@@ -55,11 +55,11 @@
 | [Graph](https://.com/fluency03/leetcode-java/blob/master/src/graph) |
 
 
-# Total: 526
+# Total: 527
 
 | Easy | Medium | Hard | - |
 |:-------:|:-------:|:----:|:---:|
-| 141 | 284 | 89 | 11 |
+| 141 | 285 | 89 | 11 |
 
 
 | Question | Solution | Difficulty |
@@ -544,6 +544,7 @@
 | [777. Swap Adjacent in LR String](https://leetcode.com/problems/swap-adjacent-in-lr-string/) | [Solution](https://.com/fluency03/leetcode-java/blob/master/src/SwapAdjacentInLRString777.java) | Medium |
 | [779. K-th Symbol in Grammar](https://leetcode.com/problems/k-th-symbol-in-grammar/) | [Solution](https://.com/fluency03/leetcode-java/blob/master/src/KthSymbolInGrammar779.java) | Medium |
 | [785. Is Graph Bipartite?](https://leetcode.com/problems/is-graph-bipartite) | [Solution](https://.com/fluency03/leetcode-java/blob/master/src/IsGraphBipartite785.java) | Medium |
+| [787. Cheapest Flights Within K Stops](https://leetcode.com/problems/cheapest-flights-within-k-stops/) | [Solution](https://.com/fluency03/leetcode-java/blob/master/src/CheapestFlightsWithinKStops787.java) | Medium |
 | [790. Domino and Tromino Tiling](https://leetcode.com/problems/domino-and-tromino-tiling/) | [Solution](https://.com/fluency03/leetcode-java/blob/master/src/DominoAndTrominoTiling790.java) | Medium |
 | [792. Number of Matching Subsequences](https://leetcode.com/problems/number-of-matching-subsequences/) | [Solution](https://.com/fluency03/leetcode-java/blob/master/src/NumberOfMatchingSubsequences792.java) | Medium |
 | [797. All Paths From Source to Target](https://leetcode.com/problems/all-paths-from-source-to-target/) | [Solution](https://.com/fluency03/leetcode-java/blob/master/src/AllPathsFromSourceToTarget797.java) | Medium |

@@ -0,0 +1,82 @@
+/**
+* There are n cities connected by m flights. Each fight starts from city u and
+* arrives at v with a price w.
+*
+* Now given all the cities and flights, together with starting city src and
+* the destination dst, your task is to find the cheapest price from src to dst
+* with up to k stops. If there is no such route, output -1.
+*
+* Example 1:
+* Input:
+* n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
+* src = 0, dst = 2, k = 1
+* Output: 200
+*
+* Explanation:
+* The graph looks like this:
+* https://s3-lc-upload.s3.amazonaws.com/uploads/2018/02/16/995.png
+*
+* The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as
+* marked red in the picture.
+*
+* Example 2:
+* Input:
+* n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
+* src = 0, dst = 2, k = 0
+* Output: 500
+*
+* Explanation:
+* The graph looks like this:
+* https://s3-lc-upload.s3.amazonaws.com/uploads/2018/02/16/995.png
+*
+* The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as
+* marked blue in the picture.
+*
+* Note:
+* The number of nodes n will be in range [1, 100], with nodes labeled from 0
+* to n - 1.
+* The size of flights will be in range [0, n * (n - 1) / 2].
+* The format of each flight will be (src, dst, price).
+* The price of each flight will be in the range [1, 10000].
+* k is in the range of [0, n - 1].
+* There will not be any duplicated flights or self cycles.
+*/
+
+public class CheapestFlightsWithinKStops787 {
+public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) {
+int[][] graph = constructGraph(n, flights);
+Queue<int[]> q = new LinkedList<>();
+q.add(new int[]{src, 0});
+int[][] dist = new int[n][K+2];
+for (int[] row: dist) Arrays.fill(row, Integer.MAX_VALUE);
+dist[src][0] = 0;
+while (!q.isEmpty()) {
+int[] cur = q.poll();
+int[] nei = graph[cur[0]];
+for (int next=0; next<n; next++) {
+int weight = nei[next];
+if (weight == 0) continue;
+int newDis = dist[cur[0]][cur[1]] + weight;
+if (cur[1] + 1 <= K+1 && newDis < dist[next][cur[1]+1]) {
+dist[next][cur[1]+1] = newDis;
+q.add(new int[]{next, cur[1]+1});
+}
+}
+}
+int res = Integer.MAX_VALUE;
+for (int r: dist[dst]) {
+res = Math.min(res, r);
+}
+return res == Integer.MAX_VALUE ? -1 : res;
+}
+
+private int[][] constructGraph(int n, int[][] flights) {
+int[][] res = new int[n][n];
+for (int[] f: flights) {
+res[f[0]][f[1]] = f[2];
+}
+return res;
+}
+
+}
+