# Count subarrays with equal number of 1’s and 0’s

Given an array arr[] of size n containing 0 and 1 only. The problem is to count the subarrays having equal number of 0’s and 1’s.

Examples:

```Input : arr[] = {1, 0, 0, 1, 0, 1, 1}
Output : 8
The index range for the 8 sub-arrays are:
(0, 1), (2, 3), (0, 3), (3, 4), (4, 5)
(2, 5), (0, 5), (1, 6)
```

The problem is closely related to Largest subarray with equal number of 0’s and 1’s.

Approach: Following are the steps:

1. Consider all 0’s in arr[] as -1.
2. Create a hash table that holds the count of each sum[i] value, where sum[i] = sum(arr[0]+..+arr[i]), for i = 0 to n-1.
3. Now start calculating cumulative sum and then we get increment count by 1 for that sum represented as index in the hash table. Sub-array by each pair of positions with same value of cumulative sum constitute a continuous range with equal number of 1’s and 0’s.
4. Now traverse the hash table and get the frequency of each element in the hash table. Let frequency be denoted as freq. For each freq > 1 we can choose any two pair of indices of sub-array by (freq * (freq – 1)) / 2 number of ways . Do the same for all freq and sum up the result that will be the number all possible sub-arrays containing equal number of 1’s and 0’s.
5. Also add freq of the sum 0 in the hash table to the final result.

Explanation:
Considering all 0’s as -1, if sum[i] == sum[j], where sum[i] = sum(arr[0]+..+arr[i]) and sum[j] = sum(arr[0]+..+arr[j]) and ‘i’ is less than ‘j’, then sum(arr[i+1]+..+arr[j]) must be 0. It can only be 0 if arr(i+1, .., j) contains equal number of 1’s and 0’s.

## C++

 `// C++ implementation to count subarrays with ` `// equal number of 1's and 0's ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// function to count subarrays with ` `// equal number of 1's and 0's ` `int` `countSubarrWithEqualZeroAndOne(``int` `arr[], ``int` `n) ` `{ ` `    ``// 'um' implemented as hash table to store ` `    ``// frequency of values obtained through ` `    ``// cumulative sum ` `    ``unordered_map<``int``, ``int``> um; ` `    ``int` `curr_sum = 0; ` ` `  `    ``// Traverse original array and compute cumulative ` `    ``// sum and increase count by 1 for this sum ` `    ``// in 'um'. Adds '-1' when arr[i] == 0 ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``curr_sum += (arr[i] == 0) ? -1 : arr[i]; ` `        ``um[curr_sum]++; ` `    ``} ` ` `  `    ``int` `count = 0; ` `    ``// traverse the hash table 'um' ` `    ``for` `(``auto` `itr = um.begin(); itr != um.end(); itr++) { ` ` `  `        ``// If there are more than one prefix subarrays ` `        ``// with a particular sum ` `        ``if` `(itr->second > 1) ` `            ``count += ((itr->second * (itr->second - 1)) / 2); ` `    ``} ` ` `  `    ``// add the subarrays starting from 1st element and ` `    ``// have equal number of 1's and 0's ` `    ``if` `(um.find(0) != um.end()) ` `        ``count += um[0]; ` ` `  `    ``// required count of subarrays ` `    ``return` `count; ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 0, 0, 1, 0, 1, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``cout << ``"Count = "` `         ``<< countSubarrWithEqualZeroAndOne(arr, n); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to count  ` `# subarrays with equal number  ` `# of 1's and 0's ` ` `  `# function to count subarrays with ` `# equal number of 1's and 0's ` `def` `countSubarrWithEqualZeroAndOne (arr, n): ` ` `  `    ``# 'um' implemented as hash table  ` `    ``# to store frequency of values ` `    ``# obtained through cumulative sum ` `    ``um ``=` `dict``() ` `    ``curr_sum ``=` `0` `     `  `    ``# Traverse original array and compute  ` `    ``# cumulative sum and increase count ` `    ``# by 1 for this sum in 'um'. ` `    ``# Adds '-1' when arr[i] == 0 ` `    ``for` `i ``in` `range``(n): ` `        ``curr_sum ``+``=` `(``-``1` `if` `(arr[i] ``=``=` `0``) ``else` `arr[i]) ` `        ``if` `um.get(curr_sum): ` `            ``um[curr_sum]``+``=``1` `        ``else``: ` `            ``um[curr_sum]``=``1` `     `  `    ``count ``=` `0` `     `  `    ``# traverse the hash table 'um' ` `    ``for` `itr ``in` `um: ` `         `  `        ``# If there are more than one  ` `        ``# prefix subarrays with a  ` `        ``# particular sum ` `        ``if` `um[itr] > ``1``: ` `            ``count ``+``=` `((um[itr] ``*` `int``(um[itr] ``-` `1``)) ``/` `2``) ` `     `  `    ``# add the subarrays starting from  ` `    ``# 1st element and have equal  ` `    ``# number of 1's and 0's ` `    ``if` `um.get(``0``): ` `        ``count ``+``=` `um[``0``] ` `     `  `    ``# required count of subarrays ` `    ``return` `int``(count) ` `     `  `# Driver code to test above ` `arr ``=` `[ ``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``1` `] ` `n ``=` `len``(arr)  ` `print``(``"Count ="``, ` `    ``countSubarrWithEqualZeroAndOne(arr, n)) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

Output:

```Count = 8
```

Time Complexity: O(n).
Auxiliary Space: O(n).
Another approach:

## C++

 `#include ` ` `  `using` `namespace` `std; ` ` `  `int` `countSubarrWithEqualZeroAndOne(``int` `arr[], ``int` `n){ ` `  ``map<``int``,``int``> mp; ` `  ``int` `sum=0; ` `  ``int` `count=0; ` `  ``for` `(``int` `i = 0; i < n; i++) { ` `            ``//Replacing 0's in array with -1 ` `            ``if` `(arr[i] == 0) ` `                ``arr[i] = -1; ` ` `  `            ``sum += arr[i]; ` ` `  `            ``//If sum = 0, it implies number of 0's and 1's are ` `            ``//equal from arr[0]..arr[i] ` `            ``if` `(sum == 0) ` `                ``count++; ` ` `  `            ``if` `(mp[sum]) ` `                ``count += mp[sum]; ` `            ``if``(mp[sum]==0) ` `                ``mp[sum]=1; ` `            ``else` `                ``mp[sum]++; ` ` `  `        ``} ` `  ``return` `count; ` `} ` ` `  `int` `main() ` `{ ` `  ``int` `arr[] = {1, 0, 0, 1, 0, 1, 1}; ` ` `  `  ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]); ` ` `  `  ``cout<<``"count="``<

## Java

 `import` `java.util.HashMap; ` `import` `java.util.Map; ` ` `  `// Java implementation to count subarrays with ` `// equal number of 1's and 0's ` `public` `class` `Main { ` ` `  `    ``// Function that returns count of sub arrays ` `    ``// with equal numbers of 1's and 0's ` `    ``static` `int` `countSubarrWithEqualZeroAndOne(``int``[] arr, ``int` `n) { ` `        ``Map myMap = ``new` `HashMap<>(); ` `        ``int` `sum = ``0``; ` `        ``int` `count = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``//Replacing 0's in array with -1 ` `            ``if` `(arr[i] == ``0``) ` `                ``arr[i] = -``1``; ` `         `  `            ``sum += arr[i]; ` `             `  `            ``//If sum = 0, it implies number of 0's and 1's are  ` `            ``//equal from arr[0]..arr[i] ` `            ``if` `(sum == ``0``) ` `                ``count++; ` `             `  `            ``if` `(myMap.containsKey(sum)) ` `                ``count += myMap.get(sum); ` ` `  `            ``if` `(!myMap.containsKey(sum)) ` `                ``myMap.put(sum, ``1``); ` `            ``else` `                ``myMap.put(sum, myMap.get(sum) + ``1``); ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// main function ` `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `arr[] = { ``1``, ``0``, ``0``, ``1``, ``0``, ``1``, ``1` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(``"Count = "` `+  ` `                            ``countSubarrWithEqualZeroAndOne(arr, n)); ` ` `  `    ``} ` ` `  `} `

## Python3

# Python3 implementation to count subarrays
# with equal number of 1’s and 0’s
def countSubarrWithEqualZeroAndOne(arr, n):
mp = dict()
Sum = 0
count = 0

for i in range(n):

# Replacing 0’s in array with -1
if (arr[i] == 0):
arr[i] = -1

Sum += arr[i]

# If Sum = 0, it implies number of
# 0’s and 1’s are equal from arr[0]..arr[i]
if (Sum == 0):
count+=1

if (Sum in mp.keys()):
count += mp[Sum]

mp[Sum] = mp.get(Sum, 0) + 1

return count

# Driver Code
arr = [1, 0, 0, 1, 0, 1, 1]

n = len(arr)

print(“count =”,
countSubarrWithEqualZeroAndOne(arr, n))

# This code is contributed by mohit kumar

Output:

```Count = 8
```

## tags:

Arrays Hash binary-string Arrays Hash