Goldman Sachs Interview Question for Software Engineer / Developers


Country: India
Interview Type: Written Test




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why we have to remove 3 from A since it is available in the B, in the above test case.

- MO August 24, 2020 | Flag Reply
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Firstly map each value of array B with index, say in "map datastructure"
3->1
4->2
...
7->5

 After that update the value of array A which are in B as the key value from "map"
A = [1,3,4,2,5,6] after updating we get C as
C = [   1,2,   4,3]
from C find the longest increasing subsequence and subtract the length of it from length of B.

ans =  ( len(B)-len(C))

- Priyank September 02, 2020 | Flag Reply
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// As I know, we dont remove 3
// A = [1,3,4,2,5,6], B = [3,4,6,5,7]
// We need find C = maximum run way from A to B = max |S = [a_i]: a_i belongs B and respect order in A and B| = [3,4,5]
// answer = |B| - |C|

int max_run_way(std::vector<int> a, std::vector<int> b) {
size_t ia = 0, ib = 0;
while (ia < a.size() && ib < b.size()) {
if (a[ia++] == b[ib]) {
++ib;
}
}
return b.size() - ib;
}

- thinhnd October 24, 2020 | Flag Reply
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0
of 0 vote

public class MinimumArrayInsertions {

    public static void main(String[] args) {
        MinimumArrayInsertions app = new MinimumArrayInsertions();

        System.out.println(app.minimumInsertions(new int[]{1, 3, 4, 2, 5, 6}, new int[]{3, 4, 6, 5, 7}));
    }

    public int minimumInsertions(int[] arrA, int[] arrB) {
        int i = 0;//min insertions
        int al = arrA.length;//element counter in A
        for (int n = 0; n < arrB.length;) {
            if (n <= al - 1) {
                //good to compare and shift until there is equivalence at this position and increment n++
                if (arrB[n] == arrA[n]) {
                    //nice
                    n += 1;
                } else {
                    //delete in A and shift and loop
                    for (int p = n; p < al - 1; p++) {
                        arrA[p] = arrA[p + 1];
                    }
                    //reduce arrA length
                    al -= 1;
                    //maintin value n and recheck again
                }
            } else {
                //concat all arrB[n] to arrB[arrB.length-1] and count insertions
                i = i + (arrB.length - n);
                //set value of n to size of B
                n = arrB.length;
            }
        }
        return i;
    }
}

- kirikanjoroge January 28, 2021 | Flag Reply


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