LeetCode #2224 — EASY

Minimum Number of Operations to Convert Time

Build confidence with an intuition-first walkthrough focused on greedy fundamentals.

The Problem

Problem Statement

You are given two strings current and correct representing two 24-hour times.

24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour 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 and correct are in the format "HH:MM"
current <= correct

Step 01

Brute Force Baseline

Problem summary: You are given two strings current and correct representing two 24-hour times. 24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

"02:30"
"04:35"

Example 2

"11:00"
"11:01"

Core Insight

What unlocks the optimal approach

Convert the times to minutes.
Use the operation with the biggest value possible at each step.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2224: Minimum Number of Operations to Convert Time
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;
    }
}

// Accepted solution for LeetCode #2224: Minimum Number of Operations to Convert Time
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
}

# Accepted solution for LeetCode #2224: Minimum Number of Operations to Convert Time
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

// Accepted solution for LeetCode #2224: Minimum Number of Operations to Convert Time
/**
 * [2224] Minimum Number of Operations to Convert Time
 *
 * You are given two strings current and correct representing two 24-hour times.
 * 24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour 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 and correct are in the format "HH:MM"
 * 	current <= correct
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-number-of-operations-to-convert-time/
// discuss: https://leetcode.com/problems/minimum-number-of-operations-to-convert-time/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn convert_time(current: String, correct: String) -> i32 {
        let operation = [60, 15, 5, 1];
        let minute = |s: String| -> i32 {
            s.split(":")
                .map(|x| x.to_string().parse().unwrap())
                .reduce(|x, y| 60 * x + y)
                .unwrap()
        };

        let mut delta_time = minute(correct) - minute(current);
        let mut result = 0;

        for k in operation {
            result += delta_time / k;
            delta_time %= k;
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2224_example_1() {
        let current = "02:30".to_string();
        let correct = "04:35".to_string();

        let result = 3;

        assert_eq!(Solution::convert_time(current, correct), result);
    }

    #[test]
    fn test_2224_example_2() {
        let current = "11:00".to_string();
        let correct = "11:01".to_string();

        let result = 1;

        assert_eq!(Solution::convert_time(current, correct), result);
    }
}

// Accepted solution for LeetCode #2224: Minimum Number of Operations to Convert Time
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2224: Minimum Number of Operations to Convert Time
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.