Longest Even Length Substring such that Sum of First and Second Half is same

Given a string ‘str’ of digits, find the length of the longest substring of ‘str’, such that the length of the substring is 2k digits and sum of left k digits is equal to the sum of right k digits.

Examples :

Input: str = "123123"
Output: 6
The complete string is of even length and sum of first and second
half digits is same

Input: str = "1538023"
Output: 4
The longest substring with same first and second half sum is "5380"


Simple Solution [ O(n3) ]
A Simple Solution is to check every substring of even length. The following is the implementation of simple approach.

C++

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// A simple C++ based program to find length of longest even length
// substring with same sum of digits in left and right 
#include<bits/stdc++.h>
using namespace std;
  
int findLength(char *str)
{
    int n = strlen(str);
    int maxlen =0; // Initialize result
  
    // Choose starting point of every substring
    for (int i=0; i<n; i++)
    {
        // Choose ending point of even length substring
        for (int j =i+1; j<n; j += 2)
        {
            int length = j-i+1;//Find length of current substr
  
            // Calculate left & right sums for current substr
            int leftsum = 0, rightsum =0;
            for (int k =0; k<length/2; k++)
            {
                leftsum += (str[i+k]-'0');
                rightsum += (str[i+k+length/2]-'0');
            }
  
            // Update result if needed
            if (leftsum == rightsum && maxlen < length)
                    maxlen = length;
        }
    }
    return maxlen;
}
  
// Driver program to test above function
int main(void)
{
    char str[] = "1538023";
    cout << "Length of the substring is " 
         << findLength(str);
    return 0;
}
  
// This code is contributed
// by Akanksha Rai

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C

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// A simple C based program to find length of longest  even length
// substring with same sum of digits in left and right 
#include<stdio.h>
#include<string.h>
  
int findLength(char *str)
{
    int n = strlen(str);
    int maxlen =0;  // Initialize result
  
    // Choose starting point of every substring
    for (int i=0; i<n; i++)
    {
        // Choose ending point of even length substring
        for (int j =i+1; j<n; j += 2)
        {
            int length = j-i+1;//Find length of current substr
  
            // Calculate left & right sums for current substr
            int leftsum = 0, rightsum =0;
            for (int k =0; k<length/2; k++)
            {
                leftsum  += (str[i+k]-'0');
                rightsum += (str[i+k+length/2]-'0');
            }
  
            // Update result if needed
            if (leftsum == rightsum && maxlen < length)
                    maxlen = length;
        }
    }
    return maxlen;
}
  
// Driver program to test above function
int main(void)
{
    char str[] = "1538023";
    printf("Length of the substring is %d", findLength(str));
    return 0;
}

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Java

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// A simple Java based program to find 
// length of longest even length substring 
// with same sum of digits in left and right 
import java.io.*;
  
class GFG {
  
static int findLength(String str)
{
    int n = str.length();
    int maxlen = 0; // Initialize result
  
    // Choose starting point of every 
    // substring
    for (int i = 0; i < n; i++)
    {
        // Choose ending point of even 
        // length substring
        for (int j = i + 1; j < n; j += 2)
        {   
            // Find length of current substr
            int length = j - i + 1;
  
            // Calculate left & right sums
            // for current substr
            int leftsum = 0, rightsum = 0;
            for (int k = 0; k < length/2; k++)
            {
                leftsum += (str.charAt(i + k) - '0');
                rightsum += (str.charAt(i + k + length/2) - '0');
            }
  
            // Update result if needed
            if (leftsum == rightsum && maxlen < length)
                    maxlen = length;
        }
    }
    return maxlen;
}
  
// Driver program to test above function
public static void main(String[] args)
{
    String str = "1538023";
    System.out.println("Length of the substring is " 
                       + findLength(str));
}
}
  
// This code is contrtibuted by Prerna Saini

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Python3

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# A simple Python 3 based
# program to find length
# of longest even length
# substring with same sum
# of digits in left and right 
  
def findLength(str):
  
    n = len(str)
    maxlen = 0 # Initialize result
  
    # Choose starting point
        # of every substring
    for i in range(0, n):
      
        # Choose ending point
                # of even length substring
        for j in range(i+1, n, 2):
                   
                        # Find length of current substr
            length = j - i + 1 
  
            # Calculate left & right 
                        # sums for current substr
            leftsum = 0
            rightsum =0
            for k in range(0,int(length/2)):
              
                leftsum += (int(str[i+k])-int('0'))
                rightsum += (int(str[i+k+int(length/2)])-int('0'))
              
  
            # Update result if needed
            if (leftsum == rightsum and maxlen < length):
                    maxlen = length
          
      
    return maxlen
  
  
# Driver program to
# test above function
str = "1538023"
print("Length of the substring is",
       findLength(str))
  
# This code is contributed by
# Smitha Dinesh Semwal

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// A simple C# based program to find 
// length of longest even length substring 
// with same sum of digits in left and right 
using System;
  
class GFG {
  
static int findLength(String str)
{
    int n = str.Length;
    int maxlen = 0; // Initialize result
  
    // Choose starting point 
    // of every substring
    for (int i = 0; i < n; i++)
    {
        // Choose ending point of 
        // even length substring
        for (int j = i + 1; j < n; j += 2)
        
            // Find length of current substr
            int length = j - i + 1;
  
            // Calculate left & right sums
            // for current substr
            int leftsum = 0, rightsum = 0;
            for (int k = 0; k < length/2; k++)
            {
                leftsum += (str[i + k] - '0');
                rightsum += (str[i + k + length/2] - '0');
            }
  
            // Update result if needed
            if (leftsum == rightsum && 
                maxlen < length)
                    maxlen = length;
        }
    }
    return maxlen;
}
  
  // Driver program to test above function
  public static void Main()
  {
    String str = "1538023";
    Console.Write("Length of the substring is " +
                                findLength(str));
  }
}
  
// This code is contrtibuted by nitin mittal

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<?php
// A simple PHP based program to find length of 
// longest even length substring with same sum 
// of digits in left and right 
  
function findLength($str)
{
    $n = strlen($str);
    $maxlen = 0; // Initialize result
  
    // Choose starting point of every substring
    for ($i = 0; $i < $n; $i++)
    {
        // Choose ending point of even 
        // length substring
        for ($j = $i + 1; $j < $n; $j += 2)
        {
            $length = $j - $i + 1; // Find length of current substr
  
            // Calculate left & right sums 
            // for current substr
            $leftsum = 0;
            $rightsum = 0;
            for ($k = 0; $k < $length / 2; $k++)
            {
                $leftsum += ($str[$i + $k] - '0');
                $rightsum += ($str[$i + $k
                              $length / 2] - '0');
            }
  
            // Update result if needed
            if ($leftsum == $rightsum && 
                $maxlen < $length)
                    $maxlen = $length;
        }
    }
    return $maxlen;
}
  
// Driver Code
$str = "1538023";
echo("Length of the substring is ");
echo(findLength($str));
  
// This code is contributed
// by Shivi_Aggarwal
?>

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Output:

Length of the substring is 4



Dynamic Programming [ O(n2) and O(n2) extra space]
The above solution can be optimized to work in O(n2) using Dynamic Programming. The idea is to build a 2D table that stores sums of substrings. The following is the implementation of Dynamic Programming approach.

C++

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// A C++ based program that uses Dynamic 
// Programming to find length of the
// longest even substring with same sum 
// of digits in left and right half
#include<bits/stdc++.h>
using namespace std;
  
int findLength(char *str)
{
    int n = strlen(str);
    int maxlen = 0; // Initialize result
  
    // A 2D table where sum[i][j] stores 
    // sum of digits from str[i] to str[j]. 
    // Only filled entries are the entries 
    // where j >= i
    int sum[n][n];
  
    // Fill the diagonal values for 
    // substrings of length 1
    for (int i =0; i<n; i++)
        sum[i][i] = str[i]-'0';
  
    // Fill entries for substrings of
    // length 2 to n
    for (int len = 2; len <= n; len++)
    {
        // Pick i and j for current substring
        for (int i = 0; i < n - len + 1; i++)
        {
            int j = i + len - 1;
            int k = len / 2;
  
            // Calculate value of sum[i][j]
            sum[i][j] = sum[i][j - k] + 
                        sum[j - k + 1][j];
  
            // Update result if 'len' is even, 
            // left and right sums are same and 
            // len is more than maxlen
            if (len % 2 == 0 && 
                sum[i][j - k] == sum[(j - k + 1)][j] && 
                len > maxlen)
                maxlen = len;
        }
    }
    return maxlen;
}
  
// Driver Code
int main(void)
{
    char str[] = "153803";
    cout << "Length of the substring is " 
         << findLength(str);
    return 0;
  
// This code is contributed
// by Mukul Singh. 

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// A C based program that uses Dynamic Programming to find length of the
// longest even substring with same sum of digits in left and right half
#include <stdio.h>
#include <string.h>
  
int findLength(char *str)
{
    int n = strlen(str);
    int maxlen = 0; // Initialize result
  
    // A 2D table where sum[i][j] stores sum of digits
    // from str[i] to str[j].  Only filled entries are
    // the entries where j >= i
    int sum[n][n];
  
    // Fill the diagonal values for sunstrings of length 1
    for (int i =0; i<n; i++)
        sum[i][i] = str[i]-'0';
  
    // Fill entries for substrings of length 2 to n
    for (int len=2; len<=n; len++)
    {
        // Pick i and j for current substring
        for (int i=0; i<n-len+1; i++)
        {
            int j = i+len-1;
            int k = len/2;
  
            // Calculate value of sum[i][j]
            sum[i][j] = sum[i][j-k] + sum[j-k+1][j];
  
            // Update result if 'len' is even, left and right
            // sums are same and len is more than maxlen
            if (len%2 == 0 && sum[i][j-k] == sum[(j-k+1)][j]
                           && len > maxlen)
                 maxlen = len;
        }
    }
    return maxlen;
}
  
// Driver program to test above function
int main(void)
{
    char str[] = "153803";
    printf("Length of the substring is %d", findLength(str));
    return 0;
}

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// A Java based program that uses Dynamic 
// Programming to find length of the longest
// even substring with same sum of digits 
// in left and right half
import java.io.*;
  
class GFG {
  
static int findLength(String str)
{
    int n = str.length();
    int maxlen = 0; // Initialize result
  
    // A 2D table where sum[i][j] stores 
    // sum of digits from str[i] to str[j]. 
    // Only filled entries are the entries
    // where j >= i
    int sum[][] = new int[n][n];
  
    // Fill the diagonal values for 
    // substrings of length 1
    for (int i = 0; i < n; i++)
        sum[i][i] = str.charAt(i) - '0';
  
    // Fill entries for substrings of
    // length 2 to n
    for (int len = 2; len <= n; len++)
    {
        // Pick i and j for current substring
        for (int i = 0; i < n - len + 1; i++)
        {
            int j = i + len - 1;
            int k = len/2;
  
            // Calculate value of sum[i][j]
            sum[i][j] = sum[i][j-k] +
                        sum[j-k+1][j];
  
            // Update result if 'len' is even,
            // left and right sums are same
            // and len is more than maxlen
            if (len % 2 == 0 && sum[i][j-k] == 
                sum[(j-k+1)][j] && len > maxlen)
                maxlen = len;
        }
    }
    return maxlen;
}
  
// Driver program to test above function
public static void main(String[] args)
{
    String str = "153803";
    System.out.println("Length of the substring is "
                       + findLength(str));
}
}
  
// This code is contributd by Prerna Saini

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# Python3 code that uses Dynamic Programming 
# to find length of the longest even substring 
# with same sum of digits in left and right half 
  
def findLength(string): 
  
    n = len(string) 
    maxlen = 0 # Initialize result 
  
    # A 2D table where sum[i][j] stores 
    # sum of digits from str[i] to str[j]. 
    # Only filled entries are the entries
    # where j >= i 
    Sum = [[0 for x in range(n)] 
              for y in range(n)]
  
    # Fill the diagonal values for 
    # substrings of length 1 
    for i in range(0, n): 
        Sum[i][i] = int(string[i]) 
  
    # Fill entries for substrings 
    # of length 2 to n 
    for length in range(2, n + 1): 
      
        # Pick i and j for current substring 
        for i in range(0, n - length + 1): 
          
            j = i + length - 1
            k = length // 2
  
            # Calculate value of sum[i][j] 
            Sum[i][j] = (Sum[i][j - k] + 
                         Sum[j - k + 1][j])
  
            # Update result if 'len' is even, 
            # left and right sums are same and 
            # len is more than maxlen 
            if (length % 2 == 0 and 
                Sum[i][j - k] == Sum[(j - k + 1)][j] and 
                length > maxlen):
                maxlen = length
          
    return maxlen 
  
# Driver Code
if __name__ == "__main__":
  
    string = "153803"
    print("Length of the substring is"
                    findLength(string)) 
      
# This code is contributed 
# by Rituraj Jain

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C#

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// A C# based program that uses Dynamic 
// Programming to find length of the longest
// even substring with same sum of digits 
// in left and right half
using System;
  
class GFG {
  
static int findLength(String str)
{
    int n = str.Length;
    int maxlen = 0; // Initialize result
  
    // A 2D table where sum[i][j] stores 
    // sum of digits from str[i] to str[j]. 
    // Only filled entries are the entries
    // where j >= i
    int[,] sum = new int[n, n];
  
    // Fill the diagonal values for 
    // substrings of length 1
    for (int i = 0; i < n; i++)
        sum[i, i] = str[i] - '0';
  
    // Fill entries for substrings of
    // length 2 to n
    for (int len = 2; len <= n; len++)
    {
        // Pick i and j for current substring
        for (int i = 0; i < n - len + 1; i++)
        {
            int j = i + len - 1;
            int k = len/2;
  
            // Calculate value of sum[i][j]
            sum[i, j] = sum[i, j-k] +
                        sum[j-k+1, j];
  
            // Update result if 'len' is even,
            // left and right sums are same
            // and len is more than maxlen
            if (len % 2 == 0 && sum[i, j-k] == 
                sum[(j-k+1), j] && len > maxlen)
                maxlen = len;
        }
    }
    return maxlen;
}
  
// Driver program to test above function
public static void Main()
{
    String str = "153803";
    Console.WriteLine("Length of the substring is "
                    + findLength(str));
}
}
  
// This code is contributd 
// by Akanksha Rai(Abby_akku)

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<?php
// A PHP based program that uses Dynamic 
// Programming to find length of the longest
// even substring with same sum of digits
// in left and right half
  
function findLength($str)
{
    $n = strlen($str);
    $maxlen = 0; // Initialize result
  
    // A 2D table where sum[i][j] stores sum 
    // of digits from str[i] to str[j]. Only 
    // filled entries are the entries where j >= i
  
    // Fill the diagonal values for
    // substrings of length 1
    for ($i = 0; $i < $n; $i++)
        $sum[$i][$i] = $str[$i] - '0';
  
    // Fill entries for substrings of
    // length 2 to n
    for ($len = 2; $len <= $n; $len++)
    {
        // Pick i and j for current substring
        for ($i = 0; $i < $n - $len + 1; $i++)
        {
            $j = $i + $len - 1;
            $k = $len / 2;
  
            // Calculate value of sum[i][j]
            $sum[$i][$j] = $sum[$i][$j - $k] + 
                           $sum[$j - $k + 1][$j];
  
            // Update result if 'len' is even, 
            // left and right sums are same and 
            // len is more than maxlen
            if ($len % 2 == 0 && 
                $sum[$i][$j - $k] == $sum[($j - $k + 1)][$j] && 
                $len > $maxlen)
                $maxlen = $len;
        }
    }
    return $maxlen;
}
  
// Driver Code
$str = "153803";
echo("Length of the substring is "); 
echo(findLength($str));
  
// This code is contributed
// by Shivi_Aggarwal
?>

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Output:

Length of the substring is 4

Time complexity of the above solution is O(n2), but it requires O(n2) extra space.



[A O(n2) and O(n) extra space solution]
The idea is to use a single dimensional array to store cumulative sum.

C++

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// A O(n^2) time and O(n) extra space solution
#include<bits/stdc++.h>
using namespace std;
  
int findLength(string str, int n)
{
    int sum[n+1]; // To store cumulative sum from first digit to nth digit
    sum[0] = 0;
  
    /* Store cumulative sum of digits from first to last digit */
    for (int i = 1; i <= n; i++)
        sum[i] = (sum[i-1] + str[i-1]  - '0'); /* convert chars to int */
  
    int ans = 0; // initialize result
  
    /* consider all even length substrings one by one */
    for (int len = 2; len <= n; len += 2)
    {
        for (int i = 0; i <= n-len; i++)
        {
            int j = i + len - 1;
  
            /* Sum of first and second half is same than update ans */
            if (sum[i+len/2] - sum[i] == sum[i+len] - sum[i+len/2])
                ans = max(ans, len);
        }
    }
    return ans;
}
  
// Driver program to test above function
int main()
{
    string str = "123123";
    cout << "Length of the substring is " << findLength(str, str.length());
    return 0;
}

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Java

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// Java implementation of O(n^2) time
// and O(n) extra space solution
class GFG {
  
static int findLength(String str, int n)
{   
    // To store cumulative sum from
    // first digit to nth digit
    int sum[] = new int[ n + 1]; 
    sum[0] = 0;
  
    /* Store cumulative sum of digits 
    from first to last digit */
    for (int i = 1; i <= n; i++)
          
        /* convert chars to int */
        sum[i] = (sum[i-1] + str.charAt(i-1
                                    - '0'); 
  
    int ans = 0; // initialize result
  
    /* consider all even length 
    substrings one by one */
    for (int len = 2; len <= n; len += 2)
    {
        for (int i = 0; i <= n-len; i++)
        {
            int j = i + len - 1;
  
            /* Sum of first and second half 
            is same than update ans */
            if (sum[i+len/2] - sum[i] == sum[i+len]
                                   - sum[i+len/2])
                ans = Math.max(ans, len);
        }
    }
    return ans;
}
  
// Driver program to test above function
public static void main(String[] args)
{
    String str = "123123";
    System.out.println("Length of the substring is " 
                    + findLength(str, str.length()));
}
}
  
// This code is contributed by Prerna Saini

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Python3

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# A O(n^2) time and O(n) extra 
# space solution in Python3
def findLength(string, n): 
      
    # To store cumulative sum
    # from first digit to nth digit 
    Sum = [0] * (n + 1
  
    # Store cumulative sum of digits 
    # from first to last digit 
    for i in range(1, n + 1): 
        Sum[i] = (Sum[i - 1] + 
              int(string[i - 1])) # convert chars to int 
  
    ans = 0 # initialize result 
  
    # consider all even length
    # substrings one by one 
    for length in range(2, n + 1, 2): 
      
        for i in range(0, n - length + 1): 
          
            j = i + length - 1
  
            # Sum of first and second half 
            # is same than update ans 
            if (Sum[i + length // 2] - 
                Sum[i] == Sum[i + length] - 
                Sum[i + length // 2]):
                ans = max(ans, length) 
      
    return ans 
  
# Driver code 
if __name__ == "__main__":
  
    string = "123123"
    print("Length of the substring is"
           findLength(string, len(string))) 
      
# This code is contributed 
# by Rituraj Jain

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C#

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// C# implementation of O(n^2) time and O(n)
// extra space solution
using System;
  
class GFG {
  
    static int findLength(string str, int n)
    
          
        // To store cumulative sum from
        // first digit to nth digit
        int []sum = new int[ n + 1]; 
        sum[0] = 0;
      
        /* Store cumulative sum of digits 
        from first to last digit */
        for (int i = 1; i <= n; i++)
              
            /* convert chars to int */
            sum[i] = (sum[i-1] + str[i-1] 
                                   - '0'); 
      
        int ans = 0; // initialize result
      
        /* consider all even length 
        substrings one by one */
        for (int len = 2; len <= n; len += 2)
        {
            for (int i = 0; i <= n-len; i++)
            {
                // int j = i + len - 1;
      
                /* Sum of first and second half 
                is same than update ans */
                if (sum[i+len/2] - sum[i] ==
                     sum[i+len] - sum[i+len/2])
                    ans = Math.Max(ans, len);
            }
        }
          
        return ans;
    }
      
    // Driver program to test above function
    public static void Main()
    {
        string str = "123123";
        Console.Write("Length of the substring"
        + " is " + findLength(str, str.Length));
    }
}
  
// This code is contributed by nitin mittal.

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PHP

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<?php
// A O(n^2) time and O(n) extra space solution
  
function findLength($str, $n)
{
    $sum[$n + 1] = array(); // To store cumulative sum from 
                            // first digit to nth digit
    $sum[0] = 0;
  
    /* Store cumulative sum of digits 
    from first to last digit */
    for ($i = 1; $i <= $n; $i++)
        $sum[$i] = ($sum[$i - 1] +
                    $str[$i - 1] - '0'); /* convert chars to int */
  
    $ans = 0; // initialize result
  
    /* consider all even length 
    substrings one by one */
    for ($len = 2; $len <= $n; $len += 2)
    {
        for ($i = 0; $i <= $n - $len; $i++)
        {
            $j = $i + $len - 1;
  
            /* Sum of first and second half is
            same than update ans */
            if ($sum[$i + $len / 2] - $sum[$i] == $sum[$i + $len] - 
                                                  $sum[$i + $len / 2])
                $ans = max($ans, $len);
        }
    }
    return $ans;
}
  
// Driver Code
$str = "123123";
echo "Length of the substring is "
     findLength($str, strlen($str));
  
// This code is contributed 
// by Akanksha Rai
?>

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Output:

Length of the substring is 6

Thanks to Gaurav Ahirwar for suggesting this method.



[A O(n2) time and O(1) extra space solution]
The idea is to consider all possible mid points (of even length substrings) and keep expanding on both sides to get and update optimal length as the sum of two sides become equal.

Below is the implementation of the above idea.

C++

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// A O(n^2) time and O(1) extra space solution
#include<bits/stdc++.h>
using namespace std;
  
int findLength(string str, int n)
{
    int ans = 0; // Initialize result
  
    // Consider all possible midpoints one by one
    for (int i = 0; i <= n-2; i++)
    {
        /* For current midpoint 'i', keep expanding substring on
           both sides, if sum of both sides becomes equal update
           ans */
        int l = i, r = i + 1;
  
        /* initialize left and right sum */
        int lsum = 0, rsum = 0;
  
        /* move on both sides till indexes go out of bounds */
        while (r < n && l >= 0)
        {
            lsum += str[l] - '0';
            rsum += str[r] - '0';
            if (lsum == rsum)
                ans = max(ans, r-l+1);
            l--;
            r++;
        }
    }
    return ans;
}
  
// Driver program to test above function
int main()
{
    string str = "123123";
    cout << "Length of the substring is " << findLength(str, str.length());
    return 0;
}

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Java

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// A O(n^2) time and O(1) extra space solution
  
class GFG {
  
    static int findLength(String str, int n) {
        int ans = 0; // Initialize result
  
        // Consider all possible midpoints one by one
        for (int i = 0; i <= n - 2; i++) {
            /* For current midpoint 'i', keep expanding substring on
           both sides, if sum of both sides becomes equal update
           ans */
            int l = i, r = i + 1;
  
            /* initialize left and right sum */
            int lsum = 0, rsum = 0;
  
            /* move on both sides till indexes go out of bounds */
            while (r < n && l >= 0) {
                lsum += str.charAt(l) - '0';
                rsum += str.charAt(r) - '0';
                if (lsum == rsum) {
                    ans = Math.max(ans, r - l + 1);
                }
                l--;
                r++;
            }
        }
        return ans;
    }
  
// Driver program to test above function
    static public void main(String[] args) {
        String str = "123123";
        System.out.println("Length of the substring is "
                + findLength(str, str.length()));
    }
}
  
// This code is contributed by Rajput-Ji

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Python 3

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# A O(n^2) time and O(n) extra 
# space solution 
def findLength(st, n): 
  
    # To store cumulative total from 
    # first digit to nth digit 
    total = [0] * (n + 1
  
    # Store cumulative total of digits
    # from first to last digit 
    for i in range(1, n + 1): 
          
        # convert chars to int
        total[i] = (total[i - 1] + 
                   int(st[i - 1]) - int('0'))
  
    ans = 0 # initialize result 
  
    # consider all even length 
    # substings one by one 
    l = 2
    while(l <= n):
  
        for i in range(n - l + 1):
      
            j = i + l - 1
  
            # total of first and second half 
            # is same than update ans 
            if (total[i + int(l / 2)] - 
                total[i] == total[i + l] -
                total[i + int(l / 2)]): 
                ans = max(ans, l) 
        l = l + 2
      
    return ans 
  
# Driver Code
st = "123123"
print("Length of the substring is"
           findLength(st, len(st)))
  
# This code is contributed by ash264

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C#

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// A O(n^2) time and O(1) extra space solution
using System;
public class GFG {
   
    static int findLength(String str, int n) {
        int ans = 0; // Initialize result
   
        // Consider all possible midpoints one by one
        for (int i = 0; i <= n - 2; i++) {
            /* For current midpoint 'i', keep expanding substring on
           both sides, if sum of both sides becomes equal update
           ans */
            int l = i, r = i + 1;
   
            /* initialize left and right sum */
            int lsum = 0, rsum = 0;
   
            /* move on both sides till indexes go out of bounds */
            while (r < n && l >= 0) {
                lsum += str[l] - '0';
                rsum += str[r] - '0';
                if (lsum == rsum) {
                    ans = Math.Max(ans, r - l + 1);
                }
                l--;
                r++;
            }
        }
        return ans;
    }
   
// Driver program to test above function
    static public void Main() {
        String str = "123123";
        Console.Write("Length of the substring is "
                + findLength(str, str.Length));
    }
}
   
// This code is contributed by Rajput-Ji

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PHP

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<?php 
// A O(n^2) time and O(1) extra space solution
  
function findLength($str, $n)
{
    $ans = 0; // Initialize result
  
    // Consider all possible midpoints one by one
    for ($i = 0; $i <= $n - 2; $i++)
    {
        /* For current midpoint 'i', 
        keep expanding substring on
        both sides, if sum of both 
        sides becomes equal update
        ans */
        $l = $i;
        $r = $i + 1;
  
        /* initialize left and right sum */
        $lsum = 0;
        $rsum = 0;
  
        /* move on both sides till
        indexes go out of bounds */
        while ($r < $n && $l >= 0)
        {
            $lsum += $str[$l] - '0';
            $rsum += $str[$r] - '0';
            if ($lsum == $rsum)
                $ans = max($ans, $r - $l + 1);
            $l--;
            $r++;
        }
    }
    return $ans;
}
  
// Driver program to test above function
  
$str = "123123";
echo "Length of the substring is " .
        findLength($str, strlen($str));
return 0;
  
// This code is contributed by Ita_c.
?>

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Output:

Length of the substring is 6

Thanks to Gaurav Ahirwar for suggesting this method.

This article is contributed by Ashish Bansal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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