Imagine you're planning a road trip on a straight highway of length l km. Along this highway, there are n roadside signs at specific positions, with the first sign at position 0 (start) and the last sign at position l (destination).
Here's the twist: each segment between adjacent signs has different travel times per kilometer. The array time[i] tells you how many minutes it takes to travel 1 km between sign i and sign i+1.
You have a special power: you can perform exactly kmerge operations. In each merge, you can:
Choose two adjacent signs at positions i and i+1 (where i > 0 and i+1 < n-1)
Combine their travel times: time[i+1] = time[i] + time[i+1]
Remove the sign at position i
Goal: After performing exactly k merges, find the minimum total travel time to go from position 0 to position l.
Think of it as optimally removing road signs to create longer segments with combined travel rates, minimizing your total journey time!
Input & Output
example_1.py — Basic Case
$Input:l = 10, n = 5, k = 1
position = [0, 2, 5, 7, 10]
time = [3, 4, 2, 1]
›Output:32
💡 Note:With k=1 merge, we can remove one intermediate sign. If we remove sign at position 5, we merge segments with times 4 and 2, giving us segments with times [3, 6, 1] and distances [2, 5, 3]. Total time = 2×3 + 5×6 + 3×1 = 6 + 30 + 3 = 39. The optimal merge gives 32.
example_2.py — Multiple Merges
$Input:l = 8, n = 4, k = 1
position = [0, 3, 6, 8]
time = [2, 3, 1]
›Output:22
💡 Note:Original segments: distances [3, 3, 2] with times [2, 3, 1]. Total = 3×2 + 3×3 + 2×1 = 6 + 9 + 2 = 17. After merging optimally with k=1, we can get a better arrangement with total time 22.
example_3.py — Edge Case
$Input:l = 5, n = 3, k = 0
position = [0, 2, 5]
time = [4, 2]
›Output:14
💡 Note:No merges allowed (k=0). Travel time = 2×4 + 3×2 = 8 + 6 = 14. This is the baseline without any optimization.
Merge Operations for Minimum Travel Time — Solution
This problem requires dynamic programming to find the optimal way to partition the road after exactly k merges. The key insight is that we're looking for the minimum cost interval partitioning where each merge operation combines adjacent segments. The optimal solution uses DP with state dp[i][j] representing minimum cost to reach position i with j merges, achieving O(n³k) time complexity.
Common Approaches
✓
Brute Force (Try All Combinations)
⏱️ Time: O(C(n,k) * k)
Space: O(n)
Generate all possible ways to perform exactly k merge operations and calculate the minimum travel time among all possibilities. This approach explores every possible sequence of merges.
Dynamic Programming (Optimal)
⏱️ Time: O(n³k)
Space: O(nk)
This problem is essentially about partitioning the road into segments optimally. We use dynamic programming where dp[i][j] represents the minimum cost to travel from position 0 to position i using exactly j merges.
Brute Force (Try All Combinations) — Algorithm Steps
Generate all possible combinations of k merge operations
For each combination, simulate the merge operations
Calculate total travel time after merges
Return the minimum travel time found
Visualization
Tap to expand
Step-by-Step Walkthrough
1
Generate Combinations
Generate all possible ways to choose k positions to merge
2
Simulate Each
For each combination, simulate the merge operations
3
Calculate Time
Calculate total travel time after merges
4
Find Minimum
Return the minimum time among all possibilities
Code -
solution.c — C
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <string.h>
static int removeIndices[1000];
static int current[1000];
static int validPositions[1000];
int calculateTime(int* removeIdx, int removeCount, int* position, int* time, int n) {
// Work on local copies
int currentTime[1000];
int currentPos[1000];
int currentTimeSize = n - 1;
int currentPosSize = n;
for (int i = 0; i < n - 1; i++) currentTime[i] = time[i];
for (int i = 0; i < n; i++) currentPos[i] = position[i];
// Sort removal indices descending so each splice doesn't shift later targets
int indices[1000];
for (int i = 0; i < removeCount; i++) indices[i] = removeIdx[i];
for (int i = 0; i < removeCount - 1; i++)
for (int j = i + 1; j < removeCount; j++)
if (indices[i] < indices[j]) {
int t = indices[i]; indices[i] = indices[j]; indices[j] = t;
}
for (int i = 0; i < removeCount; i++) {
int idx = indices[i]; // index into currentPos (1 .. currentPosSize-2)
// The traveler at position idx hands its time to the previous segment.
// time array index for the segment ENDING at position idx is (idx - 1).
currentTime[idx - 1] += currentTime[idx];
// Remove element idx from currentTime (size = currentPosSize - 1)
for (int j = idx; j < currentTimeSize - 1; j++)
currentTime[j] = currentTime[j + 1];
currentTimeSize--;
// Remove element idx from currentPos
for (int j = idx; j < currentPosSize - 1; j++)
currentPos[j] = currentPos[j + 1];
currentPosSize--;
}
int totalTime = 0;
for (int i = 0; i < currentTimeSize; i++) {
int distance = currentPos[i + 1] - currentPos[i];
totalTime += distance * currentTime[i];
}
return totalTime;
}
void generateCombinations(int arrSize, int k, int start, int currentSize,
int* minTime, int* position, int* time, int n) {
if (currentSize == k) {
int timeResult = calculateTime(current, k, position, time, n);
if (timeResult < *minTime) *minTime = timeResult;
return;
}
for (int i = start; i < arrSize; i++) {
current[currentSize] = validPositions[i];
generateCombinations(arrSize, k, i + 1, currentSize + 1, minTime, position, time, n);
}
}
int minimumTravelTime(int l, int n, int k, int* position, int* time) {
if (k == 0) {
int total = 0;
for (int i = 0; i < n - 1; i++)
total += (position[i + 1] - position[i]) * time[i];
return total;
}
int validCount = 0;
for (int i = 1; i < n - 1; i++)
validPositions[validCount++] = i;
int minTime = INT_MAX;
generateCombinations(validCount, k, 0, 0, &minTime, position, time, n);
return minTime;
}
// Parse "[0,2,5,7,10]" into arr[], returns count
int parseArray(const char* s, int* arr) {
int count = 0;
const char* p = s;
while (*p) {
// skip non-digit, non-minus
if (*p == '-' || (*p >= '0' && *p <= '9')) {
arr[count++] = (int)strtol(p, (char**)&p, 10);
} else {
p++;
}
}
return count;
}
int main() {
int l, n, k;
scanf("%d %d %d", &l, &n, &k);
static int position[1000];
static int timeArr[1000];
char line[10000];
// consume leftover newline
scanf(" ");
fgets(line, sizeof(line), stdin);
parseArray(line, position);
fgets(line, sizeof(line), stdin);
parseArray(line, timeArr);
printf("%d\n", minimumTravelTime(l, n, k, position, timeArr));
return 0;
}
Time & Space Complexity
Time Complexity
⏱️
O(C(n,k) * k)
We have C(n,k) ways to choose k positions to merge, and each simulation takes O(k) time
n
2n
⚠ Quadratic Growth
Space Complexity
O(n)
Space for storing the current state during simulation
n
2n
⚡ Linearithmic Space
26.0K Views
MediumFrequency
~35 minAvg. Time
847 Likes
Ln 1, Col 1
Smart Actions
💡Explanation
AI Ready
💡 SuggestionTabto acceptEscto dismiss
// Output will appear here after running code
Code Editor Closed
Click the red button to reopen
Algorithm Visualization
Pinch to zoom • Tap outside to close
Test Cases
0 passed
0 failed
3 pending
Select Compiler
Choose a programming language
Compiler list would appear here...
AI Editor Features
Header Buttons
💡
Explain
Get a detailed explanation of your code. Select specific code or analyze the entire file. Understand algorithms, logic flow, and complexity.
🔧
Fix
Automatically detect and fix issues in your code. Finds bugs, syntax errors, and common mistakes. Shows you what was fixed.
💡
Suggest
Get improvement suggestions for your code. Best practices, performance tips, and code quality recommendations.
💬
Ask AI
Open an AI chat assistant to ask any coding questions. Have a conversation about your code, get help with debugging, or learn new concepts.
Smart Actions (Slash Commands)
🔧
/fix Enter
Find and fix issues in your code. Detects common problems and applies automatic fixes.
💡
/explain Enter
Get a detailed explanation of what your code does, including time/space complexity analysis.
🧪
/tests Enter
Automatically generate unit tests for your code. Creates comprehensive test cases.
📝
/docs Enter
Generate documentation for your code. Creates docstrings, JSDoc comments, and type hints.
⚡
/optimize Enter
Get performance optimization suggestions. Improve speed and reduce memory usage.
AI Code Completion (Copilot-style)
👻
Ghost Text Suggestions
As you type, AI suggests code completions shown in gray text. Works with keywords like def, for, if, etc.
Tabto acceptEscto dismiss
💬
Comment-to-Code
Write a comment describing what you want, and AI generates the code. Try: # two sum, # binary search, # fibonacci
💡
Pro Tip: Select specific code before using Explain, Fix, or Smart Actions to analyze only that portion. Otherwise, the entire file will be analyzed.