int_as_array_multiply (python, cpp)

Author: jsgoller1

elements-of-programming-interviews/cpp/int_as_array_multiply.cc

@@ -1,10 +1,37 @@
+#include <algorithm>
 #include <vector>
 
 #include "test_framework/generic_test.h"
 using std::vector;
 vector<int> Multiply(vector<int> num1, vector<int> num2) {
-  // TODO - you fill in here.
-  return {};
+  int neg = ((num1[0] < 0) ^ (num2[0] < 0)) ? -1 : 1;
+  num1[0] = abs(num1[0]);
+  num2[0] = abs(num2[0]);
+
+  // multiplying a m digit number by an n digit number
+  // will produce a number with no more than m+n digits.
+  vector<int> result(num1.size() + num2.size(), 0);
+  for (int i = num1.size() - 1; i >= 0; i--) {
+    for (int j = num2.size() - 1; j >= 0; j--) {
+      // This is a clever technique for handling overflow and carry;
+      // each digit is treated as a ones place integer; any multiplication that
+      // exceeds 10 (max possible is 9x9=81) adds the 10s place overflow to the
+      // next digit and uses the ones place for the present digit.
+      result[i + j + 1] += num1[i] * num2[j];
+      result[i + j] += result[i + j + 1] / 10;
+      result[i + j + 1] %= 10;
+    }
+  }
+
+  // Then just remove the leading zeroes, and return the result.
+  result = {
+      find_if_not(result.begin(), result.end(), [](int a) { return a == 0; }),
+      result.end()};
+  if (result.empty()) {
+    return {0};
+  }
+  result[0] *= neg;
+  return result;
 }
 
 int main(int argc, char* argv[]) {

elements-of-programming-interviews/problem_mapping.js

@@ -214,15 +214,15 @@ problem_mapping = {
         },
         "5.03 Multiply two arbitrary-precision integers": {
             "C++: int_as_array_multiply.cc": {
-                "passed": 0,
+                "passed": 1000,
                 "total": 1000
             },
             "Java: IntAsArrayMultiply.java": {
                 "passed": 0,
                 "total": 1000
             },
             "Python: int_as_array_multiply.py": {
-                "passed": 0,
+                "passed": 1000,
                 "total": 1000
             }
         },

elements-of-programming-interviews/python/int_as_array_multiply.py

@@ -4,8 +4,28 @@
 
 
 def multiply(num1: List[int], num2: List[int]) -> List[int]:
-    # TODO - you fill in here.
-    return []
+    neg = -1 if (num1[0] < 0) ^ (num2[0] < 0) else 1
+    num1[0] = abs(num1[0])
+    num2[0] = abs(num2[0])
+
+    total = 0
+    place = 1
+    for digit1 in num1[::-1]:
+        curr_place = place
+        curr_total = 0
+        for digit2 in num2[::-1]:
+            curr_total += digit1 * digit2 * curr_place
+            curr_place *= 10
+        total += curr_total
+        place *= 10
+
+    solution = [] if total else [0]
+    while total:
+        solution.append(total % 10)
+        total = total // 10
+    solution[-1] *= neg
+
+    return solution[::-1]
 
 
 if __name__ == '__main__':