diff --git a/src/problem/p743_network_delay_time.rs b/src/problem/p743_network_delay_time.rs
index 9773336..8f50ef2 100644
--- a/src/problem/p743_network_delay_time.rs
+++ b/src/problem/p743_network_delay_time.rs
@@ -1,44 +1,9 @@
 /**
  * [743] Network Delay Time
- *
- * You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.
- * We will send a signal from a given node k. Return the minimum time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.
- *  
- * <strong class="example">Example 1:
- * <img alt="" src="https://assets.leetcode.com/uploads/2019/05/23/931_example_1.png" style="width: 217px; height: 239px;" />
- * Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
- * Output: 2
- *
- * <strong class="example">Example 2:
- *
- * Input: times = [[1,2,1]], n = 2, k = 1
- * Output: 1
- *
- * <strong class="example">Example 3:
- *
- * Input: times = [[1,2,1]], n = 2, k = 2
- * Output: -1
- *
- *  
- * Constraints:
- *
- * 	1 <= k <= n <= 100
- * 	1 <= times.length <= 6000
- * 	times[i].length == 3
- * 	1 <= ui, vi <= n
- * 	ui != vi
- * 	0 <= wi <= 100
- * 	All the pairs (ui, vi) are unique. (i.e., no multiple edges.)
- *
  */
 pub struct Solution {}
 
-// problem: https://leetcode.cn/problems/network-delay-time/
-// discuss: https://leetcode.cn/problems/network-delay-time/discuss/?currentPage=1&orderBy=most_votes&query=
-
 // submission codes start here
-
-use std::cmp::{max, min};
 use std::collections::HashSet;
 
 impl Solution {
@@ -91,7 +56,7 @@ impl Solution {
                     continue;
                 }
 
-                distances[i] = min(distances[i], value + matrix[index][i]);
+                distances[i] = distances[i].min(value + matrix[index][i]);
             }
         }
 
@@ -102,10 +67,14 @@ impl Solution {
                 continue;
             }
 
-            result = max(result, distances[i]);
+            result = result.max(distances[i]);
         }
 
-        return if result == i32::MAX { -1 } else { result };
+        if result == i32::MAX {
+            -1
+        } else {
+            result
+        }
     }
 }