->left == NULL));
}

void programFailure(){
cout<<"Invalid pyramid"<<endl;
exit(1);
}

bool isInvalid(node *n){
if((n->left == NULL) ^ (n->right == NULL)){
return true;
}
return false;
}

typedef priority_queue<int> mpq;

void findMaxSum(node* n, int sum, mpq &q){
if(n == NULL){
programFailure();
}
if(isInvalid(n)){
programFailure();
}
sum += n->data;
if(n->isLeaf()){
q.push(sum);
sum -= n->data;
return;
}

findMaxSum(n->left, sum, q);
findMaxSum(n->right, sum, q);
sum -= n->data;
}

int main(){
node *n = new node(55);
n->left = new node(94);
n->right = new node(48);
n->left->left = new node(95);
n->left->right = new node(30);
n->right->left = n->left->right; // 48 -> 30
n->left->left->left = new node(77);
n->left->left->right = new node(71);
n->left->right->left = n->left->left->right; // 30 -> 71
n->left->right->right = new node(26);
n->right->right = new node(96);
n->right->right->left = n->left->right->right;
n->right->right->right = new node(67);

mpq q;
int sum = 0;
findMaxSum(n, sum, q);
cout<<q.top()<<endl;
return 0;
}``````

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

Can you please explain the reason for down vote? That might help me improve.

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

Very nice solution!
1. The function should return the value and not expect main to collect it from the queue (exposing the implementation is generally a bad idea)
2. decreasing sum doesn't have any effect because there is no use of sum afterwards
3. you don't really need the queue to solve this (which I guess is why you got the down vote). It adds space complexity and time complexity for the insertions. It would make sense if the question asked for all the sums sorted but all you need here is the maximal value. Just recursively select the son with the maximal sum and return that value plus the root's value.

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

``````def findMaxPath(a):
maxTotal = 0
i = 0
j = 0
for row in a:
midval = a[i][j]

leftval = rightval = 0
if j > 1:
leftval = a[i][j-1]

if i != 0:
rightval = a[i][j+1]

if midval >= leftval and midval >= rightval:
max = midval
if i > 0:
print("remained at the same column")
else:
print("began journey")
elif leftval >= rightval:
max = leftval
j -= 1
print("walk one step left")
else:
max = rightval
j += 1
print("walk one step right")

maxTotal += max
print(max)
i += 1
print("walk down")

print(maxTotal)

a = [[55], [94,48], [95, 30, 96], [77, 71, 26, 67]]
findMaxPath(a)``````

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

static class Result {
int value;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int [][] a = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j=0; j <= i; j++)
a[i][j] = in.nextInt();
}
int k = findMaxSum(a);
System.out.println(k);
}
private static int findMaxSum(int[][] a) {
int n = a.length;
int sum = 0;
Result result = new Result();
findMaxSumHelper(a, 0, 0, sum, result);
return result.value;
}

private static void findMaxSumHelper(int[][] a, int i, int j, int sum, Result result) {
if (i >= a.length) {
if (result.value < sum)
result.value = sum;
return;
}
int origSum = sum + a[i][j];
findMaxSumHelper(a, i+1, j, origSum, result);
findMaxSumHelper(a, i+1, j+1, origSum, result);
}

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

``````static class Result {
int value;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int [][] a = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j=0; j <= i; j++)
a[i][j] = in.nextInt();
}
int k = findMaxSum(a);
System.out.println(k);
}
private static int findMaxSum(int[][] a) {
int n = a.length;
int sum = 0;
Result result = new Result();
findMaxSumHelper(a, 0, 0, sum, result);
return result.value;
}

private static void findMaxSumHelper(int[][] a, int i, int j, int sum, Result result) {
if (i >= a.length) {
if (result.value < sum)
result.value = sum;
return;
}
int origSum = sum + a[i][j];
findMaxSumHelper(a, i+1, j, origSum, result);
findMaxSumHelper(a, i+1, j+1, origSum, result);
}``````

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

It's basically finding max sum path in binary tree.

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