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Formatted question description: https://leetcode.ca/all/2224.html
2224. Minimum Number of Operations to Convert Time (Easy)
You are given two strings current
and correct
representing two 24hour times.
24hour times are formatted as "HH:MM"
, where HH
is between 00
and 23
, and MM
is between 00
and 59
. The earliest 24hour time is 00:00
, and the latest is 23:59
.
In one operation you can increase the time current
by 1
, 5
, 15
, or 60
minutes. You can perform this operation any number of times.
Return the minimum number of operations needed to convert current
to correct
.
Example 1:
Input: current = "02:30", correct = "04:35" Output: 3 Explanation: We can convert current to correct in 3 operations as follows:  Add 60 minutes to current. current becomes "03:30".  Add 60 minutes to current. current becomes "04:30".  Add 5 minutes to current. current becomes "04:35". It can be proven that it is not possible to convert current to correct in fewer than 3 operations.
Example 2:
Input: current = "11:00", correct = "11:01" Output: 1 Explanation: We only have to add one minute to current, so the minimum number of operations needed is 1.
Constraints:
current
andcorrect
are in the format"HH:MM"
current <= correct
Similar Questions:
Solution 1. Greedy
 Compute the time difference in seconds
 Greedily using
60, 15, 5, 1
operations. For each operationop
, we usediff / op
number of operations to turndiff
todiff % op
.

// OJ: https://leetcode.com/problems/minimumnumberofoperationstoconverttime/ // Time: O(1) // Space: O(1) class Solution { int getTime(string &s) { return stoi(s.substr(0, 2)) * 60 + stoi(s.substr(3)); } public: int convertTime(string current, string correct) { int diff = getTime(correct)  getTime(current), ops[4] = {60,15,5,1}, ans = 0; for (int op : ops) { ans += diff / op; diff %= op; } return ans; } };

class Solution: def convertTime(self, current: str, correct: str) > int: a = int(current[:2]) * 60 + int(current[3:]) b = int(correct[:2]) * 60 + int(correct[3:]) ans, d = 0, b  a for i in [60, 15, 5, 1]: ans += d // i d %= i return ans ############ # 2224. Minimum Number of Operations to Convert Time # https://leetcode.com/problems/minimumnumberofoperationstoconverttime/ class Solution: def convertTime(self, current: str, correct: str) > int: if current == correct: return 0 h1, h2 = int(current[:2]), int(correct[:2]) m1, m2 = int(current[3:]), int(correct[3:]) res = h2  h1 if m1 > m2: m2 += 60 res = 1 while m2 > m1: d = m2  m1 if d >= 15: res += 1 m2 = 15 elif d >= 5: res += 1 m2 = 5 else: res += 1 m2 = 1 return res

class Solution { public int convertTime(String current, String correct) { int a = Integer.parseInt(current.substring(0, 2)) * 60 + Integer.parseInt(current.substring(3)); int b = Integer.parseInt(correct.substring(0, 2)) * 60 + Integer.parseInt(correct.substring(3)); int ans = 0, d = b  a; for (int i : Arrays.asList(60, 15, 5, 1)) { ans += d / i; d %= i; } return ans; } }

func convertTime(current string, correct string) int { parse := func(s string) int { h := int(s[0]  '0') * 10 + int(s[1]  '0') m := int(s[3]  '0') * 10 + int(s[4]  '0') return h * 60 + m } a, b := parse(current), parse(correct) ans, d := 0, b  a for _, i := range []int{60, 15, 5, 1} { ans += d / i d %= i } return ans }
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https://leetcode.com/problems/minimumnumberofoperationstoconverttime/discuss/1908782/