Maximum possible time that can be formed from four digits in C++

C++Server Side ProgrammingProgramming

In this tutorial, we will be discussing a program to find maximum possible time that can be formed from four digits.

For this we will be provided with an array consisting 4 digits. Our task is to find the maximum time (24 hour format) that can formed using those four digits.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
//returning updated frequency map
map<int, int> getFrequencyMap(int arr[], int n) {
   map<int, int> hashMap;
   for (int i = 0; i < n; i++) {
      hashMap[arr[i]]++;
   }
   return hashMap;
}
//checking if the digit is present in frequency map
bool hasDigit(map<int, int>* hashMap, int digit) {
   if ((*hashMap)[digit]) {
      (*hashMap)[digit]--;
      return true;
   }
   return false;
}
//returning maximum time in 24 hour format
string getMaxtime_value(int arr[], int n) {
   map<int, int> hashMap = getFrequencyMap(arr, n);
   int i;
   bool flag;
   string time_value = "";
   flag = false;
   for (i = 2; i >= 0; i--) {
      if (hasDigit(&hashMap, i)) {
         flag = true;
         time_value += (char)i + 48;
         break;
      }
   }
   if (!flag)
      return "-1";
   flag = false;
   if (time_value[0] == '2') {
      for (i = 3; i >= 0; i--) {
         if (hasDigit(&hashMap, i)) {
            flag = true;
            time_value += (char)i + 48;
            break;
         }
      }
   }
   else {
      for (i = 9; i >= 0; i--) {
         if (hasDigit(&hashMap, i)) {
            flag = true;
            time_value += (char)i + 48;
            break;
         }
      }
   }
   if (!flag)
      return "-1";
   time_value += ":";
   flag = false;
   for (i = 5; i >= 0; i--) {
      if (hasDigit(&hashMap, i)) {
         flag = true;
         time_value += (char)i + 48;
         break;
      }
   }
   if (!flag)
      return "-1";
   flag = false;
   for (i = 9; i >= 0; i--) {
      if (hasDigit(&hashMap, i)) {
         flag = true;
         time_value += (char)i + 48;
         break;
      }
   }
   if (!flag)
      return "-1";
   return time_value;
}
int main() {
   int arr[] = { 0, 0, 0, 9 };
   int n = sizeof(arr) / sizeof(int);
   cout << (getMaxtime_value(arr, n));
   return 0;
}

Output

09:00
raja
Published on 09-Sep-2020 12:44:22
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