leetcode

commit de513a87e12819bf2ad5c47b203921cbfb5fff6f
parent 0adb20e2b5d21cd803e5214412eec724c4a797d2
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Sat,  3 Jun 2023 16:06:20 -0500

Completed find peak element in dart

Diffstat:
A
find-peak-element/find-peak-element.dart
 | 
26
++++++++++++++++++++++++++

1 file changed, 26 insertions(+), 0 deletions(-)
diff --git a/find-peak-element/find-peak-element.dart b/find-peak-element/find-peak-element.dart
@@ -0,0 +1,26 @@
+//Given a list of integers nums return a value in the list which
+//is a local peak defined as n[i] > n[i-1] and n[i] > n[i+1].
+
+//To do this I iterated through the list checking if the left and right values
+//were higher or lower than the current. If both are higher then return i.
+
+//Time complexity is O(n)
+
+//Time: 227ms Beats: 93.75%
+//Memory: 143.7MB Beats: 12.50%
+class Solution {
+  int findPeakElement(List<int> nums) {
+      if(nums.length <= 1){
+          return 0;
+      }
+      for(int i = 1 ; i < nums.length - 1 ; ++i){
+          if(nums[i] > nums[i - 1] && nums[i] > nums[i + 1]){
+              return i;
+          }
+      }
+      if(nums[0] < nums[1]){
+          return nums.length - 1;
+      }
+      return 0;
+  }
+}