added ONE IMPORTANT QUESTION OF SUBARRAYS

Author: mdfaizanahmed786

Arrays/1-D Arrays/Patterns/SubArrays/SubArraysWithEqual0sand1s.txt

@@ -0,0 +1,89 @@
+Count total subarrays with equal number of 0s and 1s
+
+
+Brute force:
+
+
+class Solution{
+  public:
+    //Function to count subarrays with 1s and 0s.
+    long long int countSubarrWithEqualZeroAndOne(int arr[], int n)
+    {
+        //Your code here
+        long long length=0;
+        for(int i=0; i<n; i++){
+            for(int j=i; j<n; j++){
+                long long zeroCount=0;
+                long long oneCount=0;
+                for(int k=i; k<=j; k++){
+                    if(arr[k]==0){
+                        zeroCount++;
+                    }
+                    else{
+                        oneCount++;
+                    }
+                    
+                }
+                    if(zeroCount==oneCount){
+                        length++;
+                    }
+                
+            }
+        }
+        
+        return length;
+    }
+};
+
+
+Efficient solution:
+Using Hashing
+
+
+
+class Solution{
+  public:
+    //Function to count subarrays with 1s and 0s.
+    long long int countSubarrWithEqualZeroAndOne(int arr[], int n)
+    {
+        //Your code here
+        unordered_map<int, int> mpp;
+        mpp[0]=1;
+        long long frequency=0;
+        int sum=0;
+        for(int i=0; i<n; i++){
+            sum=sum+(arr[i]==1 ? 1 : -1);
+            
+            if(mpp.find(sum)!=mpp.end()){
+                 frequency+=mpp[sum]; 
+            }
+            
+            mpp[sum]++;
+        }
+        
+        return frequency;
+    }
+    
+    
+};
+
+Algroithm:
+1. Create a hash table with key as sum and value as frequency of that sum.
+2. Initialize sum=0 and frequency=0.
+3. Traverse the array and for every element in the array, do the following:
+    1. If the element is 0, then decrease the sum by 1.
+    2. Else, increase the sum by 1.
+    3. If the sum is 0, then increment the frequency by 1.
+    4. If the sum already exists in the hash table, then increment the frequency by the value of that sum in the hash table.
+    5. Increment the value of sum in the hash table by 1.
+4. Return the value of frequency.
+
+TC: O(n)
+SC: O(n)
+
+
+Intuition:
+1. If the intial sum of some part is k, and then if we encouter the same sum later, that means the sum of the elements in between is 0.
+
+k(sum)  (equal number of 0s and 1s)
+                                 k(sum)