2021/winter interview training for facebook/google

puzzles/training/array_inversion_count.cpp

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+/*
+VIM: let b:cf5build="clang -std=c++20 -O2 -pthread -lstdc++ -I. {SRC} -o {OUT}"
+VIM: let b:cf5run="{OUT}"
+*/
+#include <iostream>
+#include <exception>
+#include <chrono>
+#include <string>
+#include <vector>
+#include <unordered_map>
+#include <unordered_set>
+#include <functional>
+
+/*
+https://app.codility.com/programmers/lessons/99-future_training/array_inversion_count/
+
+An array A consisting of N integers is given. An inversion is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].
+
+Write a function:
+
+    int solution(vector<int> &A);
+
+that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.
+
+For example, in the following array:
+ A[0] = -1 A[1] = 6 A[2] = 3
+ A[3] =  4 A[4] = 7 A[5] = 4
+
+there are four inversions:
+   (1,2)  (1,3)  (1,5)  (4,5)
+
+so the function should return 4.
+
+Write an efficient algorithm for the following assumptions:
+
+        N is an integer within the range [0..100,000];
+        each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
+
+*/
+
+int solution(const std::vector<int>& v)
+{
+    std::vector<size_t> iv(v.size(),0);
+    for (size_t i=0; i<v.size(); ++i){
+        iv[i] = i;
+    }
+    std::stable_sort(iv.begin(), iv.end(), [&v](size_t op1, size_t op2){
+        return v[op1] < v[op2];
+    });
+
+    size_t count = 0;
+    for (size_t i=0; i < iv.size(); ++i){
+        if ( i < iv[i]){
+            count += iv[i] - i;
+        }
+    }
+
+    return count;
+}
+
+int main ( void )
+{try{
+    auto begin = std::chrono::high_resolution_clock::now();
+
+    std::cout << solution({-1, 6, 3, 4, 7, 4}) << std::endl;
+
+    std::cout << solution({0, 0, 0, 1, 1, 1}) << std::endl;
+    std::cout << solution({0, 1, 0, 1, 1, 1}) << std::endl;
+    std::cout << solution({0}) << std::endl;
+    std::cout << solution({}) << std::endl;
+
+
+    auto end = std::chrono::high_resolution_clock::now();
+    std::chrono::duration<double> seconds = end - begin;
+    std::cout << "Time: " << seconds.count() << std::endl;
+    return 0;
+}
+catch ( const std::exception& e )
+{
+    std::cerr << std::endl
+            << "std::exception(\"" << e.what() << "\")." << std::endl;
+    return 2;
+}
+catch ( ... )
+{
+    std::cerr  << std::endl
+            << "unknown exception." << std::endl;
+    return 1;
+}}