Update README.md

Author: 15-Ab

README.md

@@ -6,21 +6,48 @@ This is my attemp to make the coding experience easir for you guys so that you c
 ## Preview 
 
 
-## Todays 31-12-23
+## Todays 01-01-24
+
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+Unitary method.
+
+# Approach
+<!-- Describe your approach to solving the problem. -->
+- I sorted both arrays - 'g' & 's'
+- Started from first element of 'g' 
+- - checked from first element for element greater than 'g[i]' in 's[j]'
+- - if found then incremented the count and nullify that 's[j]' so that that cookie cann't be considered in next child
+- Ran this loop for all elements of 'g'
+---
+Have a look at the code , still have any confusion then please let me know in the comments
+Keep Solving.:)
+ 
+# Complexity
+- Time complexity : $$O(gs)$$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity : $$O(1)$$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+# Code
 ```
 class Solution {
-    public int maxLengthBetweenEqualCharacters(String s) {
-        int msl = -1; // maximum substring length
-        HashMap<Character, Integer> m = new HashMap<>();
-        for( int i = 0; i < s.length(); i++){
-            if( m.containsKey(s.charAt(i))){
-                msl = Math.max( msl, i - m.get(s.charAt(i)));
-            }
-            else{
-                m.put(s.charAt(i), i+1);
+    public int findContentChildren(int[] g, int[] s) {
+        // sorting both the arrays
+        Arrays.sort(g);
+        Arrays.sort(s);
+        int c = 0;      // to store the number of contended children
+        for( int i = 0; i < g.length; i++){
+            for( int j = 0; j < s.length; j++){
+                if( s[j] >= g[i]){
+                    c++;
+                    s[j] = 0;
+                    break;
+                }
             }
         }
-        return msl;
+        return c;
     }
 }
 ```