Update README.md

Author: 15-Ab

README.md

@@ -5,69 +5,45 @@ This is my attemp to make the coding experience easier for you guys so that you
 
 ## Always here to assist you guys.
 
-## Today's 02-01-24 [Problem Link](https://leetcode.com/problems/convert-an-array-into-a-2d-array-with-conditions/solutions/4490329/daily-02-01-24/)
+## Today's 03-01-24 [Problem Link](https://leetcode.com/problems/number-of-laser-beams-in-a-bank/solutions/?envType=daily-question&envId=2024-01-03)
 
 # Intuition
 <!-- Describe your first thoughts on how to solve this problem. -->
-Unitary way to think.
-
+Basic multiplication.
 # Approach
 <!-- Describe your approach to solving the problem. -->
-- I counted the frequencies of every number in the given array
-- - used HashMap for that
-- The maximum frequency of all element will be the number of answer rows
-- - as in every row there should be unique elements
-- Created and initialised answerlist with empty sub-lists
-- Now, according to their frequencies adding elements to their rows
-- - elements with frequency '1' will be present in only first row
-- - elements with frequency '2' will be present in first and second row
-- - ... and so on
+- 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 : $$O(cm)$$
+- Time complexity : $$O(l)$$
 <!-- Add your time complexity here, e.g. $$O(n)$$ -->
-
-- Space complexity : $$O(c)$$
+$$l$$ : length of array
+- Space complexity : $$O(1)$$
 <!-- Add your space complexity here, e.g. $$O(n)$$ -->
-$$c$$ : number of unique elements in array
-$$m$$ : maximum frequency of any element
 
 # Code
 ```
 class Solution {
-    public List<List<Integer>> findMatrix(int[] nums) {
-        HashMap<Integer, Integer> m = new HashMap<>();
-        int mf = 0; // to store maximum frequency of any integer : it will determine the number of rows
-    
-        // adding numbers in hashmap along with its frequencies
-        for( int i = 0; i < nums.length; i++){
-            m.putIfAbsent(nums[i], 0);
-            m.put( nums[i], m.get(nums[i]) + 1 );
-            mf = Math.max(mf, m.get(nums[i]));
-        }
-        List<List<Integer>> l = new ArrayList<>(); // to store the answer list
-        // the element with maximum frequency will be present in every sub-list of answerlist
-        // making 'mf' rows in answer list
-        for( int i = 0; i < mf; i++){              
-            List<Integer> t = new ArrayList<>();
-            l.add(t);
-        }
-        
-        // according to their frequencies adding elements to their rows
-        // elements with frequency '1' will be present in only first row
-        // elements with frequency '2' will be present in first and second row
-        // ... and so on
-
-        for( int c : m.keySet() ){
-            for( int f = 0; f < m.get(c); f++){
-                l.get(f).add(c);
+    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 l;
+        return jawab;
     }
 }
 ```