Coupang Interview Question for Software Engineer / Developers


Country: United States
Interview Type: Phone Interview




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

- NoOne April 22, 2017 | Flag Reply
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)))+" ");
		}
	}

- rajendra April 23, 2017 | Flag Reply
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.

- L April 21, 2017 | Flag Reply
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;
}

- Arun April 21, 2017 | Flag Reply
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;

}

- Arun April 21, 2017 | Flag Reply
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);
		}

}

- rajendra April 21, 2017 | Flag Reply


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