Minimum Time Difference in C++

Suppose we have a list of 24-hour clock time points in "Hour:Minutes" format, we have to find the minimum minutes difference between any two time points in the list. So if the input is like [“12:30”,”15:17”], so this will return 167.

To solve this, we will follow these steps −

• Define an array called ok of size 24*60 + 1, and initially all are false.
• n := size of tp
• for i in range 0 to n – 1
  • hr := the hour part from the time
  • min := minute part of the string
  • time := hr * 60 + min
  • if ok[time] is true, then return 0, otherwise set ok[time] as true.
• last := 0, first := inf, ret := inf, prev := -inf
• for i in range 0 to 24*60
  • if ok[i] is true, then
    • last := max of i, last
    • first := min of i and first
    • if prev is not –inf, then ret := min of ret and last – prev
    • prev := i
• return minimum of ret and 24*60+first-last.

Let us see the following implementation to get better understanding &mnus;

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; class Solution {    public:    int findMinDifference(vector<string>& tp) {       vector <bool> ok(24 * 60 + 1, false);       int n = tp.size();       for(int i = 0; i < n; i++){          int hr = stoi(tp[i].substr(0, 2));          int min = stoi(tp[i].substr(3, 2));          int time = hr * 60 + min;          if(ok[time]) return 0;          ok[time] = true;       }       int last = 0;       int first = INT_MAX;       int ret = INT_MAX;       int prev = INT_MIN;       for(int i = 0; i <= 24 * 60; i++){          if(ok[i]){             last = max(i, last);             first = min(i, first);             if(prev != INT_MIN) ret = min(ret, last - prev);             prev = i;          }       }       return min(ret, 24 * 60 + first - last);    } }; main(){    vector<string> v = {"12:30","15:17"};    Solution ob;    cout << (ob.findMinDifference(v)); }

Input

["12:30","15:17"]

Output

167