Update README.md

Author: 15-Ab

README.md

@@ -31,44 +31,23 @@ 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<>();
@@ -95,20 +74,6 @@ 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
     }
 }
-```
+```