Minimum Speed to Arrive on Time

Minimum Speed to Arrive on Time - Problem

You are given a floating-point number hour, representing the amount of time you have to reach the office. To commute to the office, you must take n trains in sequential order. You are also given an integer array dist of length n, where dist[i] describes the distance (in kilometers) of the i^th train ride.

Each train can only depart at an integer hour, so you may need to wait in between each train ride.

For example: If the 1st train ride takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2nd train ride at the 2 hour mark.

Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1 if it is impossible to be on time.

Tests are generated such that the answer will not exceed 10⁷ and hour will have at most two digits after the decimal point.

Input & Output

Example 1 — Basic Case

$ Input: dist = [1,1,2], hour = 4.0

› Output: 1

💡 Note: At speed 1: Train 1 takes ceil(1/1)=1 hour, Train 2 takes ceil(1/1)=1 hour, Train 3 takes 2/1=2 hours. Total: 1+1+2=4.0 hours ≤ 4.0

Example 2 — Need Higher Speed

$ Input: dist = [1,1,100000], hour = 2.01

› Output: 10000000

💡 Note: Need very high speed for last train. At speed 10^7: ceil(1/10^7) + ceil(1/10^7) + 100000/10^7 = 1+1+0.01 = 2.01 hours

Example 3 — Impossible Case

$ Input: dist = [1,1,1], hour = 2.0

› Output: -1

💡 Note: Need at least 2 integer hours for first 2 trains, plus time for last train. Minimum time is 2+ε > 2.0, so impossible

Constraints

n == dist.length
1 ≤ n ≤ 10⁵
1 ≤ dist[i] ≤ 10⁵
1 ≤ hour ≤ 10⁹
There will be at most two digits after the decimal point in hour.

Visualization

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Asked in

G Google 12 a Amazon 8 M Microsoft 6 f Facebook 5

The key insight is to use binary search on the answer space. Since if speed X works, any speed > X also works (monotonic property), we can binary search to find the minimum working speed. The optimal approach is binary search on answer with Time: O(n × log(10⁷)), Space: O(1).

Common Approaches

✓ Binary Search on Answer

⏱️ Time: O(n × log(10^7)) Space: O(1)

Instead of trying every speed linearly, use binary search on the answer space. The key insight is that if speed X works, then any speed > X also works (monotonic property).

Brute Force Linear Search

⏱️ Time: O(answer × n) Space: O(1)

Start with speed 1 and increment until we find a speed that allows us to reach on time. For each speed, calculate total time including waiting between trains.

Binary Search on Answer — Algorithm Steps

Step 1: Set search range [1, 10^7]
Step 2: For middle speed, check if we can reach on time
Step 3: If yes, try smaller speeds (left half), else try larger speeds (right half)

Visualization

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Step-by-Step Walkthrough

Initialize Range

Set left=1, right=10^7

Test Middle

Check if mid speed works

Narrow Range

Adjust bounds based on result

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int canReachOnTime(int* dist, int n, double hour, int speed) {
    double totalTime = 0;
    
    for (int i = 0; i < n - 1; i++) {
        // All trains except last must wait for next integer hour
        totalTime += ceil((double) dist[i] / speed);
    }
    // Last train doesn't need to wait
    totalTime += (double) dist[n - 1] / speed;
    
    return totalTime <= hour;
}

int solution(int* dist, int n, double hour) {
    // If we have more trains than hours (minus 1), impossible
    if (n - 1 >= hour) {
        return -1;
    }
    
    // Binary search on speed
    int left = 1, right = 10000000;
    int result = -1;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (canReachOnTime(dist, n, hour, mid)) {
            result = mid;
            right = mid - 1;  // Try to find smaller speed
        } else {
            left = mid + 1;   // Need larger speed
        }
    }
    
    return result;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int dist[1000];
    int n;
    parseArray(line, dist, &n);
    
    double hour;
    scanf("%lf", &hour);
    
    printf("%d\n", solution(dist, n, hour));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × log(10^7))

Binary search takes log(10^7) ≈ 23 iterations, each validation takes O(n)

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra variables for binary search bounds

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search on Answer — Algorithm Steps

Visualization

Code -

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