leetcode

commit 726c43dd49b4408de500d11ea32b8202f891ceaa
parent 1e4f4a018cf65c0b3be3cfe4539f2186e591a6e5
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Tue,  2 May 2023 08:33:01 -0500

Completed sign of the product using dart in linear time

Diffstat:
A
sign-of-product-of-array/sign-of-productV2.dart
 | 
26
++++++++++++++++++++++++++

1 file changed, 26 insertions(+), 0 deletions(-)
diff --git a/sign-of-product-of-array/sign-of-productV2.dart b/sign-of-product-of-array/sign-of-productV2.dart
@@ -0,0 +1,26 @@
+//Find the sign of the product of a list.
+//To do this I checked if values were negative. If they are
+//then multiple the "total" variable that tracks the sign of the list.
+//If the list contains a 0 then it automatically returns 0.
+//The time complexity of this code is O(n) where n is the length of the list.
+//Time: 237ms Beats: 100%
+//Memory: 141.6MB Beats: 100%
+
+class Solution {
+  int arraySign(List<int> nums) {
+      int total = 1;
+      for(int i = 0 ; i < nums.length ; ++i){
+    
+          if(nums[i] < 0){
+              total *= -1;
+          }
+          else if(nums[i] == 0){
+              return 0;
+          }
+      }
+      if(total == total.abs()){
+          return 1;
+      }
+      return -1;
+  }
+}