Task Arrangement -- incomplete

Task schedule: Given a sequence of task like A B C, and a coldtime, which means you need to wait for that much time to start next[same] task. Now---

Input: string , n

Output: the best task finishing sequence.

Eg" Input: AAABBB, 2

Output: AB_AB_AB ("_" means do nothing and wait.) Source: 点击打开链接

Maybe, first see a problem easier than this one. (This version was terrible wrong!)

#include <string>#include <iostream>#include <unordered_map>#include <vector>using namespace std;/*  Give a task and some cool down time k (same task cool down distance).  calcualte the total sum of time to finish the task.*/int totalTime(string str) {  int count = str.size();  unordered_map<char, int> charToIndex;  for(int i = 0; i < str.size(); ++i) {    char tmp = str[i];    char index = i;    if(charToIndex.find(tmp) == charToIndex.end()) {      charToIndex.insert({tmp, index});      continue;    } else {      if(index - charToIndex[tmp] > 2) {        charToIndex[tmp] = index;        continue;      } else {        if(index - charToIndex[tmp] == 1) count += 2;        else count += 1;        charToIndex[tmp] = index;      }    }  }  return count;}int main(void) {  cout << totalTime("AABCB") << endl;}

 The index need to be shifted once found one repeation.

// task cool down time k#include "header.h"using namespace std;int taskCoolDown(vector<int>& nums, int k) {  int totalTime = 0;  int shift = 0;  unordered_map<int, int> timeToIndex;  for(int i = 0; i < nums.size(); ++i) {    if(timeToIndex.find(nums[i]) == timeToIndex.end()) {      timeToIndex.insert({nums[i], i + shift});      totalTime++;    } else {      int tmp = timeToIndex[nums[i]];      if(k + 1 - (i + shift - tmp) > 0) totalTime += k + 2 - (i + shift - tmp);      else totalTime += 1;      if(k + 1 - (i - tmp) > 0) shift += k + 1 - (i - tmp);      else shift = 0;      timeToIndex[nums[i]] = i + shift;    }  }  return totalTime;}int main(void) {  vector<int> nums{1, 1, 2, 1};  vector<int> nums_1{1, 2, 3, 1, 2, 3};  cout << taskCoolDown(nums, 2) << endl;}