Longest Even Length Substring such that Sum of First and Second Half is same
Last Updated :
23 Jul, 2025
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++
// 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
// 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;
}
// 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 contributed by Prerna Saini
# 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
// 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 contributed by nitin mittal
<?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
?>
<script>
// A simple Javascript based program to find
// length of longest even length substring
// with same sum of digits in left and right
function findLength(str)
{
let n = str.length;
let maxlen = 0; // Initialize result
// Choose starting point of every
// substring
for (let i = 0; i < n; i++)
{
// Choose ending point of even
// length substring
for (let j = i + 1; j < n; j += 2)
{
// Find length of current substr
let length = j - i + 1;
// Calculate left & right sums
// for current substr
let leftsum = 0, rightsum = 0;
for (let k = 0; k < Math.floor(length/2); k++)
{
leftsum += (str[i+k] - '0');
rightsum += (str[i+k+Math.floor(length/2)] - '0');
}
// Update result if needed
if (leftsum == rightsum && maxlen < length)
maxlen = length;
}
}
return maxlen;
}
let str = "1538023";
document.write("Length of the substring is " + findLength(str));
// This code is contributed by rag2127
</script>
OutputLength 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++
// 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.
// 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 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
int main(void)
{
char str[] = "153803";
printf("Length of the substring is %d", findLength(str));
return 0;
}
// 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 contributed by Prerna Saini
# 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
// 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 contributed
// by Akanksha Rai(Abby_akku)
<?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
?>
<script>
// A javascript 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)
{
var n = str.length;
var 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
var sum = Array(n).fill(0).map(x => Array(n).fill(0));
// Fill the diagonal values for
// substrings of length 1
for (i = 0; i < n; i++)
sum[i][i] = str.charAt(i).charCodeAt(0) - '0'.charCodeAt(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++)
{
var j = i + len - 1;
var k = parseInt(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
var str = "153803";
document.write("Length of the substring is "
+ findLength(str));
// This code contributed by shikhasingrajput
</script>
OutputLength 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++
// 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 then 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;
}
// Java implementation of O(n^2) time
// and O(n) extra space solution
import java.util.*;
import java.io.*;
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 then 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
# 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 then 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
// 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 then 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.
<?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 + $len - 1;
/* Sum of first and second half is
same then update ans */
if ($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
?>
<script>
// javascript implementation of O(n^2) time
// and O(n) extra space solution
function findLength(str , n)
{
// To store cumulative sum from
// first digit to nth digit
var sum = Array.from({length: n+1}, (_, i) => 0);
sum[0] = 0;
/* Store cumulative sum of digits
from first to last digit */
for (var i = 1; i <= n; i++)
/* convert chars to var */
sum[i] = (sum[i-1] + str.charAt(i-1).charCodeAt(0)
- '0'.charCodeAt(0));
var ans = 0; // initialize result
/* consider all even length
substrings one by one */
for (var len = 2; len <= n; len += 2)
{
for (i = 0; i <= n-len; i++)
{
var j = i + len - 1;
/* Sum of first and second half
is same then update ans */
if (sum[i+parseInt(len/2)] - sum[i] == sum[i+len]
- sum[i+parseInt(len/2)])
ans = Math.max(ans, len);
}
}
return ans;
}
// Driver program to test above function
var str = "123123";
document.write("Length of the substring is "
+ findLength(str, str.length));
// This code is contributed by 29AjayKumar
</script>
OutputLength of the substring is 6
[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++
// 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;
}
// A O(n^2) time and O(1) extra space solution
import java.util.*;
import java.io.*;
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
# 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):
# 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
// 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
<?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.
?>
<script>
// A O(n^2) time and O(1) extra space solution
function findLength(str,n)
{
let ans = 0; // Initialize result
// Consider all possible midpoints one by one
for (let 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 */
let l = i, r = i + 1;
/* initialize left and right sum */
let 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
let str="123123";
document.write("Length of the substring is "
+ findLength(str, str.length));
// This code is contributed by avanitrachhadiya2155
</script>
OutputLength of the substring is 6
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