# 16.09.2024 [539. Minimum Time Difference]

Min difference in a list of times `hh:mm` #medium

16.09.2024

539. Minimum Time Difference medium
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Problem TLDR

Min difference in a list of times hh:mm #medium

Intuition

The main problem is how to handle the loop:

    // 12:00
    //  1:00
    // 23:00

One way is to repeat the array twice. (Actually, only the first value matters).

Approach

• let’s use window iterator

• we can use a bucket sort

Complexity

• Time complexity:
  𝑂(𝑛𝑙𝑜𝑔(𝑛)) or 𝑂(𝑛+𝑚), where m is minutes = 24 * 60

• Space complexity:
  𝑂(𝑛) or 𝑂(𝑚)

Code

    fun findMinDifference(timePoints: List<String>) =
        timePoints.map { it.split(":")
            .let { it[0].toInt() * 60 + it[1].toInt() }}
        .sorted().let { it + (it[0] + 24 * 60) }
        .windowed(2).minOf { it[1] - it[0] }

    pub fn find_min_difference(time_points: Vec<String>) -> i32 {
        let mut times: Vec<_> = time_points.iter().map(|s| {
            s[0..2].parse::<i32>().unwrap() * 60 + s[3..5].parse::<i32>().unwrap()
        }).collect();
        times.sort_unstable(); times.push(times[0] + 60 * 24);
        times.windows(2).map(|w| w[1] - w[0]).min().unwrap()
    }

    int findMinDifference(vector<string>& timePoints) {
        vector<bool> times(24 * 60 + 1);
        for (int i = 0; i < timePoints.size(); i++) {
            int t = std::stoi(timePoints[i].substr(0, 2)) * 60 +
                std::stoi(timePoints[i].substr(3, 5));
            if (times[t]) return 0; else times[t] = true;
        }
        int res = times.size(); int j = -1; int first = -1;
        for (int i = 0; i < times.size(); i++) if (times[i]) {
            if (j >= 0) res = min(res, i - j); else first = i;
            j = i;
        }
        return min(res, first + 24 * 60 - j);
    }