Update code

Author: rongyi

bookcode/triangle.cc

@@ -0,0 +1,25 @@
+#include "../xxx.h"
+
+int cnt(vector<int> &edges) {
+  int sz = edges.size();
+  int ret = 0;
+  for (int i = 0; i < sz; i++) {
+    for (int j = i + 1; j < sz; j++) {
+      for (int k = j + 1; k < sz; k++) {
+        int sum = edges[i] + edges[j] + edges[k];
+        int max_len = max({edges[i], edges[j], edges[k]});
+        int tw_sum = sum - max_len;
+        if (tw_sum > max_len) {
+          ret++;
+        }
+      }
+    }
+  }
+
+  return ret;
+}
+
+int main(int argc, char *argv[]) {
+  vector<int> edges;
+  return 0;
+}

cc/rearranging-fruits.cc

@@ -0,0 +1,48 @@
+// https://leetcode.com/problems/rearranging-fruits/
+#include "xxx.hpp"
+
+class Solution {
+public:
+  long long minCost(vector<int> &basket1, vector<int> &basket2) {
+    map<int, int> cnt;
+    for (auto &num : basket1) {
+      cnt[num]++;
+    }
+    for (auto &num : basket2) {
+      // not plus, minus
+      // means the one with zero value is kept in this array
+      cnt[num]--;
+    }
+    vector<int> exch;
+    for (auto &kv : cnt) {
+      // odd count, can not split
+      if (kv.second & 1) {
+        return -1;
+      }
+      for (int i = 0; i < abs(kv.second) / 2; i++) {
+        exch.push_back(kv.first);
+      }
+    }
+    int sz = exch.size();
+    nth_element(exch.begin(), exch.begin() + sz / 2, exch.end());
+    int minx = min(*min_element(basket1.begin(), basket1.end()),
+                   *min_element(basket2.begin(), basket2.end()));
+
+    // Within each pair, there are two exchange methods:
+    // 1. Direct exchange, the cost is the smaller number.
+    // 2. Indirect exchange, choose another small number xxx as the
+    // "intermediate", swap ai with x ans x with bj, the cost is 2x
+    return accumulate(exch.begin(), exch.begin() + sz / 2, 0ll,
+                      [&](long long acc, int cur) -> long long {
+                        return acc + min(2 * minx, cur);
+                      });
+  }
+};
+
+int main(int argc, char *argv[]) {
+  Solution so;
+  vector<int> input1{4, 2, 2, 2};
+  vector<int> input2{1, 4, 1, 2};
+  cout << so.minCost(input1, input2) << endl;
+  return 0;
+}