# Cheapest Flights Within K Stops Problem

## Description

LeetCode Problem 787.

There are n cities connected by some number of flights. You are given an array flights where flights[i] = [from_i, to_i, price_i] indicates that there is a flight from city from_i to city to_i with cost price_i.

You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.

Example 1:

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Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
Output: 200
Explanation: The graph is shown.
The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture.

Example 2:

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Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
Output: 500
Explanation: The graph is shown.
The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.

Constraints:

- 1 <= n <= 100
- 0 <= flights.length <= (n * (n - 1) / 2)
- flights[i].length == 3
- 0 <= from_i, to_i < n
- from_i != to_i
- 1 <= price_i <= 10^4
- There will not be any multiple flights between two cities.
- 0 <= src, dst, k < n
- src != dst

## Sample C++ Code

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class Solution {
public:
int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
vector<vector<int>> dp(K + 2, vector<int> (n, INT_MAX));
for (int i = 0; i <= K + 1; i++)
dp[i][src] = 0;
for (int i = 1; i <= K + 1; i++) {
for (auto& f : flights) {
if (dp[i-1][f[0]] != INT_MAX)
dp[i][f[1]] = min(dp[i][f[1]], dp[i-1][f[0]] + f[2]);
}
}
return dp[K+1][dst] == INT_MAX ? - 1 : dp[K+1][dst];
}
};