Runtime: 2419 ms (Top 5.20%) | Memory: 274.6 MB (Top 9.24%)

Author: AnasImloul

scripts/algorithms/M/Minimum Time to Finish the Race/Minimum Time to Finish the Race.cpp

@@ -1,9 +1,10 @@
+// Runtime: 2419 ms (Top 5.20%) | Memory: 274.6 MB (Top 9.24%)
 class Solution {
 public:
     int minimumFinishTime(vector<vector<int>>& tires, int changeTime, int numLaps) {
         int n = tires.size();
         // to handle the cases where numLaps is small
-		// without_change[i][j]: the total time to run j laps consecutively with tire i
+        // without_change[i][j]: the total time to run j laps consecutively with tire i
         vector<vector<int>> without_change(n, vector<int>(20, 2e9));
         for (int i = 0; i < n; i++) {
             without_change[i][1] = tires[i][0];
@@ -13,22 +14,22 @@ class Solution {
                 without_change[i][j] = without_change[i][j-1] * tires[i][1];
             }
             // since we define it as the total time, rather than just the time for the j-th lap
-			// we have to make it prefix sum
+            // we have to make it prefix sum
             for (int j = 2; j < 20; j++) {
                 if ((long long)without_change[i][j-1] + without_change[i][j] >= 2e9)
                     break;
                 without_change[i][j] += without_change[i][j-1];
             }
         }
-        
-		// dp[x]: the minimum time to finish x laps
+
+        // dp[x]: the minimum time to finish x laps
         vector<int> dp(numLaps+1, 2e9);
         for (int i = 0; i < n; i++) {
             dp[1] = min(dp[1], tires[i][0]);
         }
         for (int x = 1; x <= numLaps; x++) {
             if (x < 20) {
-				// x is small enough, so an optimal solution might never changes tires!
+                // x is small enough, so an optimal solution might never changes tires!
                 for (int i = 0; i < n; i++) {
                     dp[x] = min(dp[x], without_change[i][x]);
                 }
@@ -37,7 +38,7 @@ class Solution {
                 dp[x] = min(dp[x], dp[j] + changeTime + dp[x-j]);
             }
         }
-        
+
         return dp[numLaps];
     }
-};
+};