# Minimum Time Difference in C++

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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