.

Author: 15-Ab

2024 January/Daily 03-01-24.md

@@ -40,4 +40,48 @@ class Solution {
         return jawab;
     }
 }
+=======
+## Today's 03-01-24 [Problem Link](https://leetcode.com/problems/number-of-laser-beams-in-a-bank/description/?envType=daily-question&envId=2024-01-03)
+
+
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+Basic multiplication.
+# Approach
+<!-- Describe your approach to solving the problem. -->
+- I kept track of number of '1' in a row
+- Now iterated over every row of array 
+- - counted the number of '1' in current row
+- - number of beams will the product of current number of device and previous number of devices
+- -  added the product to answer
+- -  now, the current one will become the previous one to next row
+---
+Have a look at the code , still have any confusion then please let me know in the comments
+Keep Solving.:)
+
+# Complexity
+- Time complexity : $$O(l)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+$$l$$ : length of array
+- Space complexity : $$O(1)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+# Code
+```
+class Solution {
+    public int numberOfBeams(String[] bank) {
+        int jawab = 0;      // to store answer
+        int picheek = 0;    // to store number of '1' in previous state
+
+        for( String r : bank){
+            int ek = (int) r.chars().filter( g -> g == '1').count(); // counting the number of '1' in current row
+            if( ek != 0){             // number of beams will the product of current number of device and previous number of devices
+                jawab += picheek*ek; // adding the product to answer
+                picheek = ek;        // now, the current one will become the previous one to next row
+            }
+        }
+        return jawab;
+    }
+}
+>>>>>>> 8326dfdfc4354991bd3c7f23df8fcaa854307652
 ```

README.md

@@ -5,6 +5,7 @@ This is my attempt to make the coding experience easier for you guys so that you
 
 ## Always here to assist you guys.
 
+<<<<<<< HEAD
 ## Today's 04-01-24 [Problem Link](https://leetcode.com/problems/minimum-number-of-operations-to-make-array-empty/description/)
 
 # Intuition
@@ -31,23 +32,44 @@ Use a HashMap (m) to store the frequency of each number in the array.
 - - - - 3*n + 2 = 3*(n) + 2*(1) -> n+1 operations 
 - Return Result :
 - - Return the total number of operations needed to make all numbers in the array divisible by 3. 
+=======
+## Today's 03-01-24 [Problem Link](https://leetcode.com/problems/number-of-laser-beams-in-a-bank/description/?envType=daily-question&envId=2024-01-03)
+
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+Basic multiplication.
+# Approach
+<!-- Describe your approach to solving the problem. -->
+- I kept track of number of '1' in a row
+- Now iterated over every row of array 
+- - counted the number of '1' in current row
+- - number of beams will the product of current number of device and previous number of devices
+- -  added the product to answer
+- -  now, the current one will become the previous one to next row
+>>>>>>> 8326dfdfc4354991bd3c7f23df8fcaa854307652
 ---
 Have a look at the code , still have any confusion then please let me know in the comments
 Keep Solving.:)
 
 # Complexity
 - Time complexity : $$O(l)$$
 <!-- Add your time complexity here, e.g. $$O(n)$$ -->
+<<<<<<< HEAD
 
 - Space complexity : $$O(u)$$
 
 $$l$$ : size of array
 $$u$$ : number of unique letters in array
+=======
+$$l$$ : length of array
+- Space complexity : $$O(1)$$
+>>>>>>> 8326dfdfc4354991bd3c7f23df8fcaa854307652
 <!-- Add your space complexity here, e.g. $$O(n)$$ -->
 
 # Code
 ```
 class Solution {
+<<<<<<< HEAD
     public int minOperations(int[] nums) {
         boolean haveone = false;
         HashMap<Integer, Integer> m = new HashMap<>();
@@ -74,6 +96,20 @@ class Solution {
             }
         }
         return operations;
+=======
+    public int numberOfBeams(String[] bank) {
+        int jawab = 0;      // to store answer
+        int picheek = 0;    // to store number of '1' in previous state
+
+        for( String r : bank){
+            int ek = (int) r.chars().filter( g -> g == '1').count(); // counting the number of '1' in current row
+            if( ek != 0){             // number of beams will the product of current number of device and previous number of devices
+                jawab += picheek*ek; // adding the product to answer
+                picheek = ek;        // now, the current one will become the previous one to next row
+            }
+        }
+        return jawab;
+>>>>>>> 8326dfdfc4354991bd3c7f23df8fcaa854307652
     }
 }
 ```