Compressed Vector Product - Facebook Top Interview Questions
Problem Statement :
You are given two integer lists a and b where each list represents a vector in run-length encoded form. For example, a vector [1, 1, 1, 1, 2, 2, 2, 2, 2] is represented as [4, 1, 5, 2]. (There are 4 ones and 5 twos.) Return the dot product of the two vectors a and b. The dot product of a vector [x1, x2, ..., xn] and [y1, y2, ..., yn] is defined to be x1 * y1 + x2 * y2 + ... + xn * yn. Constraints 1 ≤ n ≤ 200,000 where n is the length of a 1 ≤ m ≤ 200,000 where m is the length of b Example 1 Input a = [4, 1, 5, 2] b = [9, 2] Output 28 Explanation a • b = [1, 1, 1, 1, 2, 2, 2, 2, 2] • [2, 2, 2, 2, 2, 2, 2, 2, 2]
Solution :
Solution in C++ :
int solve(vector<int>& a, vector<int>& b) {
int a1, b1;
long long int sum = 0;
a1 = b1 = 0;
while (a1 < a.size()) {
if (a[a1] == b[b1]) {
sum += a[a1] * a[a1 + 1] * b[b1 + 1];
a1 += 2;
b1 += 2;
} else if (a[a1] > b[b1]) {
a[a1] -= b[b1];
sum += b[b1] * a[a1 + 1] * b[b1 + 1];
b1 += 2;
} else {
b[b1] -= a[a1];
sum += a[a1] * a[a1 + 1] * b[b1 + 1];
a1 += 2;
}
}
return sum;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] a, int[] b) {
int A = 0;
int B = 0;
int ans = 0;
while (A < a.length && B < b.length) {
int m = Math.min(a[A], b[B]);
ans += m * a[A + 1] * b[B + 1];
a[A] -= m;
if (a[A] == 0)
A += 2;
b[B] -= m;
if (b[B] == 0)
B += 2;
}
return ans;
}
}
Solution in Python :
class Solution:
def solve(self, a, b):
ans = 0
while a and b:
a_val = a.pop()
a_count = a.pop()
b_val = b.pop()
b_count = b.pop()
ans += (a_val * b_val) * min(b_count, a_count)
if b_count > a_count:
b.append(abs(b_count - a_count))
b.append(b_val)
elif a_count > b_count:
a.append(abs(b_count - a_count))
a.append(a_val)
return ans
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