## Coupang Interview Question for Software Engineer / Developers

• 0
of 0 votes

Country: United States
Interview Type: Phone Interview

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

Forced to give downvote to all. Think about it. What you are going to do when we have 0's lets say? One zero?

input: [1 2 0 5 6]
output: [ 0 0 60 0 0 ]

Hmm.
That was the trick all along.

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

``````void printArrayMultiplicationBI(int[] array) {
BigInteger multiplication = new BigInteger("1");
for (int a : array) {
multiplication = multiplication.multiply(new BigInteger(Integer.toString(a)));
}
for (int a : array) {
System.out.print(multiplication.divide(new BigInteger(Integer.toString(a)))+" ");
}
}``````

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

function result(a []int, n int){
max=1;
for i=0;i<n;i++{
max=max*a[i];
}

a[0]=max;
for i=1;i<n;i++{
a[i]=max/a[i];

}
return a

}

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

Multiply all the numbers in first iteration.
In second iteration, divide the above number with current element.

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

public static void result(int array[], int k){
int result =1;
//Time complexity O(n)
for(int i = 0 ; i < k-1 ; i++){
result *=array[i];
}
for(int i = k; i < array.length ; i++){
result *=array[i];
}
array[k-1] = result;
}

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

``````public static void result(int array[], int k){
int result =1;
//Time complexity O(n)
for(int i = 0 ; i < k-1 ; i++){
result *=array[i];
}
for(int i = k; i < array.length ; i++){
result *=array[i];
}
array[k-1] = result;``````

}

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

``````static void printArrayMultiplication(int[] array) {
int multiplication = 1;
for (int a : array) {
multiplication *= a;
}
for (int a : array) {
System.out.println(multiplication / a);
}``````

}

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