# Find numbers of balancing positions in string

Given a string, the task is to find the number of such balancing positions in the string from where the left and the right part of that string contains same characters. The frequency of characters doesn’t matters.

Examples:

```Input : str[] = abaaba
Output : Number of balancing positions : 3
Explanations : All 3 balancing positions are as :
ab|aaba, aba|aba, abaa|ba

Input : str[] = noon
Output : Number of balancing positions : 1
Explanations : Balancing position is :
no|on
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach : If we try to solve this problem by the naive approach, we have to process for all n positions of string and at each position, we must check whether the left and right parts of our string from that position have same characters or not.
The process of finding whether the position is balancing or not (frequency of both parts need not be same) can be done in O(n^2) time for a single position( where we should check if each element in left part is present in right and vice-versa). This whole process will lead an algorithm of time complexity O(n^3).

Efficient Approach: Idea of efficient algorithm came from this article. The main difference is that we should not care about equal frequency, and using traversing the string.
We first fill right[] with counts of all characters. Then we traverse the string from left to right. For every character, we increment its count in left[] and decrement count in right[]. For any point being traversed, if all characters that have non-zero value in left also have non-zero value in right, and vice versa is also true, then we increment result.

## C++

```// C++ program to find number of balancing
// points in string
#include<bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 256;

// function to return number of balancing points
int countBalance(char *str)
{
int n = strlen(str); // string length

// hash array for storing hash of string
// intialized by 0 being global
int leftVisited[MAX_CHAR] = {0};
int rightVisited[MAX_CHAR] = {0};

// process string initially for rightVisited
for (int i=0; i<n; i++)
rightVisited[str[i]]++;

// check for balancing points
int res = 0;
for (int i=0; i<n; i++)
{
// for every position inc left hash
// & dec rightVisited
leftVisited[str[i]]++;
rightVisited[str[i]]--;

// check whether  both hash have same
// character or not
int j;
for (j=0; j<MAX_CHAR; j++)
{
// Either both leftVisited[j] and
// rightVisited[j] should have none
// zero value or both should have
// zero value
if ( (leftVisited[j] == 0 &&
rightVisited[j] != 0) ||
(leftVisited[j] != 0 &&
rightVisited[j] == 0)
)
break;
}

// if both have same character increment
// count
if (j == MAX_CHAR)
res++;
}
return res;
}

//driver program
int main()
{
char str[] = "abaababa";
cout << countBalance(str);
return 0;
}
```

## Java

```// Java program to find number of balancing
// points in string

class GFG
{
static final int MAX_CHAR = 256;

// method to return number of balancing points
static int countBalance(String s)
{
char[] str=s.toCharArray();
int n = str.length; // string length

// hash array for storing hash of string
// intialized by 0 being global
int[] rightVisited = new int[MAX_CHAR];
int[] leftVisited = new int[MAX_CHAR];

// process string initially for rightVisited
for (int i=0; i<n; i++)
rightVisited[str[i]]++;

// check for balancing points
int res = 0;
for (int i=0; i<n; i++)
{
// for every position inc left hash
// & dec rightVisited
leftVisited[str[i]]++;
rightVisited[str[i]]--;

// check whether both hash have same
// character or not
int j;
for (j=0; j<MAX_CHAR; j++)
{
// Either both leftVisited[j] and
// rightVisited[j] should have none
// zero value or both should have
// zero value
if ( (leftVisited[j] == 0 &&
rightVisited[j] != 0) ||
(leftVisited[j] != 0 &&
rightVisited[j] == 0)
)
break;
}

// if both have same character increment
// count
if (j == MAX_CHAR)
res++;
}
return res;
}

// Driver Method
public static void main(String[] args)
{
String str = "abaababa";
System.out.println(countBalance(str));
}
}
/* This code is contributed by Mr. Somesh Awasthi */
```

Output:

```5
```

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3 Average Difficulty : 3/5.0
Based on 2 vote(s)