LeetCode 1575: Count All Possible Routes
LeetCode 1575 Solution Explanation
Explanation:
To solve this problem, we can use dynamic programming. We define a 3D dp array where dp[i][j][k] represents the number of ways to reach city j with k fuel units starting from city i. We initialize the dp array with base cases where dp[i][finish][0] is set to 1 for all i such that locations[i] == finish.
Next, we iterate over all possible cities, fuel units, and steps remaining. For each step, we calculate the number of ways to reach the current city with the current fuel units by considering all possible transitions from the previous cities with the remaining fuel units. We update the dp array accordingly.
Finally, the answer will be the sum of dp[start][finish][0] to dp[start][finish][fuel] modulo 10^9 + 7. :
LeetCode 1575 Solutions in Java, C++, Python
class Solution {
public int countRoutes(int[] locations, int start, int finish, int fuel) {
int n = locations.length;
int MOD = 1000000007;
int[][][] dp = new int[n][n][fuel + 1];
for (int i = 0; i < n; i++) {
if (i != finish) {
dp[i][finish][0] = 1;
}
}
for (int f = 0; f <= fuel; f++) {
for (int s = 0; s < n; s++) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i != j && f >= Math.abs(locations[i] - locations[j])) {
dp[i][s][f] = (dp[i][s][f] + dp[j][s][f - Math.abs(locations[i] - locations[j])]) % MOD;
}
}
}
}
}
int result = 0;
for (int f = 0; f <= fuel; f++) {
result = (result + dp[start][start][f]) % MOD;
}
return result;
}
}Interactive Code Editor for LeetCode 1575
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