Largest Time for Given Digits in C++

Suppose we have an array of 4 digits, we have to find the largest 24-hour time that can be made. We know that the smallest 24-hour time is 00:00, and the largest time is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight. We have to return the answer as a string of length 5. If there is no valid time to be returned, then return an empty string.

So, if the input is like [1,2,3,4], then the output will be "23:41"

To solve this, we will follow these steps −

Define a function isValid(), this will take string a,
if a[0] > '2', then −
- return false
if a[0] is same as '2' and a[1] > '3', then −
- return false
if a[3] > '5', then −
- return false
return true
Define a function dfs(), this will take an array A, res, cur,
if size of cur is same as 5, then −
- if isValid(cur) and cur > res, then −
  - res := cur
- return
for initialize i := 0, when i < 4, update (increase i by 1), do −
- if A[i] is not equal to -1, then −
  - tmp := A[i]
  - cur := cur + A[i] + ASCII of '0'
  - if size of cur is same as 2, then −
    - cur := cur concatenate with ':'
  - A[i] := -1
  - dfs(A, res, cur)
  - A[i] := tmp
  - delete last element from cur
  - if size of cur is same as 2, then −
    - delete last element from cur
From the main method do the following −
res := empty string, tmp := empty string
dfs(A, res, tmp)
return res

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   void dfs(vector<int>& A, string& res, string& cur) {
      if (cur.size() == 5) {
         if (isValid(cur) && cur > res)
            res = cur;
            return;
         }
         for (int i = 0; i < 4; ++i) {
            if (A[i] != -1) {
               int tmp = A[i];
               cur += A[i] + '0';
            if (cur.size() == 2)
               cur += ':';
               A[i] = -1;
               dfs(A, res, cur);
               A[i] = tmp;
               cur.pop_back();
               if (cur.size() == 2)
                  cur.pop_back();
            }
         }
   }
   bool isValid(const string a) {
      if (a[0] > '2')
         return false;
         if (a[0] == '2' && a[1] > '3')
            return false;
         if (a[3] > '5')
            return false;
         return true;
   }
   string largestTimeFromDigits(vector<int>& A) {
      string res = "", tmp = "";
      dfs(A, res, tmp);
      return res;
   }
};
main(){
Solution ob;
vector<int> v = {1,2,3,4};
cout << (ob.largestTimeFromDigits(v));
}