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Solution
Since query(i, j) gets frequently called, it should run as fast as possible, ideally in O(1) time.
I. When there is 0 between i and j, the query returns 0.
To quickly find out 0 between i and j, initial another array to keep track of at the current index i, where is the closest 0 to the left of i (including i).

II. To get the product of any given range without zero in between, build an array of cumulative products from the first positive number in the non-zero part up to current idx i. So that if there is no zero in between, query(i, j) equals to product[j] / product[i - 1] (O(1) time).
e.g. For A = [2, 2, 3, 4, 0, 4, 5, 6]
product array P = [2, 4, 12, 48 ,0, 4, 20 ,120]
query(1, 3) = P[3] / P[1 - 1] = 48 / 2 = 24

Remember to discuss with the interviewer if the product can get integer overflow

public class MultiplicationQueries {
    int[] array;
    int[] zeroToLeft;
    int[] product;

    public MultiplicationQueries(int[] a) {
        array = a;
        if(array.length > 0) {
            zeroToLeft = new int[array.length];
            product = new int[array.length];
            zeroToLeft[0] = array[0] == 0 ? 0: -1;
            product[0] = array[0] == 0 ? 0: array[0];
        }
        for(int i = 1; i < array.length; i++) {
            zeroToLeft[i] = array[i] == 0 ? i: zeroToLeft[i - 1];
            product[i] = array[i] == 0 ? 0: array[i - 1] == 0 ? array[i]: array[i] * product[i - 1];
        }
    }

    public int query(int i, int j) {
        if(i > j || i >= array.length || j >= array.length || i < 0 || j < 0) return -1;
        if(i == j) return array[i];
        return zeroToLeft[j] >= i ? 0: (i == 0 || array[i - 1] == 0) ? product[j]: product[j] / product[i - 1];
    }
}

Solution to problem number 2 in python.

def flatten(seq):
    r = []
    for x in seq:
        if type(x) == list:
            r += flatten(x)
        else:
            r.append(x)
    return r

product Ai....Aj

/*Apple Map Team 
1. Given an array A and some queries, query(i, j) returns the result of Ai*...*Aj, in other words the multiplication from Ai to Aj. 
The numbers in A are non-negative. 
Implement query(i, j). 
*/
public class RetrunResultOfAiAj {

	public static int findProd(int i,int j, int[] a){
		if(a==null){
			return 0;
		}
		if(i>a.length-1 || j>a.length-1 || i<0 || j<0){
			return 0;
		}
//		if(i==j){
//			return a[i]*a[j];
//		}
		int m = j-i+1;
		int product = 1;
		if(m>=1){
			product = a[i] * findProd(i+1,j,a);
		}
		return product;
	}
	public static void main(String[] args){
		int a[] = {1,2,3,4,5,6,7};
		int b[] = {1,2,3,0,5,6,7};
		int c[] = {1};
		int d[] = {};
		int e[] = {1,1,1,1,1,1,0};
		System.out.println(RetrunResultOfAiAj.findProd(1, 4, a));
		System.out.println(RetrunResultOfAiAj.findProd(1, 4, b));
		System.out.println(RetrunResultOfAiAj.findProd(1, 4, c));
		System.out.println(RetrunResultOfAiAj.findProd(1, 4, d));
		System.out.println(RetrunResultOfAiAj.findProd(1, 4, e));
		
		System.out.println(RetrunResultOfAiAj.findProd(1, 14, b));
//		System.out.println(RetrunResultOfAiAj.findProd(1, 1, b)); wont work if i and j is same.
	}
}

public static int findProd(int i,int j, int[] a){
        int product=1;

        if(i<0)return 0;
        if(j>a.length)return 0;

        for(int k=i;k<=j;k++){
            product*=a[k];
        }

        return product;
    }