Split array to three subarrays such that sum of first and third subarray is equal and maximum
Last Updated :
19 Oct, 2022
Given an array of N integers, the task is to print the sum of the first subarray by splitting the array into exactly three subarrays such that the sum of the first and third subarray elements are equal and the maximum.
Note: All the elements must belong to a subarray and the subarrays can also be empty.
Examples:
Input: a[] = {1, 3, 1, 1, 4}
Output: 5
Split the N numbers to [1, 3, 1], [] and [1, 4]
Input: a[] = {1, 3, 2, 1, 4}
Output: 4
Split the N numbers to [1, 3], [2, 1] and [4]
METHOD 1
A naive approach is to check for all possible partitions and use the prefix-sum concept to find out the partitions. The partition which gives the maximum sum of the first subarray will be the answer.
An efficient approach is as follows:
- Store the prefix sum and suffix sum of the N numbers.
- Hash the suffix sum’s index using a unordered_map in C++ or Hash-map in Java.
- Iterate from the beginning of the array, and check if the prefix sum exists in the suffix array beyond the current index i.
- If it does, then check for the previous maximum value and update accordingly.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int sumFirst( int a[], int n)
{
unordered_map< int , int > mp;
int suf = 0;
for ( int i = n - 1; i >= 0; i--)
{
suf += a[i];
mp[suf] = i;
}
int pre = 0;
int maxi = -1;
for ( int i = 0; i < n; i++)
{
pre += a[i];
if (mp[pre] > i)
{
if (pre > maxi)
{
maxi = pre;
}
}
}
if (maxi == -1)
return 0;
else
return maxi;
}
int main()
{
int a[] = { 1, 3, 2, 1, 4 };
int n = sizeof (a) / sizeof (a[0]);
cout << sumFirst(a, n);
return 0;
}
|
Java
import java.util.HashMap;
import java.util.Map;
class GfG {
static int sumFirst( int a[], int n)
{
HashMap<Integer, Integer> mp = new HashMap<>();
int suf = 0 ;
for ( int i = n - 1 ; i >= 0 ; i--)
{
suf += a[i];
mp.put(suf, i);
}
int pre = 0 , maxi = - 1 ;
for ( int i = 0 ; i < n; i++)
{
pre += a[i];
if (mp.containsKey(pre) && mp.get(pre) > i)
{
if (pre > maxi)
{
maxi = pre;
}
}
}
if (maxi == - 1 )
return 0 ;
else
return maxi;
}
public static void main(String[] args)
{
int a[] = { 1 , 3 , 2 , 1 , 4 };
int n = a.length;
System.out.println(sumFirst(a, n));
}
}
|
Python3
def sumFirst(a, n):
mp = {i: 0 for i in range ( 7 )}
suf = 0
i = n - 1
while (i > = 0 ):
suf + = a[i]
mp[suf] = i
i - = 1
pre = 0
maxi = - 1
for i in range (n):
pre + = a[i]
if (mp[pre] > i):
if (pre > maxi):
maxi = pre
if (maxi = = - 1 ):
return 0
else :
return maxi
if __name__ = = '__main__' :
a = [ 1 , 3 , 2 , 1 , 4 ]
n = len (a)
print (sumFirst(a, n))
|
C#
using System;
using System.Collections.Generic;
class GfG {
static int sumFirst( int [] a, int n)
{
Dictionary< int , int > mp
= new Dictionary< int , int >();
int suf = 0;
for ( int i = n - 1; i >= 0; i--)
{
suf += a[i];
mp.Add(suf, i);
if (mp.ContainsKey(suf))
{
mp.Remove(suf);
}
mp.Add(suf, i);
}
int pre = 0, maxi = -1;
for ( int i = 0; i < n; i++)
{
pre += a[i];
if (mp.ContainsKey(pre) && mp[pre] > i)
{
if (pre > maxi)
{
maxi = pre;
}
}
}
if (maxi == -1)
return 0;
else
return maxi;
}
public static void Main(String[] args)
{
int [] a = { 1, 3, 2, 1, 4 };
int n = a.Length;
Console.WriteLine(sumFirst(a, n));
}
}
|
Javascript
<script>
function sumFirst(a, n)
{
var mp = new Map();
var suf = 0;
for ( var i = n - 1; i >= 0; i--)
{
suf += a[i];
mp.set(suf, i);
}
var pre = 0;
var maxi = -1;
for ( var i = 0; i < n; i++)
{
pre += a[i];
if (mp.get(pre) > i)
{
if (pre > maxi)
{
maxi = pre;
}
}
}
if (maxi == -1)
return 0;
else
return maxi;
}
var a = [1, 3, 2, 1, 4];
var n = a.length;
document.write( sumFirst(a, n));
</script>
|
Time Complexity: O(n) where n is the size of the given array
Auxiliary Space: O(n)
METHOD 2
Approach: We will use two pointers concept where one pointer will start from the front and the other from the back. In each iteration, the sum of the first and last subarray is compared and if they are the same then the sum is updated in the answer variable.
Algorithm:
- Initialize front_pointer to 0 and back_pointer to n-1.
- Initialize prefixsum to arr[ front_pointer ] and suffixsum to arr[back_pointer].
- The summations are compared.
- If prefixsum > suffixsum ,back_pointer is decremented by 1 and suffixsum+= arr[ back_pointer ].
- If prefixsum < suffixsum, front_pointer is incremented by 1 and prefixsum+= arr[ front_pointer ]
- If they are the same then the sum is updated in the answer variable and both the pointers are moved by one step and both prefixsum and suffixsum are updated accordingly.
- The above step is continued until the front_pointer is no less than the back_pointer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int sumFirst( int a[], int n)
{
int front_pointer = 0;
int back_pointer = n - 1;
int prefixsum = a[front_pointer];
int suffixsum = a[back_pointer];
int answer = 0;
while (front_pointer < back_pointer)
{
if (prefixsum == suffixsum)
{
answer = max(answer, prefixsum);
front_pointer++;
back_pointer--;
prefixsum += a[front_pointer];
suffixsum += a[back_pointer];
}
else if (prefixsum > suffixsum)
{
back_pointer--;
suffixsum += a[back_pointer];
}
else
{
front_pointer++;
prefixsum += a[front_pointer];
}
}
return answer;
}
int main()
{
int arr[] = { 1, 3, 2, 1, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << sumFirst(arr, n);
}
|
Java
import java.util.*;
class GFG{
public static int sumFirst( int a[], int n)
{
int front_pointer = 0 ;
int back_pointer = n - 1 ;
int prefixsum = a[front_pointer];
int suffixsum = a[back_pointer];
int answer = 0 ;
while (front_pointer < back_pointer)
{
if (prefixsum == suffixsum)
{
answer = Math.max(answer, prefixsum);
front_pointer++;
back_pointer--;
prefixsum += a[front_pointer];
suffixsum += a[back_pointer];
}
else if (prefixsum > suffixsum)
{
back_pointer--;
suffixsum += a[back_pointer];
}
else
{
front_pointer++;
prefixsum += a[front_pointer];
}
}
return answer;
}
public static void main(String[] args)
{
int arr[] = { 1 , 3 , 2 , 1 , 4 };
int n = arr.length;
System.out.print(sumFirst(arr, n));
}
}
|
Python3
import math
def sumFirst(a, n):
front_pointer = 0
back_pointer = n - 1
prefixsum = a[front_pointer]
suffixsum = a[back_pointer]
answer = 0
while (front_pointer < back_pointer):
if (prefixsum = = suffixsum):
answer = max (answer, prefixsum)
front_pointer + = 1
back_pointer - = 1
prefixsum + = a[front_pointer]
suffixsum + = a[back_pointer]
elif (prefixsum > suffixsum):
back_pointer - = 1
suffixsum + = a[back_pointer]
else :
front_pointer + = 1
prefixsum + = a[front_pointer]
return answer
arr = [ 1 , 3 , 2 , 1 , 4 ]
n = len (arr)
print (sumFirst(arr, n))
|
C#
using System;
class GFG{
static int sumFirst( int [] a, int n)
{
int front_pointer = 0;
int back_pointer = n - 1;
int prefixsum = a[front_pointer];
int suffixsum = a[back_pointer];
int answer = 0;
while (front_pointer < back_pointer)
{
if (prefixsum == suffixsum)
{
answer = Math.Max(answer, prefixsum);
front_pointer++;
back_pointer--;
prefixsum += a[front_pointer];
suffixsum += a[back_pointer];
}
else if (prefixsum > suffixsum)
{
back_pointer--;
suffixsum += a[back_pointer];
}
else
{
front_pointer++;
prefixsum += a[front_pointer];
}
}
return answer;
}
static void Main()
{
int [] arr = { 1, 3, 2, 1, 4 };
int n = arr.Length;
Console.WriteLine(sumFirst(arr, n));
}
}
|
Javascript
<script>
function sumFirst(a, n)
{
var front_pointer = 0;
var back_pointer = n - 1;
var prefixsum = a[front_pointer];
var suffixsum = a[back_pointer];
var answer = 0;
while (front_pointer < back_pointer)
{
if (prefixsum == suffixsum)
{
answer = Math.max(answer, prefixsum);
front_pointer++;
back_pointer--;
prefixsum += a[front_pointer];
suffixsum += a[back_pointer];
}
else if (prefixsum > suffixsum)
{
back_pointer--;
suffixsum += a[back_pointer];
}
else
{
front_pointer++;
prefixsum += a[front_pointer];
}
}
return answer;
}
var arr = [ 1, 3, 2, 1, 4 ];
var n = arr.length;
document.write(sumFirst(arr, n));
</script>
|
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...